We have now tentatively improved the upper bound of the de Bruijn-Newman constant to . Among the technical improvements in our approach, we now are able to use Taylor expansions to efficiently compute the approximation to for many values of in a given region, thus speeding up the computations in the barrier considerably. Also, by using the heuristic that behaves somewhat like the partial Euler product , we were able to find a good location to place the barrier in which is larger than average, hence easier to keep away from zero.

The main remaining bottleneck is that of computing the Euler mollifier bounds that keep bounded away from zero for larger values of beyond the barrier. In going below we are beginning to need quite complicated mollifiers with somewhat poor tail behavior; we may be reaching the point where none of our bounds will succeed in keeping bounded away from zero, so we may be close to the natural limits of our methods.

Participants are also welcome to add any further summaries of the situation in the comments below.

]]>Clearly, a univalent function on the unit disk is a conformal map from to the image ; in particular, is simply connected, and not all of (since otherwise the inverse map would violate Liouville’s theorem). In the converse direction, the Riemann mapping theorem tells us that every open simply connected proper subset of the complex numbers is the image of a univalent function on . Furthermore, if contains the origin, then the univalent function with this image becomes unique once we normalise and . Thus the Riemann mapping theorem provides a one-to-one correspondence between open simply connected proper subsets of the complex plane containing the origin, and univalent functions with and . We will focus particular attention on the univalent functions with the normalisation and ; such functions will be called schlicht functions.

One basic example of a univalent function on is the Cayley transform , which is a Möbius transformation from to the right half-plane . (The slight variant is also referred to as the Cayley transform, as is the closely related map , which maps to the upper half-plane.) One can square this map to obtain a further univalent function , which now maps to the complex numbers with the negative real axis removed. One can normalise this function to be schlicht to obtain the Koebe function

which now maps to the complex numbers with the half-line removed. A little more generally, for any we have the *rotated Koebe function*

that is a schlicht function that maps to the complex numbers with the half-line removed.

Every schlicht function has a convergent Taylor expansion

for some complex coefficients with . For instance, the Koebe function has the expansion

and similarly the rotated Koebe function has the expansion

Intuitively, the Koebe function and its rotations should be the “largest” schlicht functions available. This is formalised by the famous Bieberbach conjecture, which asserts that for any schlicht function, the coefficients should obey the bound for all . After a large number of partial results, this conjecture was eventually solved by de Branges; see for instance this survey of Korevaar or this survey of Koepf for a history.

It turns out that to resolve these sorts of questions, it is convenient to restrict attention to schlicht functions that are *odd*, thus for all , and the Taylor expansion now reads

for some complex coefficients with . One can transform a general schlicht function to an odd schlicht function by observing that the function , after removing the singularity at zero, is a non-zero function that equals at the origin, and thus (as is simply connected) has a unique holomorphic square root that also equals at the origin. If one then sets

it is not difficult to verify that is an odd schlicht function which additionally obeys the equation

Conversely, given an odd schlicht function , the formula (4) uniquely determines a schlicht function .

For instance, if is the Koebe function (1), becomes

which maps to the complex numbers with two slits removed, and if is the rotated Koebe function (2), becomes

De Branges established the Bieberbach conjecture by first proving an analogous conjecture for odd schlicht functions known as Robertson’s conjecture. More precisely, we have

Theorem 1 (de Branges’ theorem)Let be a natural number.

- (i) (Robertson conjecture) If is an odd schlicht function, then
- (ii) (Bieberbach conjecture) If is a schlicht function, then

It is easy to see that the Robertson conjecture for a given value of implies the Bieberbach conjecture for the same value of . Indeed, if is schlicht, and is the odd schlicht function given by (3), then from extracting the coefficient of (4) we obtain a formula

for the coefficients of in terms of the coefficients of . Applying the Cauchy-Schwarz inequality, we derive the Bieberbach conjecture for this value of from the Robertson conjecture for the same value of . We remark that Littlewood and Paley had conjectured a stronger form of Robertson’s conjecture, but this was disproved for by Fekete and Szegö.

To prove the Robertson and Bieberbach conjectures, one first takes a logarithm and deduces both conjectures from a similar conjecture about the Taylor coefficients of , known as the *Milin conjecture*. Next, one continuously enlarges the image of the schlicht function to cover all of ; done properly, this places the schlicht function as the initial function in a sequence of univalent maps known as a Loewner chain. The functions obey a useful differential equation known as the Loewner equation, that involves an unspecified forcing term (or , in the case that the image is a slit domain) coming from the boundary; this in turn gives useful differential equations for the Taylor coefficients of , , or . After some elementary calculus manipulations to “integrate” this equations, the Bieberbach, Robertson, and Milin conjectures are then reduced to establishing the non-negativity of a certain explicit hypergeometric function, which is non-trivial to prove (and will not be done here, except for small values of ) but for which several proofs exist in the literature.

The theory of Loewner chains subsequently became fundamental to a more recent topic in complex analysis, that of the Schramm-Loewner equation (SLE), which is the focus of the next and final set of notes.

** — 1. The area theorem and its consequences — **

We begin with the area theorem of Grönwall.

Theorem 2 (Grönwall area theorem)Let be a univalent function with a convergent Laurent expansionThen

*Proof:* By shifting we may normalise . By hypothesis we have for any ; by replacing with and using a limiting argument, we may assume without loss of generality that the have some exponential decay as (in order to justify some of the manipulations below).

Let be a large parameter. If , then and . The area enclosed by the simple curve is equal to

crucially, the error term here goes to zero as . Meanwhile, by the change of variables formula (using monotone convergence if desired to work in compact subsets of the annulus initially) and Plancherel's theorem, the area of the region is

Comparing these bounds we conclude that

sending to infinity, we obtain the claim.

Exercise 3Let be a univalent function with Taylor expansionShow that the area of is equal to . (In particular, has finite area if and only if .)

Corollary 4 (Bieberbach inequality)

- (i) If is an odd schlicht function, then .
- (ii) If is a schlicht function, then .

*Proof:* For (i), we apply Theorem 2 to the univalent function defined by , which has a Laurent expansion , to give the claim. For (ii), apply part (i) to the square root of with first term .

Exercise 5Show that equality occurs in Corollary 4(i) if and only if takes the form for some , and in Corollary 4(ii) if and only if takes the form of a rotated Koebe function for some .

The Bieberbach inequality can be rescaled to bound the second coefficient of univalent functions:

Exercise 6 (Rescaled Bieberbach inequality)If is a univalent function, show thatWhen does equality hold?

The Bieberbach inequality gives a useful lower bound for the image of a univalent function, known as the Koebe quarter theorem:

Corollary 7 (Koebe quarter theorem)Let be a univalent function. Then contains the disk .

*Proof:* By applying a translation and rescaling, we may assume without loss of generality that is a schlicht function, with Taylor expansion

Our task is now to show that for every , the equation has a solution in . If this were not the case, then the function is invertible on , with inverse being univalent and having the Taylor expansion

Applying Exercise 6 we then have

while from the Bieberbach inequality one also has . Hence by the triangle inequality , which is incompatible with the hypothesis .

Exercise 8Show that the radius is best possible in Corollary 7 (thus, does not contain any disk with ) if and only if takes the form for some complex numbers and real .

Remark 9The univalence hypothesis is crucial in the Koebe quarter theorem. Consider for instance the functions defined by . These are locally univalent functions (since is holomorphic with non-zero derivative) and , , but avoids the point .

Exercise 10 (Koebe distortion theorem)Let be a schlicht function, and let have magnitude .

- (i) Show that
(

Hint:compose on the right with a Möbius automorphism of that sends to and then apply the rescaled Bieberbach inequality.)- (ii) Show that
(

Hint:use (i) to control the radial derivative of .)- (iii) Show that
- (iv) Show that
(This cannot be directly derived from (ii) and (iii). Instead, compose on the right with a Mobius automorphism that sends to and to , rescale it to be schlicht, and apply (iii) to this function at .)

- (v) Show that the space of schlicht functions is a normal family. In other words, if is any sequence of schlicht functions, then there is a subsequence that converges locally uniformly on compact sets.
- (vi) (Qualitative Bieberbach conjecture) Show that for each natural number there is a constant such that whenever is a schlicht function with Taylor expansion

Exercise 11 (Conformal radius)If is a non-empty simply connected open subset of that is not all of , and is a point in , define theconformal radiusof at to be the quantity , where is any conformal map from to that maps to (the existence and uniqueness of this radius follows from the Riemann mapping theorem). Thus for instance the disk has conformal radius around .

- (i) Show that the conformal radius is strictly monotone in : if are non-empty simply connected open subsets of , and , then the conformal radius of around is strictly greater than that of .
- (ii) Show that the conformal radius of a disk around an element of the disk is given by the formula .
- (iii) Show that the conformal radius of around lies between and , where is the radius of the maximal disk that is contained in .

We can use the distortion theorem to obtain a nice criterion for when univalent maps converge to a given limit, known as the Carathéodory kernel theorem.

Theorem 12 (Carathéodory kernel theorem)Let be a sequence of simply connected open proper subsets of containing the origin, and let be a further simply connected open proper subset of containing . Let and be the conformal maps with and (the existence and uniqueness of these maps are given by the Riemann mapping theorem). Then the following are equivalent:

- (i) converges locally uniformly on compact sets to .
- (ii) For every subsequence of the , is the set of all such that there is an open connected set containing and that is contained in for all sufficiently large .

If conclusion (ii) holds, is known as the *kernel* of the domains .

*Proof:* Suppose first that converges locally uniformly on compact sets to . If , then for some . If , then the holomorphic functions converge uniformly on to the function , which is not identically zero but has a zero in . By Hurwitz’s theorem we conclude that also has a zero in for all sufficiently large ; indeed the same argument shows that one can replace by any element of a small neighbourhood of to obtain the same conclusion, uniformly in . From compactness we conclude that for sufficiently large , has a zero in for all , thus for sufficiently large . Since is open connected and contains and , we see that is contained in the set described in (ii).

Conversely, suppose that is a subsequence of the and is such that there is an open connected set containing and that is contained in for sufficiently large . The inverse maps are holomorphic and bounded, hence form a normal family by Montel’s theorem. By refining the subsequence we may thus assume that the converge locally uniformly to a holomorphic limit . The function takes values in , but by the open mapping theorem it must in fact map to . In particular, . Since converges to , and converges locally uniformly to , we conclude that converges to , thus and hence . This establishes the derivation of (ii) from (i).

Now suppose that (ii) holds. It suffices to show that every subsequence of has a further subsequence that converges locally uniformly on compact sets to (this is an instance of the *Urysohn subsequence principle*). Then (as contains ) in particular there is a disk that is contained in the for all sufficiently large ; on the other hand, as is not all of , there is also a disk which is *not* contained in the for all sufficiently large . By Exercise 11, this implies that the conformal radii of the around zero is bounded above and below, thus is bounded above and below.

By Exercise 10(v), and rescaling, the functions then form a normal family, thus there is a subsequence of the that converges locally uniformly on compact sets to some limit . Since is positive and bounded away from zero, is also positive, so is non-constant. By Hurwitz’s theorem, is therefore also univalent, and thus maps to some region . By the implication of (ii) from (i) (with replaced by ) we conclude that is the set of all such that there is an open connected set containing and that is contained in for all sufficiently large ; but by hypothesis, this set is also . Thus , and then by the uniqueness part of the Riemann mapping theorem, as desired.

The condition in Theorem 12(ii) indicates that “converges” to in a rather complicated sense, in which large parts of are allowed to be “pinched off” from and disappear in the limit. This is illustrated in the following explicit example:

Exercise 13 (Explicit example of kernel convergence)Let be the function from (5), thus is a univalent function from to with the two vertical rays from to , and from to , removed. For any natural number , let and let , and define the transformed functions .

- (i) Show that is a univalent function from to with the two vertical rays from to , and from to , removed, and that and .
- (ii) Show that converges locally uniformly to the function , and that this latter map is a univalent map from to the half-plane . (
Hint:one does not need to compute everything exactly; for instance, any terms of the form can be written using the notation instead of expanded explicitly.)- (iii) Explain why these facts are consistent with the Carathéodory kernel theorem.

As another illustration of the theorem, let be two distinct convex open proper subsets of containing the origin, and let be the associated conformal maps from to respectively with and . Then the alternating sequence does not converge locally uniformly to any limit. The set is the set of all points that lie in a connected open set containing the origin that eventually is contained in the sequence ; but if one passes to the subsequence , this set of points enlarges to , and so the sequence does not in fact have a kernel.

However, the kernel theorem simplifies significantly when the are monotone increasing, which is already an important special case:

Corollary 14 (Monotone increasing case of kernel theorem)Let the notation and assumptions be as in Theorem 12. Assume furthermore thatand that . Then converges locally uniformly on compact sets to .

Loewner observed that the kernel theorem can be used to approximate univalent functions by functions mapping into slit domains. More precisely, define a *slit domain* to be an open simply connected subset of formed by deleting a half-infinite Jordan curve connecting some finite point to infinity; for instance, the image of the Koebe function is a slit domain.

Theorem 15 (Loewner approximation theorem)Let be a univalent function. Then there exists a sequence of univalent functions whose images are slit domains, and which converge locally uniformly on compact subsets to .

*Proof:* First suppose that extends to a univalent function on a slightly larger disk for some . Then the image of the unit circle is a Jordan curve enclosing the region in the interior. Applying the Jordan curve theorem (and the Möbius inversion ), one can find a half-infinite Jordan curve from to infinity that stays outside of . For any , one can concatenate this curve with the arc to obtain another half-infinite Jordan curve , whose complement is a slit domain which has as kernel (why?). If we let be the conformal maps from to with and , we conclude from the Carathéodory kernel theorem that converges locally uniformly on compact sets to .

If is just univalent on , then it is the locally uniform limit of the dilations , which are univalent on the slightly larger disks . By the previous arguments, each is in turn the locally uniform limit of univalent functions whose images are slit domains, and the claim now follows from a diagonalisation argument.

** — 2. Loewner chains — **

The material in this section is based on these lecture notes of Contreras.

An important tool in analysing univalent functions is to study one-parameter families of univalent functions, parameterised by a time parameter , in which the images are increasing in ; roughly speaking, these families allow one to study an arbitrary univalent function by “integrating” along such a family from back to . Traditionally, we normalise these families into (radial) Loewner chains, which we now define:

Definition 16 (Loewner chain)A (radial) Loewner chain is a family of univalent maps with and (so in particular is schlicht), such that for all . (In these notes we use the prime notation exclusively for differentiation in the variable; we will use later for differentiation in the variable.)

A key example of a Loewner chain is the family

of dilated Koebe functions; note that the image of each is the slit domain , which is clearly monotone increasing in . More generally, we have the rotated Koebe chains

Whenever one has a family of simply connected proper open subsets of containing with for , and . By definition, is then the conformal radius of around , which is a strictly increasing function of by Exercise 11. If this conformal radius is equal to at and increases continuously to infinity as , then one can reparameterise the variable so that , at which point one obtains a Loewner chain.

From the Koebe quarter theorem we see that each image in a Loewner chain contains the disk . In particular the increase to fill out all of : .

Let be a Loewner chain, Let . The relation is sometimes expressed as the assertion that is *subordinate* to . It has the consequence that one has a composition law of the form

for a univalent function , uniquely defined as , noting taht is well-defined on . By construction, we have and

as well as the composition laws

for . We will refer to the as *transition functions*.

From the Schwarz lemma, we have

for , with strict inequality when . In particular, if we introduce the function

for and , then (after removing the singularity at infinity and using (10)) we see that is a holomorphic map to the right half-plane , normalised so that

Define a Herglotz function to be a holomorphic function , thus is a Herglotz function for all . A key family of examples of a Herglotz function are the Möbius transforms for . In fact, all other Herglotz functions are basically just averages of this one:

Exercise 17 (Herglotz representation theorem)Let be a Herglotz function, normalised so that .

- (i) For any , show that
for . (

Hint:The real part of is harmonic, and so has a Poisson kernel representation. Alternatively, one can use a Taylor expansion of .)- (ii) Show that there exists a (Radon) probability measure on such that
for all . (One will need a measure-theoretic tool such as Prokhorov’s theorem, the Riesz representation theorem, or the Helly selection principle.) Conversely, show that every probability measure on generates a Herglotz function with by the above formula.

- (iii) Show that the measure constructed on (ii) is unique.

This has a useful corollary, namely a version of the Harnack inequality:

Exercise 18 (Harnack inequality)Let be a Herglotz function, normalised so that . Show thatfor all .

This gives some useful Lipschitz regularity properties of the transition functions and univalent functions in the variable:

Lemma 19 (Lipschitz regularity)Let be a compact subset of , and let . Use to denote a quantity bounded in magnitude by , where depends only on .

- (i) For any and , one has
- (ii) For any and , one has

One can make the bounds much more explicit if desired (see e.g. Lemma 2.3 of these notes of Contreras), but for our purposes any Lipschitz bound will suffice.

*Proof:* To prove (i), it suffices from (11) and the Schwarz-Pick lemma (Exercise 13 from Notes 2) to establish this claim when . We can also assume that since the claim is trivial when . From the Harnack inequality one has

for , which by (12) and some computation gives

Now we prove (ii). We may assume without loss of generality that is convex. From Exercise 10 (normalising to be schlicht) we see that for , and hence has a Lipschitz constant of on . Since , the claim now follows from (13).

As a first application of this we show that every schlicht function starts a Loewner chain.

Lemma 20Let be schlicht. Then there exists a Loewner chain with .

*Proof:* This will be similar to the proof of Theorem 15. First suppose that extends to be univalent on for some , then is a Jordan curve. Then by Carathéodory’s theorem (Theorem 20 of Notes 2) (and the Möbius inversion ) one can find a conformal map from the exterior of to the exterior of that sends infinity to infinity. If we define for to be the region enclosed by the Jordan curve , then the are increasing in with conformal radius going to infinity as . If one sets to be the conformal maps with and , then (by the uniqueness of Riemann mapping) and by the Carathéodory kernel theorem, converges locally uniformly to as . In particular, the conformal radii are continuous in . Reparameterising in one can then obtain the required Loewner chain.

Now suppose is only univalent of . As in the proof of Theorem 15, one can express as the locally uniform limit of schlicht functions , each of which extends univalently to some larger disk . By the preceding discussion, each of the extends to a Loewner chain . From the Lipschitz bounds (and the Koebe distortion theorem) one sees that these chains are locally uniformly equicontinuous in and , uniformly in , and hence by Arzela-Ascoli we can pass to a subsequence that converges locally uniformly in to a limit ; one can also assume that the transition functions converge locally uniformly to limits . It is then not difficult by Hurwitz theorem to verify the limiting relations (9), (11), and that is a Loewner chain with as desired.

Suppose that are close to each other: . Then one heuristically has the approximations

and hence by (12) and some rearranging

and hence on applying , (9), and the Newton approximation

This suggests that the should obey the *Loewner equation*

for some Herglotz function . This is essentially the case:

Theorem 21 (Loewner equation)Let be a Loewner chain. Then, for outside of an exceptional set of Lebesgue measure zero, the functions are differentiable in time for each , and obey the equation (14) for all and , and some Herglotz function for each with . Furthermore, the maps are measurable for every .

*Proof:* Let be a countable dense subset of . From Lemma 19, the function is Lipschitz continuous, and thus differentiable almost everywhere, for each . Thus there exists a Lebesgue measure zero set such that is differentiable in outside of for each . From the Koebe distortion theorem is also locally Lipschitz (hence locally uniformly equicontinuous) in the variable, so in fact is differentiable in outside of for all . Without loss of generality we may assume contains zero.

Let , and let . Then as approaches from below, we have

uniformly; from (9) and Newton approximation we thus have

which implies that

Also we have

and hence by (12)

Taking limits, we see that the function is Herglotz with , giving the claim. It is also easy to verify the measurability (because derivatives of Lipschitz functions are measurable)

Example 22The Loewner chain (7) solves the Loewner equation with the Herglotz function . With the rotated Koebe chains (8), we instead have .

Although we will not need it in this set of notes, there is also a converse implication that for every family of Herglotz functions depending measurably on , one can associate a Loewner chain.

Let us now Taylor expand a Loewner chain at each time as

as , we have . As is differentiable in almost every for each , and is locally uniformly continuous in , we see from the Cauchy integral formulae that the are also differentiable almost everywhere in . If we similarly write

for all outside of , then , and we obtain the equations

and so forth. For instance, for the Loewner chain (7) one can verify that and for solve these equations. For (8) one instead has and .

We have the following bounds on the first few coefficients of :

Exercise 23Let be a Herglotz function with . Let be the measure coming from the Herglotz representation theorem.

- (i) Show that for all . In particular, for all . Use this to give an alternate proof of the Harnack inequality.
- (ii) Show that .

We can use this to establish the first two cases of the Bieberbach conjecture:

Theorem 24 ( cases of Bieberbach)If is schlicht, then and .

The bound is not new, and indeed was implicitly used many times in the above arguments, but we include it to illustrate the use of the equations (15), (16).

*Proof:* By Lemma 20, we can write (and ) for some Loewner chain .

We can write (15) as . On the other hand, from the Koebe distortion theorem applied to the schlicht functions , we have , so in particular goes to zero at infinity. We can integrate from to infinity to obtain

From Harnack’s inequality we have , giving the required bound .

In a similar vein, writing (16) as

we obtain

As , we may integrate from to infinity to obtain the identity

Taking real parts using Exercise 23(ii) and (17), we have

Since , we thus have

where . By Cauchy-Schwarz, we have , and from the bound , we thus have

Replacing by the schlicht function (which rotates by ) and optimising in , we obtain the claim .

Exercise 25Show that equality in the above bound is only attained when is a rotated Koebe function.

The Loewner equation (14) takes a special form in the case of slit domains. Indeed, let be a slit domain not containing the origin, with conformal radius around , and let be the Loewner chain with . We can parameterise so that the sets have conformal radius around for every , in which case we see that must be the unique conformal map from to with and . For instance, for the chain (7) we would have .

Theorem 26 (Loewner equation for slit domains)In the above situation, we have the Loewner equation holding with

*Proof:* Let be a time where the Loewner equation holds. For , the image is with a Jordan arc removed, where is a point on the boundary of the sphere. The boundary of is thus not quite a Jordan curve, so Carathéodory's theorem (Theorem 20 from Notes 2) does not directly apply. But one can for instance work with a branch of the square root of to make the boundary a Jordan curve, and square again, to conclude that at least extends continuously to be a map from to , with an arc on the boundary mapping (in two-to-one fashion) to the arc . From this and (12), we see that converges to zero outside of the arc , which by the Herglotz representation theorem implies that the measure associated to is supported on the arc . From (13) we see that must converge to the point as approaches , and so converges vaguely to the Dirac mass at . Since converges locally uniformly to , we conclude the formula (18).

In fact one can show that extends to a continuous function , and that the Loewner equation holds for all , but this is a bit trickier to show (it requires some further distortion estimates on conformal maps, related to the arguments used to prove Carathéodory’s theorem in the previous notes) and will not be done here. One can think of the function as “driving force” that incrementally enlarges the slit via the Loewner equation; this perspective is often used when studying the Schramm-Loewner evolution, which is the topic of the next (and final) set of notes.

** — 3. The Bieberbach conjecture — **

We now turn to the resolution of the Bieberbach (and Robertson) conjectures. We follow the simplified treatment of de Branges’ original proof, due to FitzGerald and Pommerenke, though we omit the proof of one key ingredient, namely the non-negativity of a certain hypergeometric function.

The first step is to work not with the Taylor coefficients of a schlicht function or with an odd schlicht function , but rather with the (normalised) logarithm of a schlicht function , as the coefficients end up obeying more tractable equations. To transfer to this setting we need the following elementary inequalities relating the coefficients of a power series with the coefficients of its exponential.

Lemma 27 (Second Lebedev-Milin inequality)Let be a formal power series with complex coefficients and no constant term, and let be its formal exponential, thus

*Proof:* If we formally differentiate (19) in , we obtain the identity

extracting the coefficient for any , we obtain the formula

By Cauchy-Schwarz, we thus have

Using and telescoping series, it thus suffices to prove the identity

But this follows from observing that

and that

for all .

Exercise 28Show that equality holds in (20) for a given if and only if there is such that for all .

Exercise 29 (First Lebedev-Milin inequality)With the notation as in the above lemma, and under the additional assumption , prove that(

Hint:using the Cauchy-Schwarz inequality as above, first show that the power series is bounded term-by-term by the power series of .) When does equality occur?

Exercise 30 (Third Lebedev-Milin inequality)With the notation as in the above lemma, show that(

Hint:use the second Lebedev-Milin inequality and (21), together with the calculus inequality for all .) When does equality occur?

Using these inequalities, one can reduce the Robertson and Bieberbach conjectures to the following conjecture of Milin, also proven by de Branges:

Theorem 31 (Milin conjecture)Let be a schlicht function. Let be the branch of the logarithm of that equals at the origin, thus one hasfor some complex coefficients . Then one has

for all .

Indeed, if

is an odd schlicht function, let be the schlicht function given by (4), then

Applying Lemma 27 with , we obtain the Robertson conjecture, and the Bieberbach conjecture follows.

Example 32If is the Koebe function (1), thenso in this case and . Similarly, for the rotated Koebe function (2) one has and again . Note how in these examples has no dependence on ; this is already a hint that this is a good choice of functions to use to analyse the problem.

To prove the Milin conjecture, we use the Loewner chain method. It suffices by Theorem 15 and a limiting argument to do so in the case that is a slit domain. Then, by Theorem 26, is the initial function of a Loewner chain that solves the Loewner equation

for all and almost every , and some function .

We can transform this into an equation for . Indeed, for non-zero we may divide by to obtain

(for any local branch of the logarithm) and hence

Since , is equal to at the origin (for an appropriate branch of the logarithm). Thus we can write

The are locally Lipschitz in (basically thanks to Lemma 19) and for almost every we have the Taylor expansions

and

Comparing coefficients, we arrive at the system of ordinary differential equations

Fix (we will not need to use any induction on here). We would like to use the system (22) to show that

The most naive attempt to do this would be to show that one has a monotonicity formula

for all , and that the expression goes to zero as , as the claim would then follow from the fundamental theorem of calculus. This turns out to not quite work; however it turns out that a slight modification of this idea does work. Namely, we introduce the quantities

where for each , is a continuously differentiable function to be chosen later. If we have the initial condition

for all , then the Milin conjecture is equivalent to asking that . On the other hand, if we impose a boundary condition

for , then we also have as , since is schlicht and hence is a normal family, implying that the are bounded in for each . Thus, to solve the Milin, Robertson, and Bieberbach conjectures, it suffices to find a choice of weights obeying the initial and boundary conditions (23), (24), and such that

for almost every (note that will be Lipschitz, so the fundamental theorem of calculus applies).

Let us now try to establish (25) using (22). We first write , and drop the explicit dependence on , thus

for . To simplify this equation, we make a further transformation, introducing the functions

(with the convention ); then we can write the above equation as

We can recover the from the by the formula

It may be worth recalling at this point that in the example of the rotated Koebe Loewner chain (2) one has , , and , for some real constant . Observe that has a simpler form than in this example, suggesting again that the decision to transform the problem to one about the rather than the is on the right track.

We now calculate

Conveniently, the unknown function no longer appears explicitly! Some simple algebra shows that

and hence by summation by parts

with the convention .

In the example of the rotated Koebe function, with , the factors and both vanish, which is consistent with the fact that vanishes in this case regardless of the choice of weights . So these two factors look to be related to each other. On the other hand, for more general choices of , these two expressions do not have any definite sign. For comparison, the quantity also vanishes when , and has a definite sign. So it is natural to see of these three factors are related to each other. After a little bit of experimentation, one eventually discovers the following elementary identity giving such a connection:

Inserting this identity into the above equation, we obtain

which can be rearranged as

We can kill the first summation by fiat, by imposing the requirement that the obey the system of differential equations

Hence if we also have the non-negativity condition

for all and , we will have obtained the desired monotonicity (25).

To summarise, in order to prove the Milin conjecture for a fixed value of , we need to find functions obeying the initial condition (23), the boundary condition (24), the differential equation (26), and the nonnegativity condition (27), with the convention . This is a significant reduction to the problem, as one just has to write down an explicit formula for such functions and verify all the properties.

Let us work out some simple cases. First consider the case . Now our task is to solve the system

for all . This is easy: we just take (indeed this is the unique choice). This gives the case of the Milin conjecture (which corresponds to the case of Bieberbach).

Next consider the case . The system is now

Again, a routine computation shows that there is a unique solution here, namely and . This gives the case of the Milin conjecture (which corresponds to the case of Bieberbach). One should compare this argument to that in Theorem 24, in particular one should see very similar weight functions emerging.

Let us now move on to . The system is now

A slightly lengthier calculation gives the unique explicit solution

to the above conditions.

These simple cases already indicate that there is basically only one candidate for the weights that will work. A calculation can give the explicit formula:

Exercise 33Let .

- (i) Show there is a unique choice of continuously differentiable functions that solve the differential equations (26) with initial condition (23), with the convention . (Use the Picard existence theorem.)
- (ii) For any , show that the expression
is equal to when is even and when is odd.

- (iii) Show that the functions
for obey the properties (23), (26), (24). (

Hint:for (23), first use (ii) to show that is equal to when is even and when is odd, then use (26).)

The Bieberbach conjecture is then reduced to the claim that

for any and . This inequality can be directly verified for any fixed ; for general it follows from general inequalities on Jacobi polynomials by Askey and Gasper, with an alternate proof given subsequently by Gasper. A further proof of (28), based on a variant of the above argument due to Weinstein that avoids explicit use of (28), appears in this article of Koepf. We will not detail these arguments here.

]]>Significant progress has been made since the last update; by implementing the “barrier” method to establish zero free regions for by leveraging the extensive existing numerical verification of the Riemann hypothesis (which establishes zero free regions for ), we have been able to improve our upper bound on from 0.48 to 0.28. Furthermore, there appears to be a bit of further room to improve the bounds further by tweaking the parameters used in the argument (we are currently using ); the most recent idea is to try to use exponential sum estimates to improve the bounds on the derivative of the approximation to that is used in the barrier method, which currently is the most computationally intensive step of the argument.

]]>In previous quarters, we proved a fundamental theorem about this concept, the Riemann mapping theorem:

Theorem 1 (Riemann mapping theorem)Let be a simply connected open subset of that is not all of . Then is conformally equivalent to the unit disk .

This theorem was proven in these 246A lecture notes, using an argument of Koebe. At a very high level, one can sketch Koebe’s proof of the Riemann mapping theorem as follows: among all the injective holomorphic maps from to that map some fixed point to , pick one that maximises the magnitude of the derivative (ignoring for this discussion the issue of proving that a maximiser exists). If avoids some point in , one can compose with various holomorphic maps and use Schwarz’s lemma and the chain rule to increase without destroying injectivity; see the previous lecture notes for details. The conformal map is unique up to Möbius automorphisms of the disk; one can fix the map by picking two distinct points in , and requiring to be zero and to be positive real.

It is a beautiful observation of Thurston that the concept of a conformal mapping has a discrete counterpart, namely the mapping of one circle packing to another. Furthermore, one can run a version of Koebe’s argument (using now a discrete version of Perron’s method) to prove the Riemann mapping theorem through circle packings. In principle, this leads to a mostly elementary approach to conformal geometry, based on extremely classical mathematics that goes all the way back to Apollonius. However, in order to *prove* the basic existence and uniqueness theorems of circle packing, as well as the convergence to conformal maps in the continuous limit, it seems to be necessary (or at least highly convenient) to use much more modern machinery, including the theory of quasiconformal mapping, and also the Riemann mapping theorem itself (so in particular we are not structuring these notes to provide a completely independent proof of that theorem, though this may well be possible).

To make the above discussion more precise we need some notation.

Definition 2 (Circle packing)A (finite)circle packingis a finite collection of circles in the complex numbers indexed by some finite set , whose interiors are all disjoint (but which are allowed to be tangent to each other), and whose union is connected. Thenerveof a circle packing is the finite graph whose vertices are the centres of the circle packing, with two such centres connected by an edge if the circles are tangent. (In these notes all graphs are undirected, finite and simple, unless otherwise specified.)

It is clear that the nerve of a circle packing is connected and planar, since one can draw the nerve by placing each vertex (tautologically) in its location in the complex plane, and drawing each edge by the line segment between the centres of the circles it connects (this line segment will pass through the point of tangency of the two circles). Later in these notes we will also have to consider some infinite circle packings, most notably the infinite regular hexagonal circle packing.

The first basic theorem in the subject is the following converse statement:

Theorem 3 (Circle packing theorem)Every connected planar graph is the nerve of a circle packing.

Of course, there can be multiple circle packings associated to a given connected planar graph; indeed, since reflections across a line and Möbius transformations map circles to circles (or lines), they will map circle packings to circle packings (unless one or more of the circles is sent to a line). It turns out that once one adds enough edges to the planar graph, the circle packing is otherwise rigid:

Theorem 4 (Koebe-Andreev-Thurston theorem)If a connected planar graph is maximal (i.e., no further edge can be added to it without destroying planarity), then the circle packing given by the above theorem is unique up to reflections and Möbius transformations.

Exercise 5Let be a connected planar graph with vertices. Show that the following are equivalent:

- (i) is a maximal planar graph.
- (ii) has edges.
- (iii) Every drawing of divides the plane into faces that have three edges each. (This includes one unbounded face.)
- (iv) At least one drawing of divides the plane into faces that have three edges each.
(

Hint:use Euler’s formula , where is the number of faces including the unbounded face.)

Thurston conjectured that circle packings can be used to approximate the conformal map arising in the Riemann mapping theorem. Here is an informal statement:

Conjecture 6 (Informal Thurston conjecture)Let be a simply connected domain, with two distinct points . Let be the conformal map from to that maps to the origin and to a positive real. For any small , let be the portion of the regular hexagonal circle packing by circles of radius that are contained in , and let be an circle packing of with all “boundary circles” tangent to , giving rise to an “approximate map” defined on the subset of consisting of the circles of , their interiors, and the interstitial regions between triples of mutually tangent circles. Normalise this map so that is zero and is a positive real. Then converges to as .

A rigorous version of this conjecture was proven by Rodin and Sullivan. Besides some elementary geometric lemmas (regarding the relative sizes of various configurations of tangent circles), the main ingredients are a rigidity result for the regular hexagonal circle packing, and the theory of quasiconformal maps. Quasiconformal maps are what seem on the surface to be a very broad generalisation of the notion of a conformal map. Informally, conformal maps take infinitesimal circles to infinitesimal circles, whereas quasiconformal maps take infinitesimal circles to infinitesimal ellipses of bounded eccentricity. In terms of Wirtinger derivatives, conformal maps obey the Cauchy-Riemann equation , while (sufficiently smooth) quasiconformal maps only obey an inequality . As such, quasiconformal maps are considerably more plentiful than conformal maps, and in particular it is possible to create piecewise smooth quasiconformal maps by gluing together various simple maps such as affine maps or Möbius transformations; such piecewise maps will naturally arise when trying to rigorously build the map alluded to in the above conjecture. On the other hand, it turns out that quasiconformal maps still have many vestiges of the rigidity properties enjoyed by conformal maps; for instance, there are quasiconformal analogues of fundamental theorems in conformal mapping such as the Schwarz reflection principle, Liouville’s theorem, or Hurwitz’s theorem. Among other things, these quasiconformal rigidity theorems allow one to create conformal maps from the limit of quasiconformal maps in many circumstances, and this will be how the Thurston conjecture will be proven. A key technical tool in establishing these sorts of rigidity theorems will be the theory of an important quasiconformal (quasi-)invariant, the *conformal modulus* (or, equivalently, the extremal length, which is the reciprocal of the modulus).

** — 1. Proof of the circle packing theorem — **

We loosely follow the treatment of Beardon and Stephenson. It is slightly more convenient to temporarily work in the Riemann sphere rather than the complex plane , in order to more easily use Möbius transformations. (Later we will make another change of venue, working in the Poincaré disk instead of the Riemann sphere.)

Define a *Riemann sphere circle* to be either a circle in or a line in together with , together with one of the two components of the complement of this circle or line designated as the “interior”. In the case of a line, this “interior” is just one of the two half-planes on either side of the line; in the case of the circle, this is either the usual interior or the usual exterior plus the point at infinity; in the last case, we refer to the Riemann sphere circle as an *exterior circle*. (One could also equivalently work with an orientation on the circle rather than assigning an interior, since the interior could then be described as the region to (say) the left of the circle as one traverses the circle along the indicated orientation.) Note that Möbius transforms map Riemann sphere circles to Riemann sphere circles. If one views the Riemann sphere as a geometric sphere in Euclidean space , then Riemann sphere circles are just circles on this geometric sphere, which then have a centre on this sphere that lies in the region designated as the interior of the circle. We caution though that this “Riemann sphere” centre does not always correspond to the Euclidean notion of the centre of a circle. For instance, the real line, with the upper half-plane designated as interior, will have as its Riemann sphere centre; if instead one designates the lower half-plane as the interior, the Riemann sphere centre will now be . We can then define a Riemann sphere circle packing in exact analogy with circle packings in , namely finite collections of Riemann sphere circles whose interiors are disjoint and whose union is connected; we also define the nerve as before. This is now a graph that can be drawn in the Riemann sphere, using great circle arcs in the Riemann sphere rather than line segments; it is also planar, since one can apply a Möbius transformation to move all the points and edges of the drawing away from infinity.

By Exercise 5, a maximal planar graph with at least three vertices can be drawn as a triangulation of the Riemann sphere. If there are at least four vertices, then it is easy to see that each vertex has degree at least three (a vertex of degree zero, one or two in a triangulation with simple edges will lead to a connected component of at most three vertices). It is a topological fact, not established here, that any two triangulations of such a graph are homotopic up to reflection (to reverse the orientation). If a Riemann sphere circle packing has the nerve of a maximal planar graph of at least four vertices, then we see that this nerve induces an explicit triangulation of the Riemann sphere by connecting the centres of any pair of tangent circles with the great circle arc that passes through the point of tangency. If was not maximal, one no longer gets a triangulation this way, but one still obtains a partition of the Riemann sphere into spherical polygons.

We remark that the triangles in this triangulation can also be described purely from the abstract graph . Define a *triangle* in to be a triple of vertices in which are all adjacent to each other, and such that the removal of these three vertices from does not disconnect the graph. One can check that there is a one-to-one correspondence between such triangles in a maximal planar graph and the triangles in any Riemann sphere triangulation of this graph.

Theorems 3, 4 are then a consequence of

Theorem 7 (Riemann sphere circle packing theorem)Let be a maximal planar graph with at least four vertices, drawn as a triangulation of the Riemann sphere. Then there exists a Riemann sphere circle packing with nerve whose triangulation is homotopic to the given triangulation. Furthermore, this packing is unique up to Möbius transformations.

Exercise 8Deduce Theorems 3, 4 from Theorem 7. (Hint:If one has a non-maximal planar graph for Theorem 3, add a vertex at the interior of each non-triangular face of a drawing of that graph, and connect that vertex to the vertices of the face, to create a maximal planar graph to which Theorem 4 or Theorem 7 can be applied. Then delete these “helper vertices” to create a packing of the original planar graph that does not contain any “unwanted” tangencies. You may use without proof the above assertion that any two triangulations of a maximal planar graph are homotopic up to reflection.)

Exercise 9Verify Theorem 7 when has exactly four vertices. (Hint:for the uniqueness, one can use Möbius transformations to move two of the circles to become parallel lines.)

To prove this theorem, we will make a reduction with regards to the existence component of Theorem 7. For technical reasons we will need to introduce a notion of non-degeneracy. Let be a maximal planar graph with at least four vertices, and let be a vertex in . As discussed above, the degree of is at least three. Writing the neighbours of in clockwise or counterclockwise order (with respect to a triangulation) as (starting from some arbitrary neighbour), we see that each is adjacent to and (with the conventions and ). We say that is *non-degenerate* if there are no further adjacencies between the , and if there is at least one further vertex in besides . Here is another characterisation:

Exercise 10Let be a maximal planar graph with at least four vertices, let be a vertex in , and let be the neighbours of . Show that the following are equivalent:

- (i) is non-degenerate.
- (ii) The graph is connected and non-empty, and every vertex in is adjacent to at least one vertex in .

We will then derive Theorem 7 from

Theorem 11 (Inductive step)Let be a maximal planar graph with at least four vertices , drawn as a triangulation of the Riemann sphere. Let be a non-degenerate vertex of , and let be the graph formed by deleting (and edges emenating from ) from . Suppose that there exists a Riemann sphere circle packing whose nerve isat least(that is, and are tangent whenever are adjacent in , although we also allow additional tangencies), and whose associated subdivision of the Riemann sphere into spherical polygons is homotopic to the given triangulation with removed. Then there is a Riemann sphere circle packing with nerve whose triangulation is homotopic to the given triangulation. Furthermore this circle packing is unique up to Möbius transformations.

Let us now see how Theorem 7 follows from Theorem 14. Fix as in Theorem 7. By Exercise 9 and induction we may assume that has at least five vertices, and that the claim has been proven for any smaller number of vertices.

First suppose that contains a non-degenerate vertex . Let be the the neighbours of . One can then form a new graph with one fewer vertex by deleting , and then connecting to (one can think of this operation as contracting the edge to a point). One can check that this is still a maximal planar graph that can triangulate the Riemann sphere in a fashion compatible with the original triangulation of (in that all the common vertices, edges, and faces are unchanged). By induction hypothesis, is the nerve of a circle packing that is compatible with this triangulation, and hence this circle packing has nerve at least . Applying Theorem 14, we then obtain the required claim for .

Now suppose that contains a degenerate vertex . Let be the neighbours of traversed in order. By hypothesis, there is an additional adjacency between the ; by relabeling we may assume that is adjacent to for some . The vertices in can then be partitioned as

where denotes those vertices in that lie in the region enclosed by the loop that does not contain , and denotes those vertices in that lie in the region enclosed by the loop that does not contain . One can then form two graphs , formed by restricting to the vertices and respectively; furthermore, these graphs are also maximal planar (with triangulations that are compatible with those of ). By induction hypothesis, we can find a circle packing with nerve , and a circle packing with nerve . Note that the circles are mutually tangent, as are . By applying a Möbius transformation one may assume that these circles agree, thus (cf. Exercise 9) , . The complement of the these three circles (and their interiors) determine two connected “interstitial” regions (that are in the shape of an arbelos, up to Möbius transformation); one can check that the remaining circles in will lie in one of these regions, and the remaining circles in lie in the other. Hence one can glue these circle packings together to form a single circle packing with nerve , which is homotopic to the given triangulation. Also, since a Möbius transformation that fixes three mutually tangent circles has to be the identity, the uniqueness of this circle packing up to Möbius transformations follows from the uniqueness for the two component circle packings , .

It remains to prove Theorem 7. To help fix the freedom to apply Möbius transformations, we can normalise the target circle packing so that is the exterior circle , thus all the other circles in the packing will lie in the closed unit disk . Similarly, by applying a suitable Möbius transformation one can assume that lies outside of the interior of all the circles in the original packing, and after a scaling one may then assume that all the circles lie in the unit disk .

At this point it becomes convenient to switch from the “elliptic” conformal geometry of the Riemann sphere to the “hyperbolic” conformal geometry of the unit disk . Recall that the Möbius transformations that preserve the disk are given by the maps

for real and (see Theorem 19 of these notes). It comes with a natural metric that interacts well with circles:

Exercise 12Define the Poincaré distance between two points of by the formulaGiven a measurable subset of , define the

hyperbolic areaof to be the quantitywhere is the Euclidean area element on .

- (i) Show that the Poincaré distance is invariant with respect to Möbius automorphisms of , thus whenever is a transformation of the form (1). Similarly show that the hyperbolic area is invariant with respect to such transformations.
- (ii) Show that the Poincaré distance defines a metric on . Furthermore, show that any two distinct points are connected by a unique geodesic, which is a portion of either a line or a circle that meets the unit circle orthogonally at two points. (
Hint:use the symmetries of (i) to normalise the points one is studying.)- (iii) If is a circle in the interior of , show that there exists a point in and a positive real number (which we call the
hyperbolic centerandhyperbolic radiusrespectively) such that . (In general, the hyperbolic center and radius will not quite agree with their familiar Euclidean counterparts.) Conversely, show that for any and , the set is a circle in .- (iv) If two circles in are externally tangent, show that the geodesic connecting the hyperbolic centers passes through the point of tangency, orthogonally to the two tangent circles.

Exercise 13 (Schwarz-Pick theorem)Let be a holomorphic map. Show that for all . If , show that equality occurs if and only if is a Möbius automorphism (1) of . (This result is known as the Schwarz-Pick theorem.)

We will refer to circles that lie in the closure of the unit disk as *hyperbolic circles*. These can be divided into the finite radius hyperbolic circles, which lie in the interior of the unit disk (as per part (iii) of the above exercise), and the horocycles, which are internally tangent to the unit circle. By convention, we view horocycles as having infinite radius, and having center at their point of tangency to the unit circle; they can be viewed as the limiting case of finite radius hyperbolic circles when the radius goes to infinity and the center goes off to the boundary of the disk (at the same rate as the radius, as measured with respect to the Poincaré distance). We write for the hyperbolic circle with hyperbolic centre and hyperbolic radius (thus either and , or and is on the unit circle); there is an annoying caveat that when there is more than one horocycle with hyperbolic centre , but we will tolerate this breakdown of functional dependence of on and in order to simplify the notation. A *hyperbolic circle packing* is a circle packing in which all circles are hyperbolic circles.

We also observe that the geodesic structure extends to the boundary of the unit disk: for any two distinct points in , there is a unique geodesic that connects them.

In view of the above discussion, Theorem 7 may now be formulated as follows:

Theorem 14 (Inductive step, hyperbolic formulation)Let be a maximal planar graph with at least four vertices , let be a non-degenerate vertex of , and let be the vertices adjacent to . Suppose that there exists a hyperbolic circle packing whose nerve is at least . Then there is a hyperbolic circle packing homotopic to such that the boundary circles , are all horocycles. Furthermore, this packing is unique up to Möbius automorphisms (1) of the disk .

Indeed, once one adjoints the exterior unit circle to , one obtains a Riemann sphere circle packing whose nerve is at least , and hence equal to since is maximal.

To prove this theorem, the intuition is to “inflate” the hyperbolic radius of the circles of until the boundary circles all become infinite radius (i.e., horocycles). The difficulty is that one cannot just arbitrarily increase the radius of any given circle without destroying the required tangency properties. The resolution to this difficulty given in the work of Beardon and Stephenson that we are following here was inspired by Perron’s method of subharmonic functions, in which one faced an analogous difficulty that one could not easily manipulate a harmonic function without destroying its harmonicity. There, the solution was to work instead with the more flexible class of subharmonic functions; here we similarly work with the concept of a *subpacking*.

We will need some preliminaries to define this concept precisely. We first need some hyperbolic trigonometry. We define a hyperbolic triangle to be the solid (and closed) region in enclosed by three distinct points in and the geodesic arcs connecting them. (Note that we allow one or more of the vertices to be on the boundary of the disk, so that the sides of the triangle could have infinite length.) Let be the space of triples with and not all of infinite. We say that a hyperbolic triangle with vertices is a *-triangle* if there are hyperbolic circles with the indicated hyperbolic centres and hyperbolic radii that are externally tangent to each other; note that this implies that the sidelengths opposite have length respectively (see Figure 3 of Beardon and Stephenson). It is easy to see that for any , there exists a unique -triangle in up to reflections and Möbius automorphisms (use Möbius transforms to fix two of the hyperbolic circles, and consider all the circles externally tangent to both of these circles; the case when one or two of the are infinite may need to be treated separately.). As a consequence, there is a well defined angle for subtended by the vertex of an triangle. We need some basic facts from hyperbolic geometry:

Exercise 15 (Hyperbolic trigonometry)

- (i) (Hyperbolic cosine rule) For any , show that the quantity is equal to the ratio
Furthermore, establish the limiting angles

(Hint: to facilitate computations, use a Möbius transform to move the vertex to the origin when the radius there is finite.) Conclude in particular that is continuous (using the topology of the extended real line for each component of ). Discuss how this rule relates to the Euclidean cosine rule in the limit as go to zero. Of course, by relabeling one obtains similar formulae for and .

- (ii) (Area rule) Show that the area of a hyperbolic triangle is given by , where are the angles of the hyperbolic triangle. (Hint: there are several ways to proceed. For instance, one can prove this for small hyperbolic triangles (of diameter ) up to errors of size after normalising as in (ii), and then establish the general case by subdividing a large hyperbolic triangle into many small hyperbolic triangles. This rule is also a special case of the Gauss-Bonnet theorem in Riemannian geometry. One can also first establish the case when several of the radii are infinite, and use that to derive finite cases.) In particular, the area of a -triangle is given by the formula
- (iii) Show that the area of the interior of a hyperbolic circle with is equal to .

Henceforth we fix as in Theorem 14. We refer to the vertices as *boundary vertices* of and the remaining vertices as *interior vertices*; edges between boundary vertices are *boundary edges*, all other edges will be called *interior edges* (including edges that have one vertex on the boundary). Triangles in that involve two boundary vertices (and thus necessarily one interior vertex) will be called *boundary triangles*; all other triangles (including ones that involve one boundary vertex) will be called *interior triangles*. To any triangle of , we can form the hyperbolic triangle with vertices ; this is an -triangle. Let denote the collection of such hyperbolic triangles; because is a packing, we see that these triangles have disjoint interiors. They also fit together in the following way: if is a side of a hyperbolic triangle in , then there will be another hyperbolic triangle in that shares that side precisely when is associated to an interior edge of . The union of all these triangles is homeomorphic to the region formed by starting with a triangulation of the Riemann sphere by and removing the triangles containing as a vertex, and is therefore homeomorphic to a disk. One can think of the collection of hyperbolic triangles, together with the vertices and edges shared by these triangles, as a two-dimensional (hyperbolic) simplicial complex, though we will not develop the full machinery of such complexes here.

Our objective is to find another hyperbolic circle packing homotopic to the existing circle packing , such at all the boundary circles (circles centred at boundary vertices) are horocycles. We observe that such a hyperbolic circle packing is completely described (up to Möbius transformations) by the hyperbolic radii of these circles. Indeed, suppose one knows the values of these hyperbolic radii. Then each hyperbolic triangle in is associated to a hyperbolic triangle whose sides and angles are known from Exercise 15. As the orientation of each hyperbolic triangle is fixed, each hyperbolic triangle is determined up to a Möbius automorphism of . Once one fixes one hyperbolic triangle, the adjacent hyperbolic triangles (that share a common side with the first triangle) are then also fixed; continuing in this fashion we see that the entire hyperbolic circle packing is determined.

On the other hand, not every choice of radii will lead to a hyperbolic circle packing with the required properties. There are two obvious constraints that need to be satisfied:

- (i) (Local constraint) The angles of all the hyperbolic triangles around any given interior vertex must sum to exactly .
- (ii) (Boundary constraint) The radii associated to boundary vertices must be infinite.

There could potentially also be a global constraint, in that one requires the circles of the packing to be disjoint – including circles that are not necessarily adjacent to each other. In general, one can easily create configurations of circles that are local circle packings but not global ones (see e.g., Figure 7 of Beardon-Stephenson). However, it turns out that one can use the boundary constraint and topological arguments to prevent this from happening. We first need a topological lemma:

Lemma 16 (Topological lemma)Let be bounded connected open subsets of with simply connected, and let be a continuous map such that and . Suppose furthermore that the restriction of to is a local homeomorphism. Then is in fact a global homeomorphism.

The requirement that the restriction of to be a local homeomorphism can in fact be relaxed to local injectivity thanks to the invariance of domain theorem. The complex numbers can be replaced here by any finite-dimensional vector space.

*Proof:* The preimage of any point in the interior of is closed, discrete, and disjoint from , and is hence finite. Around each point in the preimage, there is a neighbourhood on which is a homeomorphism onto a neighbourhood of . If one deletes the closure of these neighbourhoods, the image under is compact and avoids , and thus avoids a neighbourhood of . From this we can show that is a covering map from to . As the base is simply connected, it is its own universal cover, and hence (by the connectedness of ) must be a homeomorphism as claimed.

Proposition 17Suppose we assign a radius to each that obeys the local constraint (i) and the boundary constraint (ii). Then there is a hyperbolic circle packing with nerve and the indicated radii.

*Proof:* We first create the hyperbolic triangles associated with the required hyperbolic circle packing, and then verify that this indeed arises from a circle packing.

Start with a single triangle in , and arbitrarily select a -triangle with the same orientation as . By Exercise 15(i), such a triangle exists (and is unique up to Möbius automorphisms of the disk). If a hyperbolic triangle has been fixed, and (say) is an adjacent triangle in , we can select to be the unique -triangle with the same orientation as that shares the side in common with (with the and vertices agreeing). Similarly for other permutations of the labels. As is a maximal planar graph with non-degenerate (so in particular the set of internal vertices is connected), we can continue this construction to eventually fix every triangle in . There is the potential issue that a given triangle may depend on the order in which one arrives at that triangle starting from , but one can check from a monodromy argument (in the spirit of the monodromy theorem) using the local constraint (i) and the simply connected nature of the triangulation associated to that there is in fact no dependence on the order. (The process resembles that of laying down jigsaw pieces in the shape of hyperbolic triangles together, with the local constraint ensuring that there is always a flush fit locally.)

Now we show that the hyperbolic triangles have disjoint interiors inside the disk . Let denote the topological space formed by taking the disjoint union of the hyperbolic triangles (now viewed as abstract topological spaces rather than subsets of the disk) and then gluing together all common edges, e.g. identifying the edge of with the same edge of if and are adjacent triangles in . This space is homeomorphic to the union of the original hyperbolic triangles , and is thus homeomorphic to the closed unit disk. There is an obvious projection map from to the union of the , which maps the abstract copy in of a given hyperbolic triangle to its concrete counterpart in in the obvious fashion. This map is continuous. It does not quite cover the full closed disk, mainly because (by the boundary condition (ii)) the boundary hyperbolic triangles touch the boundary of the disk at the vertices associated to and but do not follow the boundary arc connecting these vertices, being bounded instead by the geodesic from the vertex to the vertex; the missing region is a lens-shaped region bounded by two circular arcs. However, by applying another homeomorphism (that does not alter the edges from to or to ), one can “push out” the edge of this hyperbolic triangle across the lens to become the boundary arc from to . If one performs this modification for each boundary triangle, one arrives at a modified continuous map from to , which now has the property that the boundary of maps to the boundary of the disk, and the interior of maps to the interior of the disk. Also one can check that this map is a local homeomorphism. By Lemma 16, is injective; undoing the boundary modifications we conclude that is injective. Thus the hyperbolic triangles have disjoint interiors. Furthermore, the arguments show that for each boundary triangle , the lens-shaped regions between the boundary arc between the vertices associated to and the corresponding edge of the boundary triangle are also disjoint from the hyperbolic triangles and from each other. On the other hand, all of the hyperbolic circles and in and their interiors are contained in the union of the hyperbolic triangles and the lens-shaped regions, with each hyperbolic triangle containing portions only of the hyperbolic circles with hyperbolic centres at the vertices of the triangle, and similarly for the lens-shaped regions. From this one can verify that the interiors of the hyperbolic circles are all disjoint from each other, and give a hyperbolic circle packing with the required properties.

In view of the above proposition, the only remaining task is to find an assignment of radii obeying both the local condition (i) and the boundary condition (ii). This is analogous to finding a harmonic function with specified boundary data. To do this, we perform the following analogue of Perron’s method. Define a *subpacking* to be an assignment of radii obeying the following

- (i’) (Local sub-condition) The angles around any given interior vertex sum to at least .

This can be compared with the definition of a (smooth) subharmonic function as one where the Laplacian is always at least zero. Note that we always have at least one subpacking, namely the one provided by the radii of the original hyperbolic circle packing . Intuitively, in each subpacking, the radius at an interior vertex is either “too small” or “just right”.

We now need a key monotonicity property, analogous to how the maximum of two subharmonic functions is again subharmonic:

- (i) Show that the angle (as defined in Exercise 15(i)) is strictly decreasing in and strictly increasing in or (if one holds the other two radii fixed). Do these claims agree with your geometric intuition?
- (ii) Conclude that whenever and are subpackings, that is also a subpacking.
- (iii) Let be such that for . Show that , with equality if and only if for all . (
Hint:increase just one of the radii . One can either use calculus (after first disposing of various infinite radii cases) or one can argue geometrically.)

As with Perron’s method, we can now try to construct a hyperbolic circle packing by taking the supremum of all the subpackings. To avoid degeneracies we need an upper bound:

Proposition 19 (Upper bound)Let be a subpacking. Then for any interior vertex of degree , one has .

The precise value of is not so important for our arguments, but the fact that it is finite will be. This boundedness of interior circles in a circle packing is a key feature of hyperbolic geometry that is not present in Euclidean geometry, and is one of the reasons why we moved to a hyperbolic perspective in the first place.

*Proof:* By the subpacking property and pigeonhole principle, there is a triangle in such that . The hyperbolic triangle associated to has area at most by (2); on the other hand, it contains a sector of a hyperbolic circle of radius and angle , and hence has area at least , thanks to Exercise 15(iv). Comparing the two bounds gives the claim.

Now define to be the (pointwise) supremum of all the subpackings. By the above proposition, is finite at every interior vertex. By Exercise 18, one can view as a monotone increasing limit of subpackings, and is thus again a subpacking (due to the continuity properties of as long as at least one of the radii stays bounded); thus is the maximal subpacking. On the other hand, if is finite at some boundary vertex, then by Exercise 18(i) one could replace that radius by a larger quantity without destroying the subpacking property, contradicting the maximality of . Thus all the boundary radii are infinite, that is to say the boundary condition (ii) holds. Finally, if the sum of the angles at an interior vertex is strictly greater than , then by Exercise 18 we could increase the radius at this vertex slightly without destroying the subpacking property at or at any other of the interior vertices, again contradicting the maximality of . Thus obeys the local condition (i), and we have demonstrated existence of the required hyperbolic circle packing.

Finally we establish uniqueness. It suffices to establish that is the unique tuple that obeys the local condition (i) and the boundary condition (ii). Suppose we had another tuple other than that obeyed these two conditions. Then by the maximality of , we have for all . By Exercise 18(iii), this implies that

for any triangle in . Summing over all triangles and using (2), we conclude that

where the inner sum is over the pairs such that forms a triangle in . But by the local condition (i) and the boundary condition (ii), the inner sum on either side is equal to for an interior vertex and for a boundary vertex. Thus the two sides agree, which by Exercise 18(iii) implies that for all . This proves Theorem 14 and thus Theorems 7, 3, 4.

** — 2. Quasiconformal maps — **

In this section we set up some of the foundational theory of quasiconformal mapping, which are generalisations of the conformal mapping concept that can tolerate some deviations from perfect conformality, while still retaining many of the good properties of conformal maps (such as being preserved under uniform limits), though with the notable caveat that in contrast to conformal maps, quasiconformal maps need not be smooth. As such, this theory will come in handy when proving convergence of circle packings to the Riemann map. The material here is largely drawn from the text of Lehto and Virtanen.

We first need the following refinement of the Riemann mapping theorem, known as Carathéodory’s theorem:

Theorem 20 (Carathéodory’s theorem)Let be a bounded simply connected domain in whose boundary is a Jordan curve, and let be a conformal map between and (as given by the Riemann mapping theorem). Then extends to a continuous homeomorphism from to .

The condition that be a Jordan curve is clearly necessary, since if is not simple then there are paths in that end up at different points in but have the same endpoint in after applying , which prevents being continuously extended to a homeomorphism.

*Proof:* We first prove continuous extension to the boundary. It suffices to show that for every point on the boundary of the unit circle, the diameters of the sets go to zero for some sequence of radii .

First observe from the change of variables formula that the area of is given by , where denotes Lebesgue measure (or the area element). In particular, this integral is finite. Expanding in polar coordinates around , we conclude that

Since diverges near , we conclude from the pigeonhole principle that there exists a sequence of radii decreasing to zero such that

and hence by Cauchy-Schwarz

If we let denote the circular arc , we conclude from this and the triangle inequality (and chain rule) that is a rectifiable curve with length going to zero as . Let denote the endpoints of this curve. Clearly they lie in . If (say) was in , then as is a homeomorphism from to , would have one endpoint in rather than , which is absurd. Thus lies in , and similarly for . Since the length of goes to zero, the distance between and goes to zero. Since is a Jordan curve, it can be parameterised homeomorphically by , and so by compactness we also see that the distance between the parameterisations of and in must also go to zero, hence (by uniform continuity of the inverse parameterisation) and are connected along by an arc whose diameter goes to zero. Combining this arc with , we obtain a Jordan curve of diameter going to zero which separates from the rest of . Sending to infinity, we see that (which decreases with ) must eventually map in the interior of this curve rather than the exterior, and so the diameter goes to zero as claimed.

The above construction shows that extends to a continuous map (which by abuse of notation we continue to call ) from to , and the proof also shows that maps to . As is a compact subset of that contains , it must surject onto . As both and are compact Hausdorff spaces, we will now be done if we can show injectivity. The only way injectivity can fail is if there are two distinct points on that map to the same point. Let be the line segment connecting with , then is a Jordan curve in that meets only at . divides into two regions; one of which must map to the interior of , which implies that there is an entire arc of which maps to the single point . But then by the Schwarz reflection principle, extends conformally across this arc and is constant in a non-isolated set, thus is constant everywhere by analytic continuation, which is absurd. This establishes the required injectivity.

This has the following consequence. Define a *Jordan quadrilateral* to be the open region enclosed by a Jordan curve with four distinct marked points on it in counterclockwise order, which we call the *vertices* of the quadrilateral. The arcs in connecting to or to will be called the *-sides*; the arcs connecting to or to will be called *-sides*. (Thus for instance each cyclic permutation of the vertices will swap the -sides and -sides, while keeping the interior region unchanged.) A key example of a Jordan quadrilateral are the (Euclidean) rectangles, in which the vertices are the usual corners of the rectangle, traversed counterclockwise. The -sides then are line segments of some length , and the -sides are line segments of some length that are orthogonal to the -sides. A *vertex-preserving conformal map* from one Jordan quadrilateral to another will be a conformal map that extends to a homeomorphism from to that maps the corners of to the respective corners of (in particular, -sides get mapped to -sides, and similarly for -sides).

Exercise 21Let be a Jordan quadrilateral with vertices .

- (i) Show that there exists and a conformal map to the upper half-plane (viewed as a subset of the Riemann sphere) that extends continuously to a homeomorphism and which maps to respectively. (Hint: first map to increasing elements of the real line, then use the intermediate value theorem to enforce .)
- (ii) Show that there is a vertex-preserving conformal map from to a rectangle (Hint: use Schwarz-Christoffel mapping.)
- (iii) Show that the rectangle in part (ii) is unique up to affine transformations. (
Hint:if one has a conformal map between rectangles that preserves the vertices, extend it via repeated use of the Schwarz reflection principle to an entire map.)

This allows for the following definition: the *conformal modulus* (or *modulus* for short, also called *module* in older literature) of a Jordan quadrilateral with vertices is the ratio , where are the lengths of the -sides and -sides of a rectangle that is conformal to in a vertex-preserving vashion.. This is a number between and ; each cyclic permutation of the vertices replaces the modulus with its reciprocal. It is clear from construction that the modulus of a Jordan quadrilateral is unaffected by vertex-preserving conformal transformations.

Now we define quasiconformal maps. Informally, conformal maps are homeomorphisms that map infinitesimal circles to infinitesimal circles; quasiconformal maps are homeomorphisms that map infinitesimal circles to curves that differ from an infinitesimal circle by “bounded distortion”. However, for the purpose of setting up the foundations of the theory, it is slightly more convenient to work with rectangles instead of circles (it is easier to partition rectangles into subrectangles than disks into subdisks). We therefore introduce

Definition 22Let . An orientation-preserving homeomorphism between two domains in is said to be-quasiconformalif one has for every Jordan quadrilateral in . (In these notes, we do not consider orientation-reversing homeomorphisms to be quasiconformal.)

Note that by cyclically permuting the vertices of , we automatically also obtain the inequality

or equivalently

for any Jordan quadrilateral. Thus it is not possible to have any -quasiconformal maps for (excluding the degenerate case when are empty), and a map is -conformal if and only if it preserves the modulus. In particular, conformal maps are -conformal; we will shortly establish that the converse claim is also true. It is also clear from the definition that the inverse of a -quasiconformal map is also -quasiconformal, and the composition of a -quasiconformal map and a -quasiconformal map is a -quasiconformal map.

It is helpful to have an alternate characterisation of the modulus that does not explicitly mention conformal mapping:

Proposition 23 (Alternate definition of modulus)Let be a Jordan quadrilateral with vertices . Then is the smallest quantity with the following property: for any Borel measurable one can find a curve in connecting one -side of to another, and which is locally rectifiable away from endpoints, such thatwhere denotes integration using the length element of (not to be confused with the contour integral ).

The reciprocal of this notion of modulus generalises to the concept of extremal length, which we will not develop further here.

*Proof:* Observe from the change of variables formula that if is a vertex-preserving conformal mapping between Jordan quadrilaterals , and is a locally rectifiable curve connecting one -side of to another, then is a locally rectifiable curve connecting one -side of to another, with

and

As a consequence, if the proposition holds for it also holds for . Thus we may assume without loss of generality that is a rectangle, which we may normalise to be with vertices , so that the modulus is . For any measurable , we have from Cauchy-Schwarz and Fubini’s theorem that

and hence by the pigeonhole principle there exists such that

On the other hand, if we set , then , and for any curve connecting the -side from to to the -side from to , we have

Thus is the best constant with the required property, proving the claim.

Here are some quick and useful consequences of this characterisation:

**Note: it would be more logical to reverse the order of the next two exercises, since Exercise 25 can be used to prove Exercise 24. I will do this at the conclusion of this course.**

- (i) If are disjoint Jordan quadrilaterals that share a common -side, and which can be glued together along this side to form a new Jordan quadrilateral , show that . If equality occurs, show that after conformally mapping to a rectangle (in a vertex preserving fashion), , are mapped to subrectangles (formed by cutting the original parallel to the -side).
- (ii) If are disjoint Jordan quadrilaterals that share a common -side, and which can be glued together along this side to form a new Jordan quadrilateral , show that . If equality occurs, show that after conformally mapping to a rectangle (in a vertex preserving fashion), , are mapped to subrectangles (formed by cutting the original parallel to the -side).

Exercise 25 (Rengel’s inequality)Let be a Jordan quadrilateral of area , let be the shortest (Euclidean) distance between a point on one -side and a point on the other -side, and similarly let be the shortest (Euclidean) distance between a point on one -side and a point on the other -side. Show thatand that equality in either case occurs if and only if is a rectangle.

Exercise 26 (Continuity from below)Suppose is a sequence of Jordan quadrilaterals which converge to another Jordan quadrilateral , in the sense that the vertices of converge to their respective counterparts in , each -side in converges (in the Hausdorff sense) to the -side of , and the similarly for -sides. Suppose also that for all . Show that converges to . (Hint:map to a rectangle and use Rengel’s inequality.)

Proposition 27 (Local quasiconformality implies quasiconformality)Let , and let be an orientation-preserving homeomorphism between complex domains which is locally -quasiconformal in the sense that for every there is a neighbourhood of in such that is -quasiconformal from to . Then is -quasiconformal.

*Proof:* We need to show that for any Jordan quadrilateral in . The hypothesis gives this claim for all quadrilaterals in the sufficiently small neighbourhood of any point in . For any natural number , we can subdivide into quadrilaterals with modulus with adjacent -sides, by first conformally mapping to a rectangle and then doing an equally spaced vertical subdivision. Similarly, each quadrilateral can be subdivided into quadrilaterals of modulus by mapping to a rectangle and performing horizontal subdivision. By the local -quasiconformality of , we will have

for all , if is large enough. By superadditivity this implies that

for each , and hence

Applying superadditivity again we obtain

giving the claim.

We can now reverse the implication that conformal maps are -conformal:

*Proof:* By covering by quadrilaterals we may assume without loss of generality that (and hence also ) is a Jordan quadrilateral; by composing on left and right with conformal maps we may assume that and are rectangles. As is -conformal, the rectangles have the same modulus, so after a further affine transformation we may assume that is the rectangle with vertices for some modulus . If one subdivides into two rectangles along an intermediate vertical line segment connecting say to for some , the moduli of these rectangles are and . Applying the -conformal map and the converse portion of Exercise 24, we conclude that these rectangles must be preserved by , thus preserves the coordinate. Similarly preserves the coordinate, and is therefore the identity map, which is of course conformal.

Next, we can give a simple criterion for quasiconformality in the continuously differentiable case:

Theorem 29Let , and let be an orientation-preserving diffeomorphism (a continuously (real) differentiable homeomorphism whose derivative is always nondegenerate) between complex domains . Then the following are equivalent:

- (i) is -quasiconformal.
- (ii) For any point and phases , one has
where denotes the directional derivative.

*Proof:* Let us first show that (ii) implies (i). Let be a Jordan quadrilateral in ; we have to show that . From the chain rule one can check that condition (ii) is unchanged by composing with conformal maps on the left or right, so we may assume without loss of generality that and are rectangles; in fact we may normalise to have vertices and to have vertices where and . From the change of variables formula (and the singular value decomposition), followed by Fubini’s theorem and Cauchy-Schwarz, we have

3 and hence , giving the claim.

Now suppose that (ii) failed, then by the singular value decomposition we can find and a phase such that

for some real with . After translations and rotations we may normalise so that

But then from Rengel’s inequality and Taylor expansion one sees that will map a unit square with vertices to a quadrilateral of modulus converging to as , contradicting (i).

Exercise 30Show that the conditions (i), (ii) in the above theorem are also equivalent to the boundfor all , where

are the Wirtinger derivatives.

We now prove a technical regularity result on quasiconformal maps.

Proposition 31 (Absolute continuity on lines)Let be a -quasiconformal map between two complex domains for some . Suppose that contains the closed rectangle with endpoints . Then for almost every , the map is absolutely continuous on .

*Proof:* For each , let denote the area of the image of the rectangle with endpoints . This is a bounded monotone function on and is hence differentiable almost everywhere. It will thus suffice to show that the map is absolutely continuous on whenever is a point of differentiability of .

Let , and let be disjoint intervals in of total length . To show absolute continuity, we need a bound on that goes to zero as uniformly in the choice of intervals. Let be a small number (that can depend on the intervals), and for each let be the rectangle with vertices , , , This rectangle has modulus , and hence has modulus at most . On the other hand, by Rengel’s inequality this modulus is at least , where is a quantity that goes to zero as (holding the intervals fixed). We conclude that

On the other hand, we have

By Cauchy-Schwarz, we thus have

sending , we conclude

giving the claim.

Exercise 32Let be a -quasiconformal map between two complex domains for some . Suppose that there is a closed set of Lebesgue measure zero such that is conformal on . Show that is -conformal (and hence conformal, by Proposition 28). (Hint:Arguing as in the proof of Theorem 29, it suffices to show that of maps the rectangle with endpoints to the rectangle with endpoints , then . Repeat the proof of that theorem, using the absolute continuity of lines at a crucial juncture to justify using the fundamental theorem of calculus.)

Recall Hurwitz’s theorem that the locally uniform limit of conformal maps is either conformal or constant. It turns out there is a similar result for quasiconformal maps. We will just prove a weak version of the result (see Theorem II.5.5 of Lehto-Virtanen for the full statement):

Theorem 33Let , and let be a sequence of -quasiconformal maps that converge locally uniformly to an orientation-preserving homeomorphism . Then is also -quasiconformal.

It is important for this theorem that we do not insist that quasiconformal maps are necessarily differentiable. Indeed for applications to circle packing we will be working with maps that are only piecewise smooth, or possibly even worse, even though at the end of the day we will recover a smooth conformal map in the limit.

*Proof:* Let be a Jordan quadrilateral in . We need to show that . By restricting we may assume . By composing with a conformal map we may assume that is a rectangle. We can write as the increasing limit of rectangles of the same modulus, then for any we have . By choosing going to infinity sufficiently rapidly, stays inside and converges to in the sense of Exercise 26, and the claim then follows from that exercise.

Another basic property of conformal mappings (a consequence of Morera’s theorem) is that they can be glued along a common edge as long as the combined map is also a homeomorphism; this fact underlies for instance the Schwarz reflection principle. We have a quasiconformal analogue:

Theorem 34Let , and let be an orientation-preserving homeomorphism. Let be a real analytic (and topologically closed) contour that lies in except possibly at the endpoints. If is -quasiconformal, then is -quasiconformal.

We will generally apply this theorem in the case when disconnects into two components, in which case can be viewed as the gluing of the restrictions of this map to the two components.

*Proof:* As in the proof of the previous theorem, we may take to be a rectangle , and it suffices to show that . We may normalise to have vertices where , and similarly normalise to be a rectangle of vertices , so we now need to show that . The real analytic contour meets in a finite number of curves, which can be broken up further into a finite horizontal line segments and graphs for various closed intervals and real analytic . For any , we can then use the uniform continuity of the to subdivide into a finite number of rectangles where on each such rectangle, meets the interior of in a bounded number of graphs whose horizontal variation is . This subdivides into a bounded number of Jordan quadrilaterals . If we let denote the distance between the -sides of , then by uniform continuity of and the triangle inequality we have

as . By Rengel’s inequality, we have

since , we conclude using superadditivity that

and hence by Cauchy-Schwarz

and thus

Summing in , we obtain

giving the desired bound after sending .

It will be convenient to study analogues of the modulus when quadrilaterals are replaced by generalisations of annuli. We define a *ring domain* to be a region bounded between two Jordan curves , where (the inner boundary) is contained inside the interior of (the outer boundary). For instance, the annulus is a ring domain for any and . In the spirit of Proposition 23, define the *modulus* of a ring domain to be the supremum of all the quantities with the following property: for any Borel measurable one can find a rectifable curve in winding once around the inner boundary , such that

We record some basic properties of this modulus:

**these exercises should be in reverse order – this will be done after the course concludes**

- (i) Show that the modulus of an annulus is given by .
- (ii) Show that if is -quasiconformal and is an ring domain in , then . In particular, the modulus is a conformal invariant. (There is also a converse to this statement that allows for a definition of -quasiconformality in terms of the modulus of ring domains; see e.g. Theorem 7.2 of Lehto-Virtanen.)
- (iii) Show that if one ring domain is contained inside another (with the inner boundary of in the interior of the inner boundary of ), then .

Exercise 36Show that every ring domain is conformal to an annulus. (There are several ways to proceed here. One is to start by using Perron’s method to construct a harmonic function that is on one of the boundaries of the annulus and on the other. Another is to apply a logarithm map to transform the annulus to a simply connected domain with a “parabolic” group of discrete translation symmetries, use the Riemann mapping theorem to map this to a disc, and use the uniqueness aspect of the Riemann mapping theorem to figure out what happens to the symmetry.) Use this to give an alternate definition of the modulus of a ring domain that is analogous to the original definition of the modulus of a quadrilateral.

As a basic application of this concept we have the fact that the complex plane cannot be quasiconformal to any proper subset:

Proposition 37Let be a -quasiconformal map for some ; then .

*Proof:* As is homeomorphic to , it is simply connected. Thus, if we assume for contradiction that , then by the Riemann mapping theorem is conformal to , so we may assume without loss of generality that .

By Exercise 35(i), the moduli of the annuli goes to infinity as , and hence (by Exercise 35(ii) (applied to ) the moduli of the ring domains must also go to infinity. However, as the inner boundary of this domain is fixed and the outer one is bounded, all these ring domains can be contained inside a common annulus, contradicting Exercise 35(iii).

For some further applications of the modulus of ring domains, we need the following result of Grötzsch:

Theorem 38 (Grötzsch modulus theorem)Let , and let be the ring domain formed from by deleting the line segment from to . [Technically, is not quite a ring domain as defined above, but one can check that the definition of modulus, and the fact that is conformal to an annulus, remains valid.] Let be another ring domain contained in whose inner boundary encloses both and . Then .

*Proof:* Let , then by Exercise 36 we can find a conformal map from to the annulus . As is symmetric around the real axis, and the only conformal automorphisms of the annulus that preserve the inner and outer boundaries are rotations (as can be seen for instance by using the Schwarz reflection principle repeatedly to extend such automorphisms to an entire function of linear growth), we may assume that obeys the symmetry . Let be the function , then is symmetric around the real axis. One can view as a measurable function on ; from the change of variables formula we have

so in particular is square-integrable. Our task is to show that ; by the definition of modulus, it suffices to show that

for any rectifiable curve that goes once around , and thus once around and in . By a limiting argument we may assume that is polygonal. By repeatedly reflecting around the real axis whenever crosses the line segment between and , we may assume that does not actually cross this segment, and then by perturbation we may assume it is contained in . But then by change of variables we have

by the Cauchy integral formula, and the claim follows.

Exercise 39Let be a sequence of -quasiconformal maps for some , such that all the are uniformly bounded. Show that the are a normal family, that is to say every sequence in contains a subsequence that converges locally uniformly. (Hint:use an argument similar to that in the proof of Proposition 37, combined with Theorem 38, to establish some equicontinuity of the .)

There are many further basic properties of the conformal modulus for both quadrilaterals and annuli; we refer the interested reader to Lehto-Virtanen for details.

** — 3. Rigidity of the hexagonal circle packing — **

We return now to circle packings. In order to understand finite circle packings, it is convenient (in order to use some limiting arguments) to consider some infinite circle packings. A basic example of an infinite circle packing is the *regular hexagonal circle packing*

where is the hexagonal lattice

and is the unit circle centred at . This is clearly an (infinite) circle packing, with two circles in this packing (externally) tangent if and only if they differ by twice a sixth root of unity. Between any three mutually tangent circles in this packing is an open region that we will call an *interstice*. It is inscribed in a *dual circle* that meets the three original circles orthogonally and can be computed to have radius ; the interstice can then be viewed as a hyperbolic triangle in this dual circle in which all three sides have infinite length. Let denote the union of all the interstices.

For every circle in this circle packing, we can form the inversion map across this circle on the Riemann sphere, defined by setting

for and , with the convention that maps to and vice versa. These are conjugates of Möbius transformations; they preserve the circle and swap the interior with the exterior. Let be the group of transformations of generated by these inversions ; this is essentially a Schottky group (except for the fact that we are are allowing for conjugate Möbius transformations in addition to ordinary Möbius transformations). Let be the union of the images of the interstitial regions under all of these transformations. We have the following basic fact:

Proposition 40has Lebesgue measure zero.

*Proof:* (Sketch) I thank Mario Bonk for this argument. Let denote all the circles formed by applying an element of to the circles in . If lies in , then it lies inside one of the circles in , and then after inverting through that circle it lies in another circle in , and so forth; undoing the inversions, we conclude that lies in infinite number of nested circles. Let be one of these circles. contains a union of six interstices bounded by and a cycle of six circles internally tangent to and consecutively externally tangent to each other. Applying the same argument used to establish the ring lemma (Lemma 41), we see that the six internal circles have radii comparable to that of , and hence has density in the disk enclosed by , which also contains . The ring lemma also shows that the radius of each circle in the nested sequence is at most times the one enclosing it for some absolute constant , so in particular the disks shrink to zero in size. Thus cannot be a point of density of , and hence by the Lebesgue density theorem this set has measure zero.

**these two lemmas should be moved to the front of the section.**

Next we need two simple geometric lemmas, due to Rodin and Sullivan.

Lemma 41 (Ring lemma)Let be a circle that is externally tangent to a chain of circles with disjoint interiors, with each externally tangent to (with the convention ). Then there is a constant depending only on , such that the radii of each of the is at least times the radius of .

*Proof:* Without loss of generality we may assume that has radius and that the radius of is maximal among the radii of the . As the polygon connecting the centers of the has to contain , we see that . This forces , for if was too small then would be so deep in the cuspidal region between and that it would not be possible for to escape this cusp and go around . A similar argument then gives , and so forth, giving the claim.

Lemma 42 (Length-area lemma)Let , and let consist of those circles in that can be connected to the circle by a path of length at most (going through consecutively tangent circles in ). Let be circle packing with the same nerve as that is contained in a disk of radius . Then the circle in associated to the circle in has radius .

The point of this bound is that when is bounded and , the radius of is forced to go to zero.

*Proof:* We can surround by disjoint chains of consecutively tangent circles , in . Each circle is associated to a corresponding circle in of some radius . The total area of these circles is at most the area of the disk of radius . Since , this implies from the pigeonhole principle that there exists for which

and hence by Cauchy-Schwarz

Connecting the centers of these circles, we obtain a polygonal path of length that goes around , and the claim follows.

We also need another simple geometric observation:

Exercise 43Let be mutually externally tangent circles, and let be another triple of mutually external circles, with the same orientation (e.g. and both go counterclockwise around their interstitial region). Show that there exists a Möbius transformation that maps each to and which maps the interstice of conformally onto the interstice of .

Now we can give a rigidity result for the hexagonal circle packing, somewhat in the spirit of Theorem 4 (though it does not immediately follow from that theorem), and also due to Rodin and Sullivan:

Proposition 44 (Rigidity of infinite hexagonal packing)Let be an infinite circle packing in with the same nerve as the hexagonal circle packing . Then is in fact equal to the hexagonl circle packing up to affine transformations and reflections.

*Proof:* By applying a reflection we may assume that and have the same orientation. For each interstitial region of there is an associated interstitial region of , and by Exercise 43 there is a Möbius transformation . These can be glued together to form a map that is initially defined (and conformal) on the interstitial regions ; we would like to extend it to the entire complex plane by defining it also inside the circles .

Now consider a circle in . It is bounded by six interstitial regions , which map to six interstitial regions that lie between the circle corresponding to and six tangent circles . By the ring lemma, all of the circles have radii comparable to the radius of . As a consequence, the map , which is defined (and piecewise Möbius) on the boundary of as a map to the boundary of , has derivative comparable in magnitude to also. By extending this map radially (in the sense of defining for and , where is the centre of , we see from Theorem 29 that we can extend to be -quasiconformal in the interior of except possibly at for some , and to a homeomorphism from to the region consisting of the union of the disks in and their interstitial regions. By many applications of Theorem 34, is now -quasiconformal on all of , and conformal in the interstitial regions . By Proposition 37, surjects onto , thus the circle packing and all of its interstitial regions cover the entire complex plane.

Next, we use a version of the Schwarz reflection principle to replace by another -quasiconformal map that is conformal on a larger region than . Namely, pick a circle in , and let be the corresponding circle in . Let and be the inversions across and respectively. Note that maps the circle to , with the interior mapping to the interior and exterior mapping to the exterior. We can then define a modified map by setting equal to on or outside , and equal to inside (with the convention that maps to ). This is still an orientation-preserving function ; by Theorem 34 it is still -quasiconformal. It remains conformal on the interstitial region , but is now also conformal on the additional interstitial region . Repeating this construction one can find a sequence of -quasiconformal maps that map each circle to their counterparts , and which are conformal on a sequence of sets that increase up to . By Exercise 39, the restriction of to any compact set forms a normal family (the fact that the circles map to the circles will give the required uniform boundedness for topological reasons), and hence (by the usual diagonalisation argument) the themselves are a normal family; similarly for . Thus, by passing to a subsequence, we may assume that the converge locally uniformly to a limit , and that also converge locally uniformly to a limit which must then invert . Thus is a homeomorphism, and thus -quasiconformal by Theorem 33. It is conformal on , and hence by Proposition 32 it is conformal. But the only conformal maps of the complex plane are the affine maps (see Proposition 15 of this previous blog post), and hence is an affine copy of as required.

By a standard limiting argument, the perfect rigidity of the infinite circle packing can be used to give approximate rigidity of finite circle packings:

Corollary 45 (Approximate rigidity of finite hexagonal packings)Let , and suppose that is sufficiently large depending on . Let and be as in Lemma 42. Let be the radius of the circle in associated to , and let be the radius of an adjacent circle . Then .

*Proof:* We may normalise and . Suppose for contradiction that the claim failed, then one can find a sequence tending to infinity, and circle packings with nerve with , such that the radius of the adjacent circle stays bounded away from . By many applications of the ring lemma, for each circle of , the corresponding circle in has radius bounded above and below by zero. Passing to a subsequence using Bolzano-Weierstrass and using the Arzela-Ascoli diagonalisation argument, we may assume that the radii of these circles converge to a positive finite limit . Applying a rotation we may also assume that the circles converge to a limit circle (using the obvious topology on the space of circles); we can also assume that the orientation of the does not depend on . A simple induction then shows that converges to a limit circle , giving a circle packing with the same nerve as . But then by Lemma 44, is an affine copy of , which among other things implies that . Thus converges to , giving the required contradiction.

A more quantitative version of this corollary was worked out by He. There is also a purely topological proof of the rigidity of the infinite hexagonal circle packing due to Schramm.

** — 4. Approximating a conformal map by circle packing — **

Let be a simply connected bounded region in with two distinct distinguished points . By the Riemann mapping theorem, there is a unique conformal map that maps to and to a positive real. However, many proofs of this theorem are rather nonconstructive, and do not come with an effective algorithm to locate, or at least approximate, this map .

It was conjectured by Thurston, and later proven by Rodin and Sullivan, that one could achieve this by applying the circle packing theorem (Theorem 3) to a circle packing in by small circles. To formalise this, we need some more notation. Let be a small number, and let be the infinite hexagonal packing scaled by . For every circle in , define the *flower* to be the union of this circle, its interior, the six interstices bounding it, and the six circles tangent to the circle (together with their interiors). Let be a circle in such that lies in its flower. For small enough, this flower is contained in . Let denote all circles in that can be reached from by a finite chain of consecutively tangent circles in , whose flowers all lie in . Elements of will be called *inner circles*, and circles in that are not an inner circle but are tangent to it will be called *border circles*. Because is simply connected, the union of all the flowers of inner circles is also simply connected. As a consequence, one can traverse the border circles by a cycle of consecutively tangent circles, with the inner circles enclosed by this cycle. Let be the circle packing consisting of the inner circles and border circles. Applying Theorem 3 followed by a Möbius transformation, one can then find a circle packing in with the same nerve and orientation as , such that all the circles in associated to border circles of are internally tangent to . Applying a Möbius transformation, we may assume that the flower containing in is mapped to the flower containing , and the flower containing is mapped to a flower containing a positive real. (From the exercise below will lie in such a flower for small enough.)

Let be the union of all the solid closed equilateral triangles formed by the centres of mutually tangent circles in , and let be the corresponding union of the solid closed triangles from . Let be the piecewise affine map from to that maps each triangle in to the associated triangle in .

Exercise 46Show that converges to as in the Hausdorff sense. In particular, lies in for sufficiently small .

Exercise 47By modifying the proof of the length-area lemma, show that all the circles in have radius that goes uniformly to zero as . (Hint: for circles deep in the interior, the length-area lemma works as is; for circles near the boundary, one has to encircle by a sequence of chains that need not be closed, but may instead terminate on the boundary of . The argument may be viewed as a discrete version of the one used to prove Theorem 20.) Using this and the previous exercise, show that converges to in the Hausdorff sense.

From Corollary 45 we see that as , the circles in corresponding to adjacent circles of in a fixed compact subset of have radii differing by a ratio of . We conclude that in any compact subset of , adjacent circles in in also have radii differing by a ratio of , which implies by trigonometry that the triangles of in are approximately equilateral in the sense that their angles are . By Theorem 29 is -quasiconformal on each such triangle, and hence by Theorem 34 it is -quasiconformal on . By Exercise 39 every sequence of has a subsequence which converges locally uniformly on , and whose inverses converge locally uniformly on ; the limit is then a homeomorphism from to that maps to and to a positive real. By Theorem 33 the limit is locally -conformal and hence conformal, hence by uniqueness of the Riemann mapping it must equal . As is the unique limit point of all subsequences of the , this implies (by the Urysohn subsequence principle) that converges locally uniformly to , thus making precise the sense in which the circle packings converge to the Riemann map.

]]>where is the identity operator and is the commutator. Among other things, this equation is fundamental in quantum mechanics, leading for instance to the Heisenberg uncertainty principle.

The operators are unbounded on spaces such as . One can ask whether the commutator equation (1) can be solved using bounded operators on a Hilbert space rather than unbounded ones. In the finite dimensional case when are just matrices for some , the answer is clearly negative, since the left-hand side of (1) has trace zero and the right-hand side does not. What about in infinite dimensions, when the trace is not available? As it turns out, the answer is still negative, as was first worked out by Wintner and Wielandt. A short proof can be given as follows. Suppose for contradiction that we can find bounded operators obeying (1). From (1) and an easy induction argument, we obtain the identity

for all natural numbers . From the triangle inequality, this implies that

Iterating this, we conclude that

for any . Bounding and then sending , we conclude that , which clearly contradicts (1). (Note the argument can be generalised without difficulty to the case when lie in a Banach algebra, rather than be bounded operators on a Hilbert space.)

It was observed by Popa that there is a quantitative version of this result:

Theorem 1Let such that

*Proof:* By multiplying by a suitable constant and dividing by the same constant, we may normalise . Write with . Then the same induction that established (2) now shows that

and hence by the triangle inequality

We divide by and sum to conclude that

giving the claim.

Again, the argument generalises easily to any Banach algebra. Popa then posed the question of whether the quantity can be replaced by any substantially larger function of , such as a polynomial in . As far as I know, the above simple bound has not been substantially improved.

In the opposite direction, one can ask for constructions of operators that are not too large in operator norm, such that is close to the identity. Again, one cannot do this in finite dimensions: has trace zero, so at least one of its eigenvalues must outside the disk , and therefore for any finite-dimensional matrices .

However, it was shown in 1965 by Brown and Pearcy that in infinite dimensions, one can construct operators with arbitrarily close to in operator norm (in fact one can prescribe any operator for as long as it is not equal to a non-zero multiple of the identity plus a compact operator). In the above paper of Popa, a quantitative version of the argument (based in part on some earlier work of Apostol and Zsido) was given as follows. The first step is to observe the following Hilbert space version of Hilbert’s hotel: in an infinite dimensional Hilbert space , one can locate isometries obeying the equation

where denotes the adjoint of . For instance, if has a countable orthonormal basis , one could set

and

where denotes the linear functional on . Observe that (4) is again impossible to satisfy in finite dimension , as the left-hand side must have trace while the right-hand side has trace .

Multiplying (4) on the left by and right by , or on the left by and right by , then gives

From (4), (5) we see in particular that, while we cannot express as a commutator of bounded operators, we can at least express it as the sum of two commutators:

We can rewrite this somewhat strangely as

and hence there exists a bounded operator such that

Moving now to the Banach algebra of matrices with entries in (which can be equivalently viewed as ), a short computation then gives the identity

for some bounded operator whose exact form will not be relevant for the argument. Now, by Neumann series (and the fact that have unit operator norm), we can find another bounded operator such that

and then another brief computation shows that

Thus we can express the operator as the commutator of two operators of norm . Conjugating by for any , we may then express as the commutator of two operators of norm . This shows that the right-hand side of (3) cannot be replaced with anything that blows up faster than as . Can one improve this bound further?

]]>In a similar fashion, the fundamental object of study in complex differential geometry are the complex manifolds, in which the model space is rather than , and the transition maps are required to be holomorphic (and not merely smooth or continuous). In the real case, the one-dimensional manifolds (curves) are quite simple to understand, particularly if one requires the manifold to be connected; for instance, all compact connected one-dimensional real manifolds are homeomorphic to the unit circle (why?). However, in the complex case, the connected one-dimensional manifolds – the ones that look locally like subsets of – are much richer, and are known as Riemann surfaces. For sake of completeness we give the (somewhat lengthy) formal definition:

Definition 1 (Riemann surface)If is a Hausdorff connected topological space, a (one-dimensional complex) atlas is a collection of homeomorphisms from open subsets of that cover to open subsets of the complex numbers , such that the transition maps defined by are all holomorphic. Here is an arbitrary index set. Two atlases , on are said to beequivalentif their union is also an atlas, thus the transition maps and their inverses are all holomorphic. A Riemann surface is a Hausdorff connected topological space equipped with an equivalence class of one-dimensional complex atlases.A map from one Riemann surface to another is

holomorphicif the maps are holomorphic for any charts , of an atlas of and respectively; it is not hard to see that this definition does not depend on the choice of atlas. It is also clear that the composition of two holomorphic maps is holomorphic (and in fact the class of Riemann surfaces with their holomorphic maps forms a category).

Here are some basic examples of Riemann surfaces.

Example 2 (Quotients of )The complex numbers clearly form a Riemann surface (using the identity map as the single chart for an atlas). Of course, maps that are holomorphic in the usual sense will also be holomorphic in the sense of the above definition, and vice versa, so the notion of holomorphicity for Riemann surfaces is compatible with that of holomorphicity for complex maps. More generally, given any discrete additive subgroup of , the quotient is a Riemann surface. There are an infinite number of possible atlases to use here; one such is to pick a sufficiently small neighbourhood of the origin in and take the atlas where and for all . In particular, given any non-real complex number , the complex torus formed by quotienting by the lattice is a Riemann surface.

Example 3Any open connected subset of is a Riemann surface. By the Riemann mapping theorem, all simply connected open , other than itself, are isomorphic (as Riemann surfaces) to the unit disk (or, equivalently, to the upper half-plane).

Example 4 (Riemann sphere)The Riemann sphere , as a topological manifold, is the one-point compactification of . Topologically, this is a sphere and is in particular connected. One can cover the Riemann sphere by the two open sets and , and give these two open sets the charts and defined by for , for , and . This is a complex atlas since the is holomorphic on .An alternate way of viewing the Riemann sphere is as the projective line . Topologically, this is the punctured complex plane quotiented out by non-zero complex dilations, thus elements of this space are equivalence classes with the usual quotient topology. One can cover this space by two open sets and and give these two open sets the charts and defined by for , . This is a complex atlas, basically because for and is holomorphic on .

Exercise 5Verify that the Riemann sphere is isomorphic (as a Riemann surface) to the projective line.

Example 6 (Smooth algebraic plane curves)Let be a complex polynomial in three variables which is homogeneous of some degree , thus

Define the complex projective plane to be the punctured space quotiented out by non-zero complex dilations, with the usual quotient topology. (There is another important topology to place here of fundamental importance in algebraic geometry, namely the Zariski topology, but we will ignore this topology here.) This is a compact space, whose elements are equivalence classes . Inside this plane we can define the (projective, degree ) algebraic curve

this is well defined thanks to (1). It is easy to verify that is a closed subset of and hence compact; it is non-empty thanks to the fundamental theorem of algebra.

Suppose that is

irreducible, which means that it is not the product of polynomials of smaller degree. As we shall show in the appendix, this makes the algebraic curve connected. (Actually, algebraic curves remain connected even in the reducible case, thanks to Bezout’s theorem, but we will not prove that theorem here.) We will in fact make the strongernonsingularityhypothesis: there is no triple such that the four numbers simultaneously vanish for . (This looks like four constraints, but is in fact essentially just three, due to the Euler identitythat arises from differentiating (1) in . The fact that nonsingularity implies irreducibility is another consequence of Bezout’s theorem, which is not proven here.) For instance, the polynomial is irreducible but singular (there is a “cusp” singularity at ). With this hypothesis, we call the curve

smooth.Now suppose is a point in ; without loss of generality we may take non-zero, and then we can normalise . Now one can think of as an inhomogeneous polynomial in just two variables , and by nondegeneracy we see that the gradient is non-zero whenever . By the (complexified) implicit function theorem, this ensures that the

affine algebraic curveis a Riemann surface in a neighbourhood of ; we leave this as an exercise. This can be used to give a coordinate chart for in a neighbourhood of when . Similarly when is non-zero. This can be shown to give an atlas on , which (assuming the connectedness claim that we will prove later) gives the structure of a Riemann surface.

Exercise 7State and prove a complex version of the implicit function theorem that justifies the above claim that the charts in the above example form an atlas, and an algebraic curve associated to a non-singular polynomial is a Riemann surface.

Exercise 8

- (i) Show that all (irreducible plane projective) algebraic curves of degree are isomorphic to the Riemann sphere. (Hint: reduce to an explicit linear polynomial such as .)
- (ii) Show that all (irreducible plane projective) algebraic curves of degree are isomorphic to the Riemann sphere. (Hint: to reduce computation, first use some linear algebra to reduce the homogeneous quadratic polynomial to a standard form, such as or .)

Exercise 9If are complex numbers, show that the projective cubic curveis nonsingular if and only if the discriminant is non-zero. (When this occurs, the curve is called an elliptic curve (in Weierstrass form), which is a fundamentally important example of a Riemann surface in many areas of mathematics, and number theory in particular. One can also define the discriminant for polynomials of higher degree, but we will not do so here.)

A recurring theme in mathematics is that an object is often best studied by understanding spaces of “good” functions on . In complex analysis, there are two basic types of good functions:

Definition 10Let be a Riemann surface. Aholomorphic functionon is a holomorphic map from to ; the space of all such functions will be denoted . Ameromorphic functionon is a holomorphic map from to the Riemann sphere , that is not identically equal to ; the space of all such functions will be denoted .

One can also define holomorphicity and meromorphicity in terms of charts: a function is holomorphic if and only if, for any chart , the map is holomorphic in the usual complex analysis sense; similarly, a function is meromorphic if and only if the preimage is discrete (otherwise, by analytic continuation and the connectedness of , will be identically equal to ) and for any chart , the map becomes a meromorphic function in the usual complex analysis sense, after removing the discrete set of complex numbers where this map is infinite. One consequence of this alternate definition is that the space of holomorphic functions is a commutative complex algebra (a complex vector space closed under pointwise multiplication), while the space of meromorphic functions is a complex field (a commutative complex algebra where every non-zero element has an inverse). Another consequence is that one can define the notion of a zero of given order , or a pole of order , for a holomorphic or meromorphic function, by composing with a chart map and using the usual complex analysis notions there, noting (from the holomorphicity of transition maps and their inverses) that this does not depend on the choice of chart. (However, one cannot similarly define the residue of a meromorphic function on this way, as the residue turns out to be chart-dependent thanks to the chain rule. Residues should instead be applied to meromorphic -forms, a concept we will introduce later.) A third consequence is analytic continuation: if two holomorphic or meromorphic functions on agree on a non-empty open set, then they agree everywhere.

On the complex numbers , there are of course many holomorphic functions and meromorphic functions; for instance any power series with an infinite radius of convergence will give a holomorphic function, and the quotient of any two such functions (with non-zero denominator) will give a meromorphic function. Furthermore, we have extremely wide latitude in how to specify the zeroes of the holomorphic function, or the zeroes and poles of the meromorphic function, thanks to tools such as the Weierstrass factorisation theorem or the Mittag-Leffler theorem (covered in previous quarters).

It turns out, however, that the situation changes dramatically when the Riemann surface is *compact*, with the holomorphic and meromorphic functions becoming much more rigid. First of all, compactness eliminates all holomorphic functions except for the constants:

Lemma 11Let be a holomorphic function on a compact Riemann surface . Then is constant.

This result should be seen as a close sibling of Liouville’s theorem that all bounded entire functions are constant. (Indeed, in the case of a complex torus, this lemma is a corollary of Liouville’s theorem.)

*Proof:* As is continuous and is compact, must attain a maximum at some point . Working in a chart around and applying the maximum principle, we conclude that is constant in a neighbourhood of , and hence is constant everywhere by analytic continuation.

This dramatically cuts down the number of possible meromorphic functions – indeed, for an abstract Riemann surface, it is not immediately obvious that there are any non-constant meromorphic functions at all! As the poles are isolated and the surface is compact, a meromorphic function can only have finitely many poles, and if one prescribes the location of the poles and the maximum order at each pole, then we shall see that the space of meromorphic functions is now finite dimensional. The precise dimensions of these spaces are in fact rather interesting, and obey a basic duality law known as the Riemann-Roch theorem. We will give a mostly self-contained proof of the Riemann-Roch theorem in these notes, omitting only some facts about genus and Euler characteristic, as well as construction of certain meromorphic -forms (also known as Abelian differentials).

** — 1. Divisors — **

To discuss the zeroes and poles of meromorphic functions, it is convenient to introduce an abstraction of the concept of “a collection of zeroes and poles”, known as a divisor.

Definition 12 (Divisor)Let be a compact Riemann surface. Adivisoron is a formal integer linear combination , where ranges over a finite collection of points in , and are integers, with the obvious additive group structure; equivalently, the space of divisors is the free abelian group with generators with (where we make the usual convention ). The number is thedegreeof the divisor; we call each theorderof the divisor at , with the convention that the order is zero for points not appearing in the sum. A divisor isnon-negative(oreffective) if all the are non-negative, and we partially order the divisors by writing if is non-negative. This makes a lattice, so we can define the maximum or minimum of two divisors. Given a non-zero meromorphic function , theprincipal divisorassociated to is the divisor , where ranges over the zeroes and poles of , and is the order of zero (or negative the order of pole) at . (Note that as zeroes and poles are isolated, and is compact, the number of zeroes and poles is automatically finite.)

Informally, one should think of as the abstraction of a zero of order at , or a pole of order if is negative.

Example 13Consider a rational functionfor some non-zero complex number and some complex numbers . This is a meromorphic function on , and is also meromorphic, so extends to a meromorphic function on the Riemann sphere . It has zeroes at and poles at , and also has a zero of order (or a pole of order ) at , as can be seen by inspection of near the origin (or the growth of near infinity), and thus

In particular, has degree zero.

Exercise 14Show that all meromorphic functions on the Riemann sphere come from rational functions as in the above example. In particular, every principal divisor on the Riemann sphere has degree zero. Give an alternate proof of this latter fact using the residue theorem. (We will generalise this fact to other Riemann surfaces shortly; see Proposition 24.)

It is easy to see (by working in a coordinate chart around ) that if are non-zero meromorphic functions, that one has the valuation axioms

for any (adopting the convention the zero function has order everywhere); thus we have

again adopting the convention that is larger than every divisor. In particular, the space of principal divisors of is a subgroup of . We call two divisors *linearly equivalent* if they differ by a principal divisor; this is clearly an equivalence relation.

The properties (2) have the following consequence. Given a divisor , let be the space of all meromorphic functions such that (including, by convention, the zero function ); thus, if , then consists of functions that have at worst a pole of order at (or a zero of order or greater, if is negative). For instance, is the space of meromorphic functions that have at most a double pole at , a single pole at , and at least a simple zero at , if are distinct points in . From (2) (and the fact that non-zero constant functions have principal divisor zero) we see that each is a vector space. We clearly have the nesting properties if , and also if then .

Remark 15In the language of vector bundles, one can identify a divisor with a certain holomorphic line bundle on , and can be identified with the space of sections of this bundle. This is arguably the more natural way to think about divisors; however, we will not adopt this language here.

If and , then is holomorphic on and hence (by Lemma 11) constant. We can thus easily compute for zero or negative divisors:

Corollary 16Let be a compact Riemann surface. Then consists only of the constant functions, and consists only of if . In particular, has dimension when and when .

Exercise 17If and are principal divisors with , show that is a constant multiple of with .

Exercise 18Let be a divisor. Show that if and only if is linearly equivalent to an effective divisor.

The situation for (i.e., has positive order at at least one point) is more interesting. We first have a simple observation from linear algebra:

Lemma 19Let be a compact Riemann surface, be a divisor, and be a point. Then has codimension at most in .

*Proof:* Let be a chart that maps to the origin, and suppose that already had order at (so that had order ). Then functions , when composed with the inverse of the chart function have Laurent expansion

for some complex coefficients (which will depend on the choice of chart). The map is clearly a linear map from to , whose kernel is , and the claim follows.

As a corollary of this lemma and Corollary 16, we see that the spaces are all finite dimensional, with the dimension increasing by zero or one each time one adds an additional pole to .

Here is another simple linear algebra relation between the dimensions of the spaces :

Lemma 20Let be a compact Riemann surface, and let be divisors. Then

*Proof:* From linear algebra we have

Since and , the claim follows.

If is a divisor and is a principal divisor, then (2) gives an isomorphism between and , by mapping to . In particular, the dimensions and of the linearly equivalent divisors are the same. If we define a *divisor class* to be a coset of the principal divisors in (that is to say, an equivalence class for linear equivalence), then we conclude that the dimension depends only on the divisor class of . The space of divisor classes is an abelian group, which is known as the *divisor class group*. (For nonsingular algebraic curves, this group also coincides with the Picard group, though the situation is more subtle if one allows singularities.)

It is now easy to understand the spaces for the Riemann sphere:

Exercise 21Show that two divisors on the Riemann sphere are equivalent if and only if they have the same degree, so that the degree map gives an isomorphism between the divisor class group of the Riemann sphere and the integers. If is a divisor on the Riemann sphere, show that is equal to . (Hint: first show that for any integer , that is the space of polynomials of degree at most .)

From the above exercise we observe in particular that

whenever has degree ; as we will see later, this is a special case of the Riemann-Roch theorem.

** — 2. Meromorphic -forms — **

To proceed further, we will introduce the concept of a *meromorphic -form* on a compact Riemann surface . To motivate this concept, observe that one can think of a meromorphic function on as a collection of meromorphic functions on open subsets of the complex plane, where ranges over a suitable atlas of . These meromorphic functions are compatible with each other in the following sense: if and are charts, then we have

for all (this condition is vacuous if do not overlap). As already noted, one can define such concepts as the order of at a pole by declaring it to be the order of at for any chart that contains in its domain, and the compatibility condition (4) ensures that this definition is well defined.

On the other hand, several other basic notions in complex analysis do not seem to be well defined for such meromorphic functions. Consider for instance the question of how to define the residue of at a pole . The natural thing to do is to again pick a chart around and use the residue of ; however one can check that this is *not* independent of the choice of chart in general, as from (4) one will find that the residues of and are related to each other, but not equal. Similarly, one encounters a difficulty integrating on a contour in , even if the contour is short enough to fit into the domain of a single chart and also avoids all the poles of ; the natural thing to do is to compute , but again this will depend on the choice of chart (substituting (4) will reveal that is not equal to in general due to an additional Jacobian factor). Finally, one encounters a difficulty trying to differentiate a meromorphic function ; on each chart one would like to just differentiate , but the resulting derivatives do not obey the compatibility condition (4), but instead (by the chain rule) obey the slightly different condition

The solution to all of these issues is to introduce a new type of object on , the *meromorphic -forms*.

Definition 22Ameromorphic -formon is a collection of expressions for each coordinate chart of , with meromorphic on , which obey the compatibility condition

for any pair , of charts and any . If all the are holomorphic, we say that is holomorphic also. The space of meromorphic -forms will be denoted .

As with meromorphic functions, we can define the order of at a point to be the order of at for some chart that contains in its domain; from (5) we see that this is well defined. Similarly we may define the divisor of . The divisor of a non-zero meromorphic -form is called a canonical divisor. (We will show later that at least one non-zero meromorphic -form is available, so that canonical divisors exist.)

Let be a meromorphic -form. Given a contour that lies in the domain of a single chart and avoids the poles of , we can define the integral to be equal to . One checks from (5) and the change of variables formula that this definition is independent of the choice of chart. One then defines for longer contours by partitioning into short contours; again, one can check that this definition is independent of the choice of partition.

The residue of at can be defined as the residue of at for a chart that contains in its domain, or equivalently (by the residue theorem) where is a sufficiently small contour winding around once anticlockwise (note that we have a consistent orientation on since invertible holomorphic maps are orientation preserving).

Meromorphic -forms are also known as *Abelian differentials*, while holomorphic -forms are Abelian differentials of the first kind. (Abelian differentials of the second kind are meromorphic -forms in which all residues vanish, while Abelian differentials of the third kind are meromorphic -forms in which all poles are simple.) To specify a meromorphic form , it suffices to prescribe for all in a single atlas of ; as long as (5) is obeyed within this atlas, it is easy to see that can then be defined uniquely using (5) for all other coordinate charts.

There are two basic ways to create meromorphic -forms. One is to start with a meromorphic function and form its differential , which when evaluated any chart of is given by the formula

the compatibility condition (5) is then clear from the chain rule. Another way is to start with an existing meromorphic -form and multiply it by a meromorphic function to give a new meromorphic -form , which when evaluated at a given chart of is given by

again, it is clear that the compatibility condition (5) holds. Conversely, given two meromorphic -forms , with not identically zero, one can form the ratio to be the unique meromorphic function such that ; it is easy to see that exists and is unique. These properties are compatible with taking divisors, thus and .

Of course, one can also add two meromorphic -forms to obtain another meromorphic -form. Thus is in fact a one-dimensional vector space over the field (here we assume that non-zero meromorphic -forms exist, a claim which we will return to later). In particular, the canonical divisor is unique up to linear equivalence.

Later on we will discuss a further way to create a meromorphic -form, by taking the gradient of a harmonic function with specific types of singularities.

Example 23The coordinate function can be viewed as a meromorphic function on the Riemann sphere (it has a simple zero at and a simple pole at ). Its derivative then has a double pole at infinity (note that in the reciprocal coordinate , transforms to ), so . Any other meromorphic -form is of the form , where is a meromorphic function (that is to say, a rational function). In particular, since meromorphic functions have divisor of degree , all meromorphic -forms on the Riemann sphere have a divisor of degree ; indeed, the canonical divisors here are precisely the divisors of degree .

We now give a key application of meromorphic -forms to the divisors of meromorphic functions:

Proposition 24Let be a compact Riemann surface.

- (i) For any meromorphic -form , the sum of all the residues of vanishes.
- (ii) Every principal divisor has degree zero.

*Proof:* We begin with (i). By evaluating at coordinate charts, the counterclockwise integral of around any small loop that avoids any pole is zero; thus is closed outside of these poles, and hence by Stokes’ theorem we conclude that the integral of around the sum of small counterclockwise loops around every pole is zero. On the other hand, by the residue theorem applied in each chart, this integral is equal to times the sum of the residues, and the claim follows.

To prove (ii), apply (i) to the meromorphic function (cf. the usual proof of the argument principle).

Exercise 25Let be a compact Riemann surface, and let be a divisor on .

- (i) If , show that .
- (ii) If , show that is equal to or , with the latter occuring if and only if is principal. Furthermore, any non-zero element of has divisor .
- (iii) If , establish the bound .

We have already discussed how algebraic curves give good examples of Riemann surfaces. In the converse direction, it is common for Riemann surfaces to map into algebraic curves, as hinted by the following exercise:

Exercise 26Let be a compact Riemann surface, and let be two non-constant meromorphic functions on . Show that there exists a non-zero polynomial of two variables with complex coefficients such that . (Hint:look at the monomials for for some large , and show that they lie in for a suitable divisor . Then use part (iii) of the previous exercise and linear algebra.) Show furthermore that one can take to be irreducible.

** — 3. The case of a complex torus — **

For the special case when the Riemann surface being studied is a complex torus , one can obtain more precise information on the dimensions by explicit computations. First observe we have a natural holomorphic -form on , namely the form , defined in any small coordinate chart on a small disk in (with ) by , and then defined for any other coordinate chart by compatibility. This form has no poles and zeroes, and so is a canonical divisor. Using this -form, we have a bijection between meromorphic functions and meromorphic -forms on which maps to ; in contrast to the situation with other Riemann surfaces with non-zero canonical divisor, this bijection does not affect the divisor. In particular, canonical divisors are principal and vice versa. Using this bijection, we can think of the differential of a meromorphic function as another meromorphic function, which we call the derivative , as per the familiar formula . Of course, with respect to the above coordinate charts, this derivative corresponds to the usual complex derivative.

We also have a fundamental meromorphic function on , or equivalently a -periodic function on , namely the Weierstrass -function

It is easy to see that the sum converges outside of , and that this is a meromorphic -periodic function on that has a double pole at every point in ; this descends to a meromorphic function on with divisor . By translation we can then create a meromorphic function with divisor for any .

Using this function and some manipulations, we can compute for most divisors :

Lemma 27Let be a complex torus, and let be a divisor.

- (i) If , then .
- (ii) If , then is equal to or . If for some distinct , then . Also, .
- (iii) If , then .

*Proof:* Part (i) and the first claim of part (ii) follows from Exercise 25. To prove the second claim of part (ii), it suffices by Exercise 25 to show that there is no meromorphic function with divisor , that is to say a simple pole at and a simple zero at . But this follows from Proposition 24(i) (and identifying meromorphic functions with meromorphic -forms) since the residue at is non-zero and there is no other residue to cancel it. The third claim comes from Exercise 25 and the observation that is principal if and only if is.

Call a divisor *good* if . We need to show that all divisors of positive degree are good. First we check that is good for a point . By Proposition 24(i) we see that the only meromorphic functions in are constant, hence , and so is good.

The Weierstrass -function at gives a an element of which is non-constant (it has a double pole at ), so by Lemma 19 we have , and so is good. Taking a derivative of to obtain a meromorphic function with a triple pole at , we obtain a further element of that is not in , and so , and so is good. Continuing to differentiate in this fashion we see that is good for any natural number .

Next, for any distinct pairs of points , we write for some complex number , and define the meromorphic function

where is some contour from to . (As locally has an antiderivative at every point, this definition does not depend on the choice of , though it is a little sensitive to the choice of .) One can check that this function is meromorphic with simple poles at and , which shows that . From Lemma 19 and the fact that is good, we conclude that is good.

Observe from Lemma 19 and Lemma 20 we see that if is a divisor and are distinct points such that are good, then is also good. We have just shown that all effective divisors of degree and are good; by induction one can now show that all effective divisors of positive degree are good.

Call a degree one divisor *very good* if is good for every . We have shown that is very good for all . We now claim that if is very good then so is is very good. First note that and cannot both be principal, since their difference is not principal. Thus by Exercise 25, at least one of or vanishes, and hence by Lemma 19. On the other hand, as is very good, is good, and so . By Lemma 19 we conclude that is good.

For any , we know that and are good, hence by Lemma 19 the intermediate divisor must also be good. Iterating this argument we see that is good for every , thus is very good. Iterating this we see that all degree one divisors are very good, giving (iii).

An alternate way to show that is good was shown to me by Redmond McNamara as follows. By subtracting a constant from one can find a meromorphic function with a double pole at and at least a single zero at . If it is exactly a single zero, multiplying by will create a function with precisely a simple pole at and at most a double pole at ; subtracting a multiple of if necessary will then give the required non-trivial element of . If has a double zero at instead, multiply by and subtract multiples of and to obtain the same result. Note that cannot have more than a double zero because of Proposition 24(ii).

As a corollary of the above proposition we obtain the complex torus case of the Riemann-Roch theorem:

valid for any divisor (regardless of degree); compare with (3). The one remaining point is to work out which degree zero divisors are principal. It turns out that there is an additional constraint beyond degree zero:

Exercise 28Suppose that is a principal divisor on a complex torus (we allow repetition). Show that using the group law on . (Hint:if is a meromorphic function with zeroes at and poles at , integrate around a parallelogram fundamental domain of (translating if necessary so that the boundary of the parallelogram avoids the zeroes and poles).)

In fact, this is the only condition:

Proposition 29A degree zero divisor is principal if and only if .

*Proof:* By the above exercise it suffices to establish the “if” direction. We may of course assume . By Lemma 27, the space is one-dimensional, thus there exists a non-zero meromorphic function with poles at , zeroes at , and no further poles (counting multiplicity). By Proposition 24(ii) must have one further zero, and by the above exercise this zero must be . The claim follows.

One can explicitly write down a formula for these meromorphic functions using theta functions, but we will not do so here.

The above proposition links the group law on a complex torus with the group law on divisors. This is part of a more general relation involving the Jacobian variety of a curve and the Abel-Jacobi theorem, but we will not discuss this further in this course.

Exercise 30Let be a complex torus. Show that the Weierstrass function obeys the differential equationfor some complex numbers depending on . Also show that the map for (with mapping to ) is a holomorphic invertible map from to the algebraic curve

which is non-singular and irreducible. (Thus, every complex torus is isomorphic to an elliptic curve. The converse is also true, but will not be established here.)

Proposition 31Let be a complex torus, and let be a function. Show that is holomorphic if and only if it takes the formfor all , and some complex numbers , with lying in the set . Furthermore, show that is either equal to the integers, or to a lattice of the form for some quadratic algebraic integer (thus obeys an equation for some integers ). In the latter case, the complex torus is said to have complex multiplication.

** — 4. The Riemann Roch theorem — **

We now leave the example of the complex torus and return to more general compact Riemann surfaces . We would like to generalise the identity (7) (or (3)) to this setting. As a first step we establish

Proposition 32 (Baby Riemann Roch theorem)Let be a canonical divisor in a compact Riemann surface , and let be an effective divisor. Then

*Proof:* Write where ranges over some finite set of points in , and are positive integers. As is a canonical divisor, we can find a meromorphic -form , not identically zero, with divisor . If , then is a holomorphic -form. The proof relies on using linear algebra to combine the following three observations that tie together , , and :

- If and , then . This follows from Proposition 24(i) and the fact that the only possible poles of are in .
- If and , then we have the stronger assertion that for each individual . This follows because the divisor of is at least , and so has no pole at . Furthermore this is a local statement: it holds even if is only defined on a small neighborhood of , rather than on all of . Finally, the claim is sharp: if then one can find and some defined locally near in for which
- For , , and is holomorphic at , then , since is also holomorphic at . Again, this is a local statement, and holds even if is only defined in a neighbourhood of .

Let us now see how these facts combine to give the proposition. Around each let us form a chart that maps to . Then for any , has a pole of order at most at the origin, and can thus be written as

for near , where are complex numbers and is holomorphic at the origin. We call the expression the *principal part* of (uniformised by ) at . If we let denote the collection of tuples with complex, then is a complex vector space of dimension . Inside this space we have the subspace of tuples that can actually arise as the principal parts of a meromorphic function in . Observe that if two functions have the same principal parts, then their difference is holomorphic and hence constant by Lemma 11. Thus, the space has dimension exactly .

As is a canonical divisor, we have a meromorphic -form with divisor . If , then is a holomorphic -form. If is a tuple in , we can define a pairing by the formula

This is a bilinear pairing from to . If , then all the components of are principal parts of some , and by Observation 3 one can then write as , which then vanishes by Observation 1. Thus whenever and . As row rank equals column rank, we conclude that there is a subspace of of dimension at least

such that whenever and . But then if , must vanish to order at least at each , hence , which is equivalent to and hence to ; this is Observation 2. One concludes that

and the claim follows by rearranging.

One can amplify this proposition if one is in possession of the following three non-trivial claims.

- There is at least one non-zero meromorphic -form; in particular, canonical divisors exist.
- Every canonical divisor has degree , where is the (topological) genus of .
- The space of holomorphic -forms has dimension . Equivalently, for any canonical divisor , . (In algebraic geometry language, this asserts that for compact Riemann surfaces, the topological genus is equal to the geometric genus.)

Example 33The Riemann sphere has genus . All meromorphic -forms, such as , have degree and so cannot be holomorphic, so there are no holomorphic -forms. Meanwhile, a complex torus has genus . All meromorphic -forms, such as , have degree . In particular, a holomorphic -form is times a holomorphic function, so by Lemma 11 the space of holomorphic -forms is one-dimensional.

Assuming these claims, the above proposition gives, for any canonical divisor , that

when is effective and (replacing by )

when is effective. Since the second right-hand side is the negative of the first, we conclude that

whenever and are both effective. In fact we have the more general

Theorem 34 (Riemann-Roch theorem)Let be a compact Riemann surface of genus , let be a canonical divisor, and let be any divisor. Then

This of course generalises (3) on the Riemann sphere (which has genus zero) and (7) on a complex torus (which has genus one).

It remains to establish the above three claims, and to obtain the Riemann-Roch theorem in full generality. I have not been able to locate particularly simple proofs of these steps that do not require significant machinery outside of complex analysis, so will only sketch some arguments justifying each of these.

To create meromorphic -forms one can take gradients of harmonic functions, in the spirit of the proof of the uniformization theorem that was (mostly) given in these 246A lecture notes. A function is said to be harmonic if, for every coordinate chart , is harmonic; as the property of being harmonic on open subsets of the complex plane is unaffected by conformal transformations, this definition does not depend on the choice of atlas that the charts are drawn from. If is harmonic, one can form a holomorphic -form on by defining

for each chart and .

For instance, on , the harmonic function gives rise to the holomorphic -form .

Exercise 35Show that this definition indeed defines a holomorphic -form (thus the are all holomorphic and obey the compatibiltiy condition (5). (The computations are slightly less tedious if one uses Wirtinger derivatives.)

Unfortunately, for compact Riemann surfaces , the same maximum principle argument used to prove Lemma 11 shows that there are no non-constant globally harmonic functions on , so we cannot use this construction directly to produce non-trivial holomorphic or meromorphic -forms on . However, one can produce harmonic functions with logarithmic singularities, a prototypical example of which is the function on the Riemann sphere, which is harmonic except at and . More generally, one has

Proposition 36 (Existence of dipole Green’s function)Let be a Riemann surface, and let be distinct points in . Then there exists a harmonic function on with the property that for any chart that maps to , is equal to plus a bounded function near , and for any chart that maps to , is equal to plus a bounded function near .

This proposition is essentially Proposition 65 of these 246A notes and can be proven using (a somewhat technical modification of) Perron’s method of subharmonic functions; we will not do so here. One can combine this proposition with the preceding construction to obtain a non-constant meromorphic -form:

Exercise 37Using the above proposition, show that if is a compact Riemann surface and are distinct points in , then there is a meromorphic -form on with poles only at , with a residue of at and a residue of at .Using this, conclude the

Riemann existence theorem: for any compact Riemann surface and distinct points in , there exists a meromorphic function on that takes different values at and and is in particular non-constant. (In other words, the meromorphic functions separate points.)

To prove the full Riemann-Roch theorem we will also need a variant of this exercise, not proven here:

Proposition 38If is a compact Riemann surface, is a point in , and , then there exists a meromorphic -form on with a pole of order at and no other poles.

The -forms constructed by this proposition can be viewed as generalisations of the Weierstrass functions (and their derivatives) to other Riemann surfaces, much as the -form constructed in Exercise 37 are generalisations of first integrals of these functions. (Indeed, one can think of as a sort of “derivative” of , formed as approaches and taking a suitable renormalised limit; more generally, one can furthermore of as a suitably renormalised limit of as approaches .) Note from Proposition 24 that the constructed by the above proposition automatically have vanishing residue at (in classical language, these are Abelian differentials of the third kind, while the are Abelian differentials of the second kind).

The first claim is now settled by Exercise 37, so we now turn to the second. A non-constant meromorphic function on can be viewed as a non-constant holomorphic map from to the Riemann sphere . By Proposition 24(ii), the number of times equals (counting multiplicity) equals the number of times equals (counting multiplicity). Calling this number (the degree of ), then by Lemma 11, and we see (by again applying Proposition 24(ii) to for any constant ) that attains each value on the Riemann sphere times (counting multiplicity). As long as one stays away from the zeroes of , the zeroes of are all simple, and vary continuously in by the inverse function theorem (or Rouché’s theorem), and hence after deleting a finite number of ramification points from (and also deleting their preimages from ), one can think of as a -fold covering map from (a punctured version) of by (a punctured version of) , that is to say is a -fold *branched cover* of . By applying a fractional linear transformation if necessary, we may assume that is not a ramification point of this cover (this is mainly for notational convenience).

One can use such a branched covering, together with some algebraic topology (which we will assume here as “black boxes”), to verify the second claim. Instead of working directly with the genus of , one can work instead with the Euler characteristic of , which is known from algebraic topology to equal . For instance, the Riemann sphere has genus and Euler characteristic , while a complex torus has genus and Euler characteristic .

If one has a -fold covering map from one surface to another , one can show that the Euler characteristics are related by the formula . With branched coverings this is not quite the case, but there is a substitute formula that takes into account the ramification points known as the Riemann-Hurwitz formula. Basically, if since we have a -fold cover from a punctured version of to a punctured version of the Riemann sphere, we have

On the other hand, if one reinserts a point from back into the punctured Riemann sphere, and also inserts all the preimages of that point back into , one can calculate that the Euler characteristic of the punctured sphere increases by , while the Euler characteristic of the punctured version of increases by the cardinality of the preimage. We conclude the Riemann-Hurwitz formula

As has degree , we have

for each and so we can rearrange the above (using and ) as

The meromorphic -form on the Riemann sphere has a double pole at and no zeroes. As is not a point of ramification, the pullback of this form then has a double pole at each of the preimages in . However, it also acquires a zero of order whenever and . Taking divisors, we conclude that the left-hand side of (8) is equal to the degree of , which is a canonical divisor. Since all canonical divisors have the same degree, this gives the second claim.

Exercise 39Let and be compact Riemann surfaces, with having higher genus than . Show that there does not exist any non-constant holomorphic map from to .

Now we discuss the third claim. It is relatively easy to show that the dimension of the space of holomorphic -forms is *upper bounded* by . Indeed, we may assume without loss of generality that there exists at least one non-zero holomorphic -form, giving an effective canonical divisor , which we have just shown to have degree . From Proposition 32 applied to we then have

and hence , giving the claimed upper bound.

The lower bound is harder. Basically, it asserts that the pairing in Proposition 32, when quotiented down to a pairing between and , is non-degenerate. This is a special case of an algebraic geometry fact known as Serre duality, which we will not prove here. It can also be proven from Hodge theory, using the fact that the first de Rham cohomology has dimension ; we do not pursue this approach here. Alternatively, one can try to explicitly construct linearly holomorphic -forms on the Riemann surface . We will not do this in general, but show how to do this in the case of a smooth algebraic curve of degree . The genus of such a curve turns out to be given by the genus-degree formula

One can sketch a proof of this using the Riemann-Hurwitz formula. For simplicity of notation let us assume that the polynomial is in “general position” in a number of senses that we will not specify precisely. We focus on the affine curve

generically this is with points deleted, and thus will have an Euler characteristic of . The projection map from to (which has Euler characteristic ) that maps to has ramification points whenever vanishes, which generically will be simple; away from these points one has a -fold covering. Bezout’s theorem shows that this happens times. A modification of the proof of (8) then gives

which gives the claim.

To construct linearly independent holomorphic -forms on one can argue as follows. Again it is convenient for notational reasons to work on the affine curve and assume that is in general position. The cases can be worked out by hand, so suppose . Taking the differential of the coordinate function gives a meromorphic -form , which generically has simple zeroes whenever the degree polynomial vanishes, and has a pole of order at the points in (i.e., the points where meets the line at infinity). This implies that for any polynomial of degree at most , the -form

is holomorphic (we have killed all the poles and removed the simple zeroes, while possibly creating new zeroes where vanishes). The space of such polynomials has dimension , giving the claim.

It remains to remove the condition that and be effective to obtain the Riemann-Roch theorem in full generality. We first prove a weaker version known as *Riemann’s inequality*:

Proposition 40 (Riemann’s inequality)Let be a compact Riemann surface of genus , and let be a divisor. Then .

*Proof:* Let be a canonical divisor. Choose a non-zero effective divisor such that . We will show that

since from Lemma 19 we have , and , Riemann’s inequality will follow after a brief calculation.

Dividing through by a meromorphic -form of divisor , we see that is the dimension of the space of meromorphic -forms with divisor at least . If with , is the space of meromorphic -forms that have poles of order at most at each , and no other poles.

As in the proof of Proposition 32, let be the space of tuples with complex; this has dimension , and there is a linear map that takes a meromorphic -form in to the tuple of its principal parts at points in . The image is constrained by Proposition 24(i), which forces the residues to sum to zero. On the other hand, by taking linear combinations of the meromorphic -forms from Exercise 37 and Propsosition 38, we see conversely that any tuple in whose residues sum to zero lies in the image of . Thus the image of has dimension . On the other hand, the kernel of is simply the space of holomorphic -forms, which has dimension . The claim follows.

Now we prove the Riemann-Roch theorem. We split into cases, depending on the dimensions of and .

First suppose that and are both positive dimensional. By Exercise 18, is linearly equivalent to an effective divisor, hence by Proposition 32 we have

and similarly (replacing by

and the claim then follows by using .

Now suppose that and are both trivial. Riemann’s inequality then gives

which (again using ) gives , and the claim again follows.

Now suppose that is trivial but is positive dimensional. From Exercise 18 and Proposition 32 as before we have

while from Riemann’s inequality and the triviality of we have

giving the claim. The final case when is trivial and is positive dimensional then follows by swapping with .

Exercise 41Let be a compact Riemann surface of genus one, and let be a point on . Show that for any points on , there is a unique point on such that is a principal divisor. Furthermore show that this defines an abelian group law on . What is this group law in the case that is an elliptic curve?

Exercise 42Let be a compact Riemann surface, and there exists a meromorphic function on with one simple pole and no other poles. Show that is an isomorphism between and the Riemann sphere. Conclude in particular that the Riemann sphere is the only genus zero compact Riemann surface (up to isomorphism, of course).

Exercise 43Let be a compact Riemann surface of genus , and let be a divisor of degree . Show that when is a canonical divisor, and otherwise.

Exercise 44 (Gap theorems)Let be a compact Riemann surface of genus .

- (i) (Weierstrass gap theorem) If is a point in , show that there are precisely positive integers with the property that there does not exist a meromorphic function on with a pole of order at , and no other poles. Show in addition that all of these integers are less than or equal to .
- (ii) (Noether gap theorem) If are a sequence of distinct points in , show that there are precisely positive integers with the property that there does not exist a meromorphic function with a simple pole at , at most a simple pole at , and no other poles. Show in addition that all of these integers are less than or equal to .

** — 5. Appendix: connectedness of irreducible algebraic curves — **

In this section we prove

Theorem 45Let be an irreducible homogeneous polynomial of degree . Then is connected.

We begin with the affine version of this theorem:

Proposition 46Let be an irreducible polynomial of degree . Then the affine curveis connected.

We observe that this theorem fails if one replaces the complex numbers by the real ones; for instance, the quadratic polynomial is irreducible, but the hyperbola it defines in is disconnected. Thus we will need properties of the complex numbers that are not true for the reals. We will rely in particular on the fundamental theorem of algebra, the removability of bounded singularities, the generalised Liouville theorem that entire functions of polynomial growth are polynomial, and the fact that the complex numbers remain connected even after removing finitely many points.

We now prove the proposition. We will use the classical approach of thinking of as a branched -fold cover over the complex numbers, possibly after some preparatory change of variables; the main difficulty is then to work around the ramification points of this cover. We turn to the details. Let be an irreducible polynomial of degree , then we can write it as

where for , lies in the space of polynoimals of one variable of degree at most . In particular is a constant. It could happen that this constant vanishes (e.g., consider the example and ); but in that case we will make a change of variables and consider instead the polynomial for a complex parameter . Now the analogue of is a non-trivial polynomial function of (because one of the must have degree exactly ), and so this quantity will be non-zero for some (in fact for all but at most values of ).

Henceforth we assume we have placed into a form where is non-zero. Then, for each , the function is a polynomial of one variable of degree exactly , so it has roots (counting multiplicity) by the fundamental theorem of algebra. Let us call these set of roots , thus

and

From Rouché’s theorem we know that the zero set varies continuously in in the following sense: for any and , each point of will stay within of some point in if is sufficiently close to . (In other words, is continuous with respect to Hausdorff distance.) We also see that the elements in grow at most polynomially in . For some values of , some of these roots in may be repeated. For instance, if , then , which has a double root at if or . However, if this occurs for some , the the degree polynomial and the degree have a common root . This only occurs when the resultant of and vanishes. (One can also use the discriminant of here in place of the resultant; the two are constant multiples of each other.) From the definition of the resultant we see that is a polynomial in , and furthermore we have a Bezout identity

where are polynomials of degree at most and in respectively, with coefficients that are polynomials in . The resultant cannot vanish identically, as this would mean that divides viewed as polynomials in , which contradicts unique factorisation and the irreducibility of since cannot divide the lower degree polynomials or . Thus the resultant can only vanish for a finite number of , and so for all but finitely many the roots of are distinct, thus has cardinality exactly and is non-vanishing at each element of .

From this and the inverse function theorem we see that for outside of a finite number of points (known as ramification points), the set varies holomorphically with . Locally, one can thus describe as a family of holomorphic functions, although the order in which one labels these points is arbitrary, as one varies around one of the ramification points, theis ordering may be permuted (consider for instance the case as goes around the origin, in which we can write for various branches of the square root function).

Now suppose that is disconnected, so it splits into two non-empty clopen subsets . At each non-ramified point , the set meets some subset of . In local coordinates, the are distinct and vary continuously with , the number of points in which meets is locally constant; since with finitely many points removed is connected, this number is then globally constant, thus there is such that has cardinality precisely . This lets us factor , where is the degree polynomial

and is the degree polynomial

The coefficients of these polynomials are functions of that vary holomorphically with outside of the ramification points; they also stay bounded as one approaches these points and grow at most polynomially. Hence (by the generalised Liouville theorem) they depend polynomially on , thus and are in fact a polynomial jointly in . But this contradicts the irreducibility of , unless or . We conclude that is connected after deleting its ramification points. But from the continuous dependence of on , the ramification points adhere to the rest of (the zeroes of are stable under small perturbations, even at points of ramification), so that is connected, proving the proposition.

Now we prove the theorem. The case can be done by hand, so assume . Let be an irreducible homogeneous polynomial of degree . Then is an irreducible polynomial of degree (it cannot be less than , as this will make contain a power of which makes it reducible since ). As a consequence, we see from the proposition that the affine part is connected; similarly if we replace the condition with and . As these three pieces of cover the whole zero locus, it will suffice to show that they intersect each other; for instance, it will suffice to show that the zero set of is not completely contained in any line. But this is clear from the proof of the proposition, which shows that (after a linear transformation) almost every vertical line meets this zero set in points.

The following exercise will be moved to a more appropriate position later (to avoid renumbering for homework exercises).

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Exercise 47Let be a holomorphic map between a compact Riemann surface and a Riemann surface . Show that is either surjective or constant.

The most recent news is that we appear to have completed the verification that is free of zeroes when and , which implies that . For very large (for instance when the quantity is at least ) this can be done analytically; for medium values of (say when is between and ) this can be done by numerically evaluating a fast approximation to and using the argument principle in a rectangle; and most recently it appears that we can also handle small values of , in part due to some new, and significantly faster, numerical ways to evaluate in this range.

One obvious thing to do now is to experiment with lowering the parameters and and see what happens. However there are two other potential ways to bound which may also be numerically feasible. One approach is based on trying to exclude zeroes of in a region of the form , and for some moderately large (this acts as a “barrier” to prevent zeroes from flowing into the region at time , assuming that they were not already there at time ). This require significantly less numerical verification in the aspect, but more numerical verification in the aspect, so it is not yet clear whether this is a net win.

Another, rather different approach, is to study the evolution of statistics such as over time. One has fairly good control on such quantities at time zero, and their time derivative looks somewhat manageable, so one may be able to still have good control on this quantity at later times . However for this approach to work, one needs an effective version of the Riemann-von Mangoldt formula for , which at present is only available asymptotically (or at time ). This approach may be able to avoid almost all numerical computation, except for numerical verification of the Riemann hypothesis, for which we can appeal to existing literature.

Participants are also welcome to add any further summaries of the situation in the comments below.

]]>The last two threads have been focused primarily on the test problem of showing that whenever . We have been able to prove this for most regimes of , or equivalently for most regimes of the natural number parameter . In many of these regimes, a certain explicit approximation to was used, together with a non-zero normalising factor ; see the wiki for definitions. The explicit upper bound

has been proven for certain explicit expressions (see here) depending on . In particular, if satisfies the inequality

then is non-vanishing thanks to the triangle inequality. (In principle we have an even more accurate approximation available, but it is looking like we will not need it for this test problem at least.)

We have explicit upper bounds on , , ; see this wiki page for details. They are tabulated in the range here. For , the upper bound for is monotone decreasing, and is in particular bounded by , while and are known to be bounded by and respectively (see here).

Meanwhile, the quantity can be lower bounded by

for certain explicit coefficients and an explicit complex number . Using the triangle inequality to lower bound this by

we can obtain a lower bound of for , which settles the test problem in this regime. One can get more efficient lower bounds by multiplying both Dirichlet series by a suitable Euler product mollifier; we have found for to be good choices to get a variety of further lower bounds depending only on , see this table and this wiki page. Comparing this against our tabulated upper bounds for the error terms we can handle the range .

In the range , we have been able to obtain a suitable lower bound (where exceeds the upper bound for ) by numerically evaluating at a mesh of points for each choice of , with the mesh spacing being adaptive and determined by and an upper bound for the derivative of ; the data is available here.

This leaves the final range (roughly corresponding to ). Here we can numerically evaluate to high accuracy at a fine mesh (see the data here), but to fill in the mesh we need good upper bounds on . It seems that we can get reasonable estimates using some contour shifting from the original definition of (see here). We are close to finishing off this remaining region and thus solving the toy problem.

Beyond this, we need to figure out how to show that for as well. General theory lets one do this for , leaving the region . The analytic theory that handles and should also handle this region; for presumably the argument principle will become relevant.

The full argument also needs to be streamlined and organised; right now it sprawls over many wiki pages and github code files. (A very preliminary writeup attempt has begun here). We should also see if there is much hope of extending the methods to push much beyond the bound of that we would get from the above calculations. This would also be a good time to start discussing whether to move to the writing phase of the project, or whether there are still fruitful research directions for the project to explore.

Participants are also welcome to add any further summaries of the situation in the comments below.

]]>246C is primarily a topics course, and tends to be a somewhat miscellaneous collection of complex analysis subjects that were not covered in the previous two installments of the series. The initial topics I have in mind to cover are

- Elliptic functions;
- The Riemann-Roch theorem;
- Circle packings;
- The Bieberbach conjecture (proven by de Branges); and
- the Schramm-Loewner equation (SLE).
- This list is however subject to change (it is the first time I will have taught on any of these topics, and I am not yet certain on the most logical way to arrange them; also I am not completely certain that I will be able to cover all the above topics in ten weeks). I welcome reference recommendations and other suggestions from readers who have taught on one or more of these topics.

As usual, I will be posting lecture notes on this blog as the course progresses.

[Update: Mar 13: removed elliptic functions, as I have just learned that this was already covered in the prior 246B course.]

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