Here is a purely algebraic form of the problem:
Problem 1 Let be a formal function of one variable . Suppose that is the formal function defined by
where we use to denote the -fold derivative of with respect to the variable .
- (i) Show that can be formally recovered from by the formula
- (ii) There is a remarkable further formal identity relating with that does not explicitly involve any infinite summation. What is this identity?
To rigorously formulate part (i) of this problem, one could work in the commutative differential ring of formal infinite series generated by polynomial combinations of and its derivatives (with no constant term). Part (ii) is a bit trickier to formulate in this abstract ring; the identity in question is easier to state if are formal power series, or (even better) convergent power series, as it involves operations such as composition or inversion that can be more easily defined in those latter settings.
To illustrate Problem 1(i), let us compute up to third order in , using to denote any quantity involving four or more factors of and its derivatives, and similarly for other exponents than . Then we have
and hence
multiplying, we have
and
and hence after a lot of canceling
Thus Problem 1(i) holds up to errors of at least. In principle one can continue verifying Problem 1(i) to increasingly high order in , but the computations rapidly become quite lengthy, and I do not know of a direct way to ensure that one always obtains the required cancellation at the end of the computation.
Problem 1(i) can also be posed in formal power series: if
is a formal power series with no constant term with complex coefficients with , then one can verify that the series
makes sense as a formal power series with no constant term, thus
For instance it is not difficult to show that . If one further has , then it turns out that
as formal power series. Currently the only way I know how to show this is by first proving the claim for power series with a positive radius of convergence using the Cauchy integral formula, but even this is a bit tricky unless one has managed to guess the identity in (ii) first. (In fact, the way I discovered this problem was by first trying to solve (a variant of) the identity in (ii) by Taylor expansion in the course of attacking another problem, and obtaining the transform in Problem 1 as a consequence.)
The transform that takes to resembles both the exponential function
and Taylor’s formula
but does not seem to be directly connected to either (this is more apparent once one knows the identity in (ii)).
Lemma 1 (Holomorphicity and harmonicity are conformal invariants) Let be a complex diffeomorphism between two Riemann surfaces .
- (i) If is a function to another Riemann surface , then is holomorphic if and only if is holomorphic.
- (ii) If are open subsets of and is a function, then is harmonic if and only if is harmonic.
Proof: Part (i) is immediate since the composition of two holomorphic functions is holomorphic. For part (ii), observe that if is harmonic then on any ball in , is the real part of some holomorphic function thanks to Exercise 62 of Notes 3. By part (i), is also holomorphic. Taking real parts we see that is harmonic on each ball in , and hence harmonic on all of , giving one direction of (ii); the other direction is proven similarly.
Exercise 2 Establish Lemma 1(ii) by direct calculation, avoiding the use of holomorphic functions. (Hint: the calculations are cleanest if one uses Wirtinger derivatives, as per Exercise 27 of Notes 1.)
Exercise 3 Let be a complex diffeomorphism between two open subsets of , let be a point in , let be a natural number, and let be holomorphic. Show that has a zero (resp. a pole) of order at if and only if has a zero (resp. a pole) of order at .
From Lemma 1(ii) we can now define the notion of a harmonic function on a Riemann surface ; such a function is harmonic if, for every coordinate chart in some atlas, the map is harmonic. Lemma 1(ii) ensures that this definition of harmonicity does not depend on the choice of atlas. Similarly, using Exercise 3 one can define what it means for a holomorphic map on a Riemann surface to have a pole or zero of a given order at a point , with the definition being independent of the choice of atlas.
In view of Lemma 1, it is thus natural to ask which Riemann surfaces are complex diffeomorphic to each other, and more generally to understand the space of holomorphic maps from one given Riemann surface to another. We will initially focus attention on three important model Riemann surfaces:
The designation of these model Riemann surfaces as elliptic, parabolic, and hyperbolic comes from Riemannian geometry, where it is natural to endow each of these surfaces with a constant curvature Riemannian metric which is positive, zero, or negative in the elliptic, parabolic, and hyperbolic cases respectively. However, we will not discuss Riemannian geometry further here.
All three model Riemann surfaces are simply connected, but none of them are complex diffeomorphic to any other; indeed, there are no non-constant holomorphic maps from the Riemann sphere to the plane or the disk, nor are there any non-constant holomorphic maps from the plane to the disk (although there are plenty of holomorphic maps going in the opposite directions). The complex automorphisms (that is, the complex diffeomorphisms from a surface to itself) of each of the three surfaces can be classified explicitly. The automorphisms of the Riemann sphere turn out to be the Möbius transformations with , also known as fractional linear transformations. The automorphisms of the complex plane are the linear transformations with , and the automorphisms of the disk are the fractional linear transformations of the form for and . Holomorphic maps from the disk to itself that fix the origin obey a basic but incredibly important estimate known as the Schwarz lemma: they are “dominated” by the identity function in the sense that for all . Among other things, this lemma gives guidance to determine when a given Riemann surface is complex diffeomorphic to a disk; we shall discuss this point further below.
It is a beautiful and fundamental fact in complex analysis that these three model Riemann surfaces are in fact an exhaustive list of the simply connected Riemann surfaces, up to complex diffeomorphism. More precisely, we have the Riemann mapping theorem and the uniformisation theorem:
Theorem 4 (Riemann mapping theorem) Let be a simply connected open subset of that is not all of . Then is complex diffeomorphic to .
Theorem 5 (Uniformisation theorem) Let be a simply connected Riemann surface. Then is complex diffeomorphic to , , or .
As we shall see, every connected Riemann surface can be viewed as the quotient of its simply connected universal cover by a discrete group of automorphisms known as deck transformations. This in principle gives a complete classification of Riemann surfaces up to complex diffeomorphism, although the situation is still somewhat complicated in the hyperbolic case because of the wide variety of discrete groups of automorphisms available in that case.
We will prove the Riemann mapping theorem in these notes, using the elegant argument of Koebe that is based on the Schwarz lemma and Montel’s theorem (Exercise 57 of Notes 4). The uniformisation theorem is however more difficult to establish; we discuss some components of a proof (based on the Perron method of subharmonic functions) here, but stop short of providing a complete proof.
The above theorems show that it is in principle possible to conformally map various domains into model domains such as the unit disk, but the proofs of these theorems do not readily produce explicit conformal maps for this purpose. For some domains we can just write down a suitable such map. For instance:
Exercise 6 (Cayley transform) Let be the upper half-plane. Show that the Cayley transform , defined by
is a complex diffeomorphism from the upper half-plane to the disk , with inverse map given by
Exercise 7 Show that for any real numbers , the strip is complex diffeomorphic to the disk . (Hint: use the complex exponential and a linear transformation to map the strip onto the half-plane .)
Exercise 8 Show that for any real numbers $latex {a<b0, a < \theta < b \}}&fg=000000$ is complex diffeomorphic to the disk . (Hint: use a branch of either the complex logarithm, or of a complex power .)
We will discuss some other explicit conformal maps in this set of notes, such as the Schwarz-Christoffel maps that transform the upper half-plane to polygonal regions. Further examples of conformal mapping can be found in the text of Stein-Shakarchi.
— 1. Maps between the model Riemann surfaces —
In this section we study the various holomorphic maps, and conformal maps, between the three model Riemann surfaces , , and .
From Exercise 19 of Notes 4, we know that the only holomorphic maps from the Riemann sphere to itself take the form of a rational function away from the zeroes of (and from ), with these singularities all being removable, and with not identically zero. We can of course reduce to lowest terms and assume that and have no common factors. In particular, if is to take values in rather than , then can have no roots (since will have a pole at these roots) and so by the fundamental theorem of calculus is constant and is a polynomial; in order for to have no pole at infinity, must then be constant. Thus the only holomorphic maps from to are the constants; in particular, the only holomorphic maps from to are the constants. In particular, is not complex diffeomorphic to or (this is also topologically obvious since the Riemann sphere is compact, and and are not).
Exercise 9 More generally, show that if is a compact Riemann surface and is a connected non-compact Riemann surface, then the only holomorphic maps from to are the constants. (Hint: use the open mapping theorem, Theorem 37 of Notes 4.)
Now we consider complex automorphisms of the Riemann sphere to itself. There are some obvious examples of such automorphisms:
More generally, given any complex numbers with , we can define the Möbius transformation (or fractional linear transformation) for , with the convention that is mapped to and is mapped to (where we adopt the further convention that for non-zero ). For , this is an affine transformation , which is clearly a composition of a translation and dilation map; for , this is a combination of translations, dilations, and the inversion map. Thus all Möbius transformations are formed from composition of the translations, dilations, and inversions, and in particular are also automorphisms of the Riemann sphere; it is also easy to see that the Möbius transformations are closed under composition, and are thus the group generated by the translations, dilations, and inversions.
One can interpret the Möbius transformations as projective linear transformations as follows. Recall that the general linear group is the group of matrices with non-vanishing determinant . Clearly every such matrix generates a Möbius transformation . However, two different elements of can generate the same Möbius transformation if they are scalar multiples of each other. If we define the projective linear group to be the quotient group of by the group of scalar invertible matrices, then we may identify the set of Möbius transformations with . The group acts on the space by the usual map
If we let be the complex projective line, that is to say the space of one-dimensional subspaces of , then acts on this space also, with the action of the scalars being trivial, so we have an action of on . We can identify the Riemann sphere with the complex projective line by identifying each with the one-dimensional subspace of , and identifying with . With this identification, one can check that the action of on has become identified with the action of the group of Möbius transformations on . (In particular, the group of Möbius transformations is isomorphic to .)
There are enough Möbius transformations available that their action on the Riemann sphere is not merely transitive, but is in fact -transitive:
Lemma 10 (-transitivity) Let be distinct elements of the Riemann sphere , and let also be three distinct elements of the Riemann sphere. Then there exists a unique Möbius transformation such that for .
Proof: We first show existence. As the Möbius transformations form a group, it suffices to verify the claim for a single choice of , for instance . If then the affine transformation will have the desired properties. If , we can use translation and inversion to find a Möbius transformation that maps to ; applying the previous case with with and then applying , we obtain the claim.
Now we prove uniqueness. By composing on the left and right with Möbius transforms we may assume that . A Möbius transformation that fixes must obey the constraints and so must be the identity, as required.
Möbius transformations are not 4-transitive, thanks to the invariant known as the cross-ratio:
Exercise 11 Define the cross-ratio between four distinct points on the Riemann sphere by the formula
if all of avoid , and extended continuously to the case when one of the points equals (e.g. .
- (i) Show that an injective map is a Möbius transform if and only if it preserves the cross-ratio, that is to say that for all distinct points . (Hint: for the “only if” part, work with the basic Möbius transforms. For the “if” part, reduce to the case when fixes three points, such as .)
- (ii) If are distinct points in , show that lie on a common extended line (i.e., a line in together with ) or circle in if and only if the cross-ratio is real. Conclude that a Möbius transform will map an extended line or circle to an extended line or circle.
As one quick application of Möbius transformations, we have
Proposition 12 is simply connected.
Proof: We have to show that any closed curve in is contractible to a point in . By deforming locally into line segments in either of the two standard coordinate charts of we may assume that is the concatenation of finitely many such line segments; in particular, cannot be a space-filling curve (as one can see from e.g. the Baire category theorem) and thus avoids at least one point in . If avoids then it lies in and can thus be contracted to a point in (and hence in ) since is convex. If avoids any other point , then we can apply a Möbius transformation to move to , contract the transformed curve to a point, and then invert the Möbius transform to contract to a point in .
Exercise 13 (Jordan curve theorem in the Riemann sphere) Let be a simple closed curve in the Riemann sphere. Show that the complement of in is the union of two disjoint simply connected open subsets of . (Hint: one first has to exclude the possibility that is space-filling. Do this by verifying that is homeomorphic to the unit circle.)
It turns out that there are no other automorphisms of the Riemann sphere than the Möbius transformations:
Proposition 14 (Automorphisms of Riemann sphere) Let be a complex diffeomorphism. Then is a Möbius transformation.
Proof: By Lemma 10 and composing with a Möbius transformation, we may assume without loss of generality that fixes . From Exercise 19 of Notes 4 we know that is a rational function (with all singularities removed); we may reduce terms so that have no common factors. Since is bijective and fixes , it has no poles in , and hence can have no roots; by the fundamental theorem of algebra, this makes constant. Similarly, has no zeroes other than , and so must be a monomial; as also fixes , it must be of the form for some natural number . But this is only injective if , in which case is clearly a Möbius transformation.
Now we look at holomorphic maps on . There are plenty of holomorphic maps from to ; indeed, these are nothing more than the entire functions, of which there are many (indeed, an entire function is nothing more than a power series with an infinite radius of convergence). There are even more holomorphic maps from to , as these are just the meromorphic functions on . For instance, any ratio of two entire functions, with not identically zero, will be meromorphic on . On the other hand, from Liouville’s theorem (Theorem 28 of Notes 3) we see that the only holomorphic maps from to are the constants. In particular, and are not complex diffeomorphic (despite the fact that they are diffeomorphic over the reals, as can be seen for instance by using the projection ).
The affine maps with and are clearly complex automorphisms on . In analogy with Proposition 14, these turn out to be the only automorphisms:
Proposition 15 (Automorphisms of complex plane) Let be a complex diffeomorphism. Then is an affine transformation for some and .
Proof: By the open mapping theorem (Theorem 37 of Notes 4), is open, and hence avoids the non-empty open set on . By the Casorati-Weierstrass theorem (Theorem 11 of Notes 4), we conclude that does not have an essential singularity at infinity. Thus extends to a holomorphic function from to , hence by Exercise 19 of Notes 4 is rational. As the only pole of is at infinity, is a polynomial; as is a diffeomorphism, the derivative has no zeroes and is thus constant by the fundamental theorem of algebra. Thus must be affine, and the claim follows.
Exercise 16 Let be an injective holomorphic map. Show that is a Möbius transformation (restricted to ).
We remark that injective holomorphic maps are often referred to as univalent functions in the literature.
Finally, we consider holomorphic maps on . There are plenty of holomorphic maps from to (indeed, these are just the power series with radius of convergence at least ), and even more holomorphic maps from to (for instance, one can take the quotient of two holomorphic functions with non-zero). There are also many holomorphic maps from to , for instance one can take any bounded holomorphic function and multiply it by a small constant. However, we have the following fundamental estimate concerning such functions, the Schwartz lemma:
Lemma 17 (Schwarz lemma) Let be a holomorphic map such that . Then we have for all . In particular, .
Furthermore, if for some , or if , then there exists a real number such that for all .
Proof: By the factor theorem (Corollary 22 of Notes 3), we may write for some holomorphic . On any circle with , we have and hence ; by the maximum principle we conclude that for all . Sending to zero, we conclude that for all , and hence and .
Finally, if for some or , then equals for some , and hence by a variant of the maximum principle (see Exercise 18 below) we see that is constant, giving the claim.
Exercise 18 (Variant of maximum principle) Let be a connected Riemann surface, and let be a point in .
- (i) If is a harmonic function such that for all , then for all .
- (ii) If is a holomorphic function such that for all , then for all .
(Hint: use Exercise 17 of Notes 3 .)
One can think of the Schwarz lemma as follows. Let denote the collection of holomorphic functions with . Inside this collection we have the rotations for defined by . The Schwarz lemma asserts that these rotations “dominate” the remaining functions in in the sense that on , and in particular ; furthermore these inequalities are strict as long as is not one of the .
As a first application of the Schwarz lemma, we characterise the automorphisms of the disk . For any , one can check that the Möbius transformation preserves the boundary of the disk (since when ), and maps the point to the origin, and thus maps the disk to itself. More generally, for any and , the Möbius transformation is an automorphism of the disk . It turns out that these are the only such automorphisms:
Theorem 19 (Automorphisms of disk) Let be a complex diffeomorphism. Then there exists and such that for all . If furthermore , then we can take , thus for .
Proof: First suppose that . By the Schwarz lemma applied to both and its inverse , we see that . But by the inverse function theorem (or the chain rule), , hence . Applying the Schwarz lemma again, we conclude that for some , as required.
In the general case, there exists such that . If one then applies the previous analysis to , where is the automorphism , we obtain the claim.
Exercise 20 (Automorphisms of half-plane) Let be a complex diffeomorphism from the upper half-plane to itself. Show that there exist real numbers with such that for . Conclude that the automorphism group of either or is isomorphic as a group to the projective special linear group formed by starting with the special linear group of real matrices of determinant , and then quotienting out by the central subgroup .
Remark 21 Collecting the various assertions above about the holomorphic maps between the elliptic, parabolic, and hyperbolic model Riemann surfaces , , , one arrives at the following rule of thumb: there are “many” holomorphic maps from “more hyperbolic” surfaces to “less hyperbolic” surfaces, but “very few” maps going in the other direction (and also relatively few automorphisms from one space to an “equally hyperbolic” surface). This rule of thumb also turns out to be accurate in the context of compact Riemann surfaces, where “higher genus” becomes the analogue of “more hyperbolic” (and similarly for “less hyperbolic” or “equally hyperbolic”). One can formalise this latter version of the rule of thumb using such results as the Riemann-Hurwitz formula and the de Franchis theorem, but these are beyond the scope of this course.
Exercise 22 Let be a non-constant holomorphic map between Riemann surfaces . If is compact and is connected, show that is surjective and is compact. Conclude in particular that there are no non-constant bounded holomorphic functions on a compact Riemann surface.
— 2. Quotients of the model Riemann surfaces —
The three model Riemann surfaces , , are all simply connected, and the uniformisation theorem will tell us that up to complex diffeomorphism, these are the only simply connected Riemann surfaces that exist. However, it is possible to form non-simply-connected Riemann surfaces from these model surfaces by the procedure of taking quotients, as follows. Let be a Riemann surface, and let be a group of complex automorphisms of . We assume that the action of on is free, which means that the non-identity transformations in have no fixed points (thus for all ). We also assume that the action is proper (viewing as a discrete group), which means that for any compact subset of , there are only finitely many automorphisms in for which intersects . If the action is both free and proper, then we see that every point has a small neighbourhood with the property that the images are all disjoint; by making small enough we can also find a holomorphic coordinate chart to some open subset of . We can then form the quotient manifold of orbits , using the coordinate charts for any defined by setting
for all . One can easily verify that is a Riemann surface, and that the quotient map defined by is a surjective holomorphic map. The ability to easily take quotients is one of the key advantages of the Riemann surface formalism; another is the converse ability to construct covers, such as the universal cover of a Riemann surface, defined in Theorem 25 below.
Exercise 23 Let be a Riemann surface, a group of complex automorphisms of acting in a proper and free fashion, and let be the quotient map. Let be a holomorphic map to another Riemann surface . Show that there exists a holomorphic map such that if and only if for all .
Remark 24 It is also of interest to quotient a Riemann surface by a group of complex automorphisms whose action is not free. In that case, the quotient space need not be a manifold, but is instead a more general object known as an orbifold. A typical example is the modular curve (where is the group of matrices with integer coefficients and determinant ); this is of great importance in analytic number theory. However, we will not study orbifolds in this course.
Since the continuous image of a connected space is always connected, we see that any quotient of a connected Riemann surface is again connected. In the converse direction, one can use this construction to describe a connected Riemann surface as a quotient of a simply connected Riemann surface:
Theorem 25 (Universal cover) Let be a connected Riemann surface. Then there exists a simply connected Riemann surface , and a group of complex automorphisms acting on in a proper and free fashion, such that is complex diffeomorphic to .
Proof: For sake of brevity we omit some of the details of the construction as exercises.
We use the following abstract construction to build the Riemann surface . Fix a base point in . For any point in , we can form the space of all continuous paths from to for some interval with . We let denote the space of equivalence classes of such paths with respect to the operation of homotopy with fixed endpoints up to reparameterisation; for instance, if was simply connected, then would simply be a point. (One could also omit the reparameterisation by restricting the domain of to be a fixed interval such as .) As is connected, all the are non-empty. We then let be the (disjoint) union of all the . This defines a set together with a projection map that sends all the homotopy classes in to for each ; this is clearly a surjective map.
This defines as a set, but we want to give the structure of a Riemann surface, and thus must create an atlas of coordinate charts. For every , let be a coordinate chart that is a diffeomorphism between some neighbourhood of and the unit disk. Given a homotopy class in and a point in , we can then associate a point in by taking a path from to in the homotopy class , and concatenating it with the path that connects to via a line segment in the disk using the coordinate chart ; the homotopy class of this concatenated path does not depend on the precise choice of and will be denoted . If we let denote all the points obtained in this fashion as varies over , then it is easy to see (Exercise!) that the are disjoint and partition the set . We can then form coordinate charts for each and by setting for all . This defines both a topology on (by declaring a subset of to be open if is open for all ) and a complex structure, as the transition maps are easily verified (Execise!) to be both continuous and holomorphic. By construction we now see that is a covering space of , with the covering map.
Let be the homotopy class of the constant curve at . It is easy to see (Exercise!) that is connected (indeed, any point in determines (more or less tautologically) a family of paths in from to ). Next, we make the stronger claim that is simply connected. It suffices to show that any closed path from to is contractible to a point. Let denote the projected curve , thus is a closed curve from to itself. From the continuity method (Exercise!) we see that for any , the restriction of to lies in the homotopy class of ; in particular, itself lies in the homotopy class of , and is thus homotopic to a point. Another application of the continuity method (Exercise!) then shows that as one continuously deforms to a point, each of the curves obtained in this deformation lifts to a closed curve in from to , in the sense that ; furthermore, varies continuously in , giving the required homotopy from to a point.
Define a deck transformation to be a holomorphic map such that (that is to say, preserves each of the “fibres” of ). Clearly the composition of two deck transformations is again a deck transformation. From Corollary 50 of Notes 4 we see that for any and , there exists a unique deck transformations that maps to . Composing that a deck transformations with the deck transformations that maps to we see that all deck transformations are invertible and are thus complex automorphisms. If we let denote the collection of all deck transformations then we see that is a group that acts freely on and transitively on each fibre . For any , and the neighbourhoods as before, one can verify (Exercise!) that each deck transformation in permutes the disjoint open sets covering , and given any two of these sets there is exactly one deck transformation that maps to . From this one can check (Exercise!) that is complex diffeomorphic to as required.
Exercise 26 Write out the steps marked “Exercise!” in the above argument.
The manifold in the above theorem is called a universal cover of , and the group is (a copy of) the fundamental group of . These objects are basically uniquely determined by :
Exercise 27 Suppose one has two simply connected Riemann surfaces and two groups of automorphisms of respectively acting in proper and free fashions. Show that the following statements are equivalent:
- (i) The quotients and are complex diffeomorphic.
- (ii) There exists a complex diffeomorphism and a group isomorphism such that
for all and . In particular and are complex diffeomorphic, and the groups and are isomorphic.
(Hint: use Exercise 23 for one direction of the implication, and Corollary 50 of Notes 4 for the other implication.)
Exercise 28 Let be a connected Riemann surface, and let be a point in . Define the fundamental group based at to be the collection of equivalence classes of closed curves from to , under the relation of homotopy with fixed endpoints up to reparameterisation.
- (i) Show that is indeed a group, with the equivalence class of the constant curves as the identity element, the inverse of a homotopy class of a curve defined as , and the product of two homotopy classes of curves as .
- (ii) If are as in Theorem 25, show that is isomorphic to .
Exercise 29 Show that the fundamental group of is isomorphic to the integers (viewed as an additive group).
If we assume for now the uniformisation theorem, we conclude that every connected Riemann surface is the quotient of one of the three model surfaces , , by a group of complex automorphisms that act freely and properly; depending on which surface is used, we call these Riemann surfaces of elliptic type, parabolic type, and hyperbolic type respectively. We can then study each of the three model types in turn:
Elliptic type: By Proposition 14, the automorphisms of are the Möbius transformations. From the quadratic formula (or the fundamental theorem of algebra) we see that every Möbius transformation has at least one fixed point (for instance, the translations fix ). Thus the only group of complex automorphisms that can act freely on is the trivial group, so the only Riemann surfaces of elliptic type are those that are complex diffeomorphic to the Riemann sphere.
Parabolic type: By Proposition 14, the automorphisms of are the affine transformations . These transformations have fixed points in if , so in order to obtain a free action we must restrict to the translations . Thus we can view as an additive subgroup of , with now being the group quotient; as the action is additive, we can also write as . In order for the action to be proper, must be a discrete subgroup of (every point isolated). We can classify all such subgroups:
Exercise 30 Let be a discrete additive subgroup of . Show that takes on one of the following three forms:
- (i) (Rank zero case) the trivial group ;
- (ii) (Rank one case) a cyclic group for some ; or
- (iii) (Rank two case) a group for some with strictly complex (i.e., not real).
We conclude that every Riemann surface of parabolic type is complex diffeomorphic to a plane , a cylinder for some , or a torus for and strictly complex.
The case of the plane is self-explanatory. Using dilation maps we see that all cylinders are complex diffeomorphic to each other; for instance, they are all diffeomorphic to . The exponential map is -periodic and thus descends to a map from to ; it is easy to see that this map is a complex diffeomorphism, thus the punctured plane can be used as a model for all Riemann surface cylinders.
The case of the tori are more interesting. One can use dilations to normalise one of the to be a specific value such as , but one cannot normalise both:
Exercise 31 Let and be two tori. Show that these two tori are complex diffeomorphic if and only if there exists an element of the special linear group (thus are integers with ) such that
(Hint: lift any such diffeomorphism to a holomorphic map from to of linear growth.)
In contrast to the cylinders , which are complex diffeomorphic to a subset of the complex plane, one cannot model a torus by a subset of ; indeed, if there were a complex diffeomorphism , then would have to be non-empty, compact, and (by the open mapping theorem) open in , which is impossible since is non-compact and connected. However, it is an important fact in algebraic geometry, classical analysis and number theory that these tori can be modeled instead by elliptic curves. The theory of elliptic curves is extremely rich, but is beyond the scope of this course and will not be discussed further here (but the Weierstrass elliptic functions used to construct the complex diffeomorphism between tori and elliptic curves may be covered in subsequent quarters).
Exercise 32 Let be a connected subset of that omits at least two points of . Show that cannot be of elliptic or parabolic type. (Hint: in addition to the open mapping theorem argument given above, one can use either the great Picard theorem, Theorem 56 of Notes 4, or the simpler Casorati-Weierstrass theorem (Theorem 11 of Notes 4).) In particular, assuming the uniformisation theorem, such sets must be of hyperbolic type. (Note this is compatible with our previous intuition that “more hyperbolic” is analogous to “higher genus” or “has more holes”.)
Hyperbolic type: Here it is convenient to model the hyperbolic Riemann surface using the upper half-plane (the Poincaré half-plane model) rather than the disk (the Poincaré disk model). By Exercise 20, a Riemann surface of hyperbolic type is then isomorphic to a quotient of by some subgroup of that acts freely and properly. Properness is easily seen to be equivalent to being a discrete subgroup of (using the topology inherited from the embedding of in ); such groups are known as Fuchsian groups. Freeness can also be described explicitly:
Exercise 33 Show that a subgroup of acts freely on if and only if it avoids all (equivalence classes) of matrices in that are elliptic in the sense that they obey the trace condition .
It turns out that in contrast to the elliptic type and parabolic type situations, there are a very large number of possible subgroups obeying these conditions, and a complete classification of them is basically hopeless. The theory of Fuchsian groups is again very rich, being a foundational topic in hyperbolic geometry, but is again beyond the scope of this course.
Remark 34 The twice-punctured plane must be of hyperbolic type by the uniformisation theorem and Exercise 32. This gives another proof of the little Picard theorem: an entire function that omits (say) the points must then lift (by Corollary 50 of Notes 4) to a holomorphic map from to , which must then be constant by Liouville’s theorem. A more complicated argument along these lines also proves the great Picard theorem. It turns out that the covering map from to can be described explicitly using the theory of elliptic functions (and specifically the modular lambda function), but this is beyond the scope of this course.
Exercise 35 Show that any annulus is of hyperbolic type, and is in fact complex diffeomorphic to for some cyclic group of dilations. (Hint: first use the complex exponential to cover the annulus by a strip, then use Exercise 7.)
Exercise 36 Let and be two annuli with and . Show that and are complex diffeomorphic if and only if . (Hint: one can either argue by lifting to the half-plane using the previous exercise, or else using the Schwarz reflection principle (adapted to circles in place of lines) repeatedly to extend a holomorphic map from to to a holomorphic map from a punctured disk to a punctured disk; one can also combine the methods by taking logarithms to lift , to strips, and then using the original Schwarz reflection principle.)
Exercise 37
- (i) Show that the punctured disk is of hyperbolic type, and is complex diffeomorphic to , where acts on by translations.
- (ii) Show that the Jowkowsky transform is a complex diffeomorphism from to the slitted extended complex plane . Conclude that is also complex diffeomorphic to .
— 3. The Riemann mapping theorem —
We are now ready to prove the Riemann mapping theorem, Theorem 4, using an argument of Koebe. To motivate the argument, let us rephrase the Schwarz lemma in the following form:
Lemma 38 (Schwarz lemma, again) Let be a Riemann surface, and let be a point in . Let denote the collection of holomorphic functions with . If contains an element that is a complex diffeomorphism, then for all ; if is a subset of the complex plane , we also have . Furthermore, in either of these two inequalities, equality holds if and only if for some real number .
Proof: Apply Lemma 17 to the map .
This lemma suggests the following strategy to prove the Riemann mapping theorem: starting with the open subset of the complex plane , pick a point in that subset, and form the collection of holomorphic maps that map to , and locate an element of this collection for which the magnitude is maximal. If the Riemann mapping theorem were true, then Lemma 38 would ensure that this would be a complex diffeomorphism, and we would be done.
It turns out to be convenient to work with the somewhat smaller collection of injective holomorphic maps (also known as univalent functions from to ). We first observe that this collection is non-empty for the sets of interest:
Proposition 39 Let be a simply connected subset of that is not all of . Then there exists an injective holomorphic map .
Proof: By applying a translation to , we may assume that avoids the origin . If in fact avoided a disk , then we could use the map to map injectively into the disk . At present, need not avoid any disk (e.g. could be the complex plane with the negative axis removed). However, as is simply connected and avoids , we can argue as in Section 4 of Notes 4 to obtain a holomorphic branch of the square root function, that is to say a holomorphic map such that for all . As is injective, must also be injective; it is also clearly non-constant, so from the open mapping theorem is open and thus contains some disk . But if lies in then cannot lie in since this would make the map non-injective; thus avoids a disk , and the claim follows.
If is a point in , then the map constructed by the above proposition need not map to the origin, but this is easily fixed by composing with a suitable automorphism of . To prove the Riemann mapping theorem, it will thus suffice to show
Proposition 40 Let be a simply connected Riemann surface, let be a point in , and let be the collection of injective holomorphic maps with . If is non-empty, then is complex diffeomorphic to .
Proof: By identifying with its image under one of the elements of , we may assume without loss of generality that is itself an open subset of , with .
Define the quantity
As contains the identity map, is at least ; from the Cauchy inequalities (Corollary 27 of Notes 2) we see that is finite. Hence there exists a sequence in with converging to as . From Montel’s theorem (Exercise 57(i) of Notes 4) we know that is a normal family, so on passing to a subsequence we may assume that the converge locally uniformly to some limit . By Hurwitz’s theorem (Exercise 41 of Notes 4), the limit is holomorphic and is either injective or constant. But from the higher order Cauchy integral formula (Theorem 25 of Notes 3), converges to , hence and so cannot be constant, and is thus injective. From the maximum principle (Exercise 18), we know that takes values in , and not just .
To conclude the proposition, we need to show that is also surjective. Here we use a variant of the argument used to prove Proposition 39. Suppose for contradiction that avoids some point in . Let be an automorphism of that sends to , then avoids the origin. As is simply connected, we can thus find a holomorphic square root of , thus
Since and hence are injective, is also. Finally, if is an automorphism of that sends to , then the map lies in . The map is related to by the formula
where is the squaring map. Observe that the map is a holomorphic map from to that maps to , and is not a rotation map (since is not a Möbius transformation). Thus by the Schwarz lemma (Lemma 17), we have
and hence by the chain rule
But this contradicts the definition of , and we are done.
Exercise 41 Let be an open connected non-empty subset of . Show that the following are equivalent:
- (i) is simply connected.
- (ii) One has for every holomorphic function and every closed curve in .
- (iii) For every holomorphic function there exists a holomorphic with .
- (iv) For every holomorphic function there exists a holomorphic with .
- (v) The complement of in the Riemann sphere is connected.
(Hint: to relate (v) to the other claims, use Exercise 43 from Notes 4.)
— 4. Schwartz-Christoffel mappings —
The Riemann mapping theorem guarantees the existence of complex diffeomorphisms for any simply connected subset of the complex plane that is not all of ; in particular, such diffeomorphisms exist if is a polygon, by which we mean the interior region of a simple closed anticlockwise polygonal path . However, the proof of the Riemann mapping theorem does not supply an easy way to compute what this map is. Nevertheless, in the case of polygons a reasonably explicit formula for (or more precisely, for the derivative of the inverse of ) may be found. Our arguments here are based on those in the text of Ahlfors.
We set up some notation. Let be a simple closed anticlockwise polygonal path (in particular, the are all distinct), let be the polygon enclosed by this path, and let be a complex diffeomorphism, the existence of which is guaranteed by the Riemann mapping theorem. (The map is only unique up to composition by an automorphism of , but this will not concern us for the present analysis.) We adopt the convention that and , and for , we let denote the counterclockwise angle subtended by the polygon at (normalised by a factor of ), in the sense that
for some real . (Note that cannot attain the values or as this would cause the polygona path to be non-simple.) It is also convenient to introduce the normalised exterior angle by (thus is positive at a convex angle of the polygon, zero at a reflex angle, and negative at a concave angle), so that
Telescoping this identity, we conclude that must be an even integer. Indeed, from Euclidean geometry we know that the sum of the exterior angles of a polygon add up to , so that
we will give an analytic proof of this fact presently.
From the Alexander numbering rule (Exercise 55 of Notes 3) we see that always lies to the left of the polygonal path . We can formalise this statement as follows. First suppose that is a non-vertex boundary point of , thus for some and . Then we can form the affine map by the formula
and the numbering rule tells us that for small enough, the half-disk
is mapped holomorphically by into . If is instead a vertex of , the situation is a little trickier; we now define the map by the formula
where we choose the branch of with branch cut at the negative imaginary axis and to be positive real on the positive real axis. Then again will map holomorphically into for small enough. (The reader is encouraged to draw a picture to understand these maps.)
Now we perform some local analysis near the boundary. We first need a version of the Schwarz reflection principle (Exercise 37 of Notes 3) for harmonic functions.
Exercise 42 (Dirichlet problem) Let be a continuous function. Show that there exists a unique function that is continuous on the closed disk , harmonic on the open disk , and equal to on the boundary . Furthermore, show that is given by the formula
for , where is the Poisson kernel
(compare with Exercise 17 of Notes 3).
Lemma 43 (Schwarz reflection for harmonic functions) Let be an open subset of symmetric around the real axis, and let be a continuous function on the region that vanishes on and is harmonic in . Let be the antisymmetric extension of , defined by setting and for . Then is harmonic.
Proof: Morally speaking, this lemma follows from the analogous reflection principle for holomorphic functions, but there is a difficulty because we do not have enough regularity on the real axis to easily build a harmonic conjugate that is continuous all the way to the real axis. Instead we shall rely on the maximum principle as follows.
It is clear that is continuous and harmonic away from the real axis, so it suffices to show for any and any small that is harmonic on .
Using Exercise 42, we can find a continuous function which agrees with on the boundary and is harmonic on the interior. From the antisymmetry of and uniqueness (or the Poisson kernel formula) we see that is also antisymmetric and thus vanishes on the real axis. The difference is then harmonic on the half-disks and and vanishes on the boundary of these half-disks, so by the maximum principle they vanish everywhere in . Thus agrees with on and is therefore harmonic on this disk as required.
Proposition 44 Let be a boundary point of (which may or may not be a vertex). Then for small enough, the maps extend holomorphically to a map from to which maps the origin to a point on the unit circle. Furthermore, this map is injective for small enough.
Proof: For any , the preimage of the closed disk is a compact subset of and thus stays a positive distance away from the boundary of . In particular, for sufficiently close to the boundary of , must exceed . We conclude that the function extends continuously to a map from to , by declaring the map to equal on the boundary. In particular, for small enough, the map also extends continously to , and equals on the real boundary of . For small enough, avoids zero on this region, and so the function will extend continuously to , and vanish on the real portion of the boundary. By taking local branches of we see that this function is also harmonic. By Lemma 43, extends harmonically to , and on taking harmonic conjugates we conclude that extends holomorphically to . Taking exponentials, we obtain a holomorphic extension of to , with . To prove injectivity, it suffices (shrinking as necessary) to show that the derivative of at is non-zero. But if this were not the case, then would have a zero of order at least two, which by the factor theorem implies that would not map to a half-plane bordering the origin, and in particular cannot map to , a contradiction.
As a corollary, we see that extends to a continuous map that maps to , and around every point in the boundary of , maps a small neighbourhood of in to a small neighbourhood of in . As is injective on , this implies that is also injective on the boundary of . The image is compact in and contains , hence is in fact a bijective continuous map between compact Hausdorff spaces and is thus a homoeomorphism. Thus we can form an inverse map , which maps holomorphically to . (This latter claim in fact works if one replaces the polygonal path by a arbitrary simple closed curve; this is a theorem of Carathéodory.)
Consider the function on the line segment from to . By Proposition 44, is smooth on this line segment, has non-zero derivative, and takes values in ; setting , we see that must traverse a simple curve from to in . As is orientation preserving, lies to the left of the line segment , and the disk lies to the left of traversed anticlockwise, we see that must traverse the anticlockwise arc from to . Following all around , we see that must be arranged anticlockwise in the unit circle in the sense that we have for all for some
Inverting, we see that for any , smoothly maps the anticlockwise arc from to to the line segment from to , with derivative nonvanishing. Thus on taking arguments
Next, we study near (and near ) for some . From Proposition 44 we see that in a sufficiently small neighbourhood of in , one has for some injective holomorphic map from a neighbourhood of in to a neighbourhood of in that maps to zero. Since maps the arc from to to the line segment from to , must map the portion of the arc from to near to a portion of the positive real axis; in particular, by the chain rule, is a positive real, call it . If we factor
noting that the third factor is close to one and the second factor lies in the upper half-plane, we have
and hence from we have the factorisation
for near in , for some that is holomorphic and non-zero in a neighbourhood of in . Differentiating using , we conclude that
for near in , for some that is also holomorphic and non-zero in a neighbourhood of in .
The function is holomorphic and non-vanishing; as is simply connected, we must therefore have for some holomorphic (by Exercise 46 of Notes 4). For any between and , we see from the previous discussion that extends holomorphically to a neighbourhood of , with non-vanishing at , so extends also. From (3) we see that the argument of is constant on the interval , and hence
is also constant on this interval. Meanwhile, from (4) we see that for near in , we have
for some holomorphic in a neighbourhood of in , where is a branch of the complex logarithm with branch cut at . From this we see that the function has a jump discontinuity with jump as crosses . As this function clearly increases by when increases by , we conclude the geometric identity (2).
Now consider the modified function defined by
Then is holomorphic on , and by the above analysis it extends continuously to . We consider the imaginary part at ,
where is a branch of the argument function with branch cut at . Writing , we see that is constant as long as is not an integer multiple of . From this, (5), and (2), we see that the function is constant on each arc . Thus the function is harmonic on , continuous on , and constant on the boundary , so by the maximum principle it is constant, which from the Cauchy-Riemann equations makes constant also. Thus we have
on for some complex constant , which on exponentiating gives
on for some non-zero complex constant . Applying the fundamental theorem of calculus, we obtain the Schwarz-Christoffel formula:
Theorem 45 (Schwarz-Christoffel for the disk) Let be a closed simple anticlockwise polygonal path, and define the exterior angles as above. Let be the polygon enclosed by this path, and let be a complex diffeomorphism. Then there exist phases , for some , a non-zero complex number , and a complex number such that
for all , where the integral is over an arbitrary curve from to , and one selects a branch of with branch cut on the negative imaginary axis . Furthermore, converges to as approaches for every .
Note that one can change the branches of here, and also modify the normalising factors , by adjusting the constant in a suitable fashion, as long one does not move the branch cut for into the disk ; one can similarly change the initial point of the curve to any other point in by adjusting . By taking log-derivatives in (6), we can also express the Schwarz-Christoffel formula equivalently as a partial fractions decomposition of :
The Schwarz-Christoffel formula does not completely describe the conformal mappings from to the disk, because it does not specify exactly what the phases and the complex constants are. As the group of automorphisms of has three degrees of freedom (one real parameter and one complex parameter ), one can for instance fix three of the phases , but in general there are no simple formulae to then reconstruct the remaining parameters in the Schwarz-Christoffel formula, although numerical algorithms exist to compute them approximately. (In the case when the polygon is a rectangle, though, the Schwarz-Christoffel formula essentially produces an elliptic integral, and the complex diffeomorphisms from the rectangle to the disk or half-space are closely tied to elliptic functions; see Section 4.5 of Stein-Shakarchi for more discussion.)
Exercise 46 (Schwarz-Christoffel in a half-space) Let be a closed simple anticlockwise polygonal path, and define the exterior angles to .
- (i) Show that extends to a homeomorphism from the closure of the upper half-plane in the Riemann sphere to , and that all lie on .
- (ii) If all of the are finite, show that after a cyclic permutation one has , and that there exists a non-zero complex number , and a complex number such that
for all , where the integral is over any curve from to .
- (iii) If one of the are infinite, show after a cyclic permutation that one has and , and there exists a non-zero complex number , and a complex number such that
for all .
Remark 47 One could try to apply the Schwarz-Christoffel formula to a closed polygonal path that is not simple. In such cases (and after choosing the parameters correctly), what tends to happen is that the map still maps the circle to the closed path, but fails to be injective.
Exercise 48 Let be a complex diffeomorphism from the half-strip to the upper half-plane , which extends to a continuous map to the closures of , in the Riemann sphere. Suppose that maps to respectively. Show that , where we take the branch of the square root that is positive on the real axis and has a branch cut at . (Hint: is not quite a polygon, so one cannot directly apply the Schwarz-Christoffel formula; however the proof of that formula will still apply.)
— 5. The uniformisation theorem (optional) —
Now we discuss a proof of the uniformisation theorem, Theorem 5, following the approach in these notes of Marshall. Unfortunately the argument is rather complicated, and we will only give a portion of the proof here. One of the many difficulties in trying to prove this theorem is the fact that the conclusion is a disjunction of three alternatives, each with a rather different complex geometry; it would be easier if there was only one target geometry that one was trying to impose on the Riemann surface . To begin separating the three geometries from each other, recall from Liouville’s theorem that there are no non-constant bounded holomorphic functions on or , but plenty of non-constant bounded holomorphic functions on . By Lemma 1, the same claims hold for Riemann surfaces that are complex diffeomorphic to or or to respectively. Note that without loss of generality we may normalise “bounded” by replacing it with “mapping into “. From this we see that the uniformisation theorem can be broken up into two simpler pieces:
Theorem 49 (Uniformisation theorem, hyperbolic case) Let be a simply connected Riemann surface that admits a non-constant holomorphic map from to . Then is complex diffeomorphic to .
Theorem 50 (Uniformisation theorem, non-hyperbolic case) Let be a simply connected Riemann surface that does not admit a non-constant holomorphic map from to . Then is complex diffeomorphic to or .
Let us now focus on the hyperbolic case of the uniformisation theorem, Theorem 49. Now we do not have the disjunction problem as there is only one target geometry to impose on ; we will be able to give a complete proof of this theorem here (in contrast to Theorem 50, where we will only give part of the proof). Let be a point in , and recall that denotes the collection of holomorphic maps that maps to . By hypothesis (and applying a suitable automorphism of ), contains at least one non-constant map. If Theorem 49 were true, then from Lemma 38 we see that would contain a “maximal” element which would exhibit the desired complex diffeomorphism between and .
It turns out that the converse statement is true: if we can locate “maximal” elements of with certain properties, then we can prove Theorem 49. More precisely, Theorem 49 can be readily deduced from the following claim.
Theorem 51 (Maximal maps into ) Let be a simply connected Riemann surface, let be a point in , and let be the collection of holomorphic maps from to that map to . Suppose that contains a non-constant map. Then contains a map with the property that for all , with equality only if for some real number . Furthermore has a simple zero at , and no other zeroes.
We have seen how Theorem 49 implies Theorem 51. Let us now demonstrate the converse implication, assuming Theorem 51 for the moment and deriving Theorem 49. Let be a simply connected Riemann surface that admits non-constant holomorphic maps from to , and pick a point in . By applying a suitable automorphism of we see that has a non-constant map, so by Theorem 51 this collection contains an element with the stated properties. If were injective, then we could apply Proposition 40 to conclude that and were complex diffeomorphic, so suppose for contradiction that was not injective. Since has a zero only at , we thus have for some distinct . Let be the automorphism
that maps to and to , then the function lies in and also has a zero at . From Theorem 51, we thus have
since vanishes, we thus have from the definition of that
Swapping the roles of and gives the reverse inequality, thus we in fact have
Applying Theorem 51 again, we conclude that
for some . But has a zero at while cannot have any zeroes other than at , a contradiction.
Remark 52 We only established that was injective in the above argument, but by inspecting the proof of Proposition 40 and using the maximality properties of we see that is also surjective, and thus supplies the required complex diffeomorphism between and .
To finish off the hyperbolic case of the uniformisation theorem, it remains to prove Theorem 51. It is convenient to work with harmonic functions instead of holomorphic functions. Observe that if were holomorphic with a simple zero at but no other zeroes, then we have local holomorphic branches of on small neighbourhoods of any point in . Taking real parts, we conclude that the function is harmonic on the punctured surface ; it is also positive since takes values in . Furthermore, the function has a logarithmic singularity at in the following sense: if was any coordinate chart on some neighbourhood of that mapped to , then as had a simple zero at , the function , defined on , stays bounded as one approaches .
Conversely, one can reconstruct from (up to a harmless phase ) by the following lemma.
Lemma 53 (Reconstructing a holomorphic function from its magnitude) Let be a simply connected Riemann surface, let be a point in , and let be harmonic. Suppose that has a logarithmic singularity at in the sense that is bounded near for some coordinate chart on a neighbourhood of that maps to . Then there exists a holomorphic function with a simple zero at and no other zeroes, such that on .
Proof: Let be as above. Call a function on an open subset of good if it is holomorphic with on (in particular this forces to be non-zero away from ), and has a simple zero at if lies in . Clearly it will suffice to find a good function on all of .
We first solve the local problem, showing that for any there exists a neighbourhood of that supports a good function . If , we can work in a chart avoiding which is diffeomorphic to a disk . If we identify with then restricted to can be viewed as a harmonic function on . As this disk is simply connected, will have a harmonic conjugate and is thus the real part of a holomorphic function on this disk. Taking to be we obtain the required good function. Now suppose instead that . Using the coordinate chart to identify with , we now have a harmonic function with bounded near zero. Applying Exercise 59 of Notes 4, we conclude that extends to a holomorphic function on , which is then the real part of a holomorphic function ; taking then gives a good function on .
Next, we make the following compatibility observation: if and are both good functions, then is constant on every connected component of (after removing any singularity at ). Indeed, by construction is holomorphic and of magnitude one, so locally there are holomorphic branches of that have vanishing real part, hence locally constant imaginary part by the Cauchy-Riemann equations. Hence is locally constant as claimed.
Now we need to glue together the local good functions into a global good functions. This is a “monodromy problem”, which can be solved using analytic continuation and the simply connected nature of by the following “monodromy theorem” argument. Let us pick a good function on some neighbourhood of . Given any other point in , we can form a path from to . We claim that for any , we can find a finite sequence and good functions for such that each contains , and such that and agree on a neighbourhood of for each , and and also agree on a neighbourhood of . The set of such is easily seen to be an open non-empty subset of . Now we claim that it is closed. Suppose that converges to a limit as . If any of the are greater than or equal to it is easy to see that , so suppose instead that the are all less than . We take a good function supported on some neighbourhood of . By continuity, will contain for some sufficiently large . We would like to append and to the sequence of good functions , one obtains from the hypothesis , but there is the issue that need not agree with at the endpoint . However, they only differ by a constant of magnitude near this endpoint, so after multiplying by an appropriate constant of magnitude one, we can conclude that as claimed.
By the continuity method, is all of , and in particular contains . Thus we can find and good functions for such that each contains , and such that and agree on a neighbourhood of for each , and and also agree on a neighbourhood of . Consider the final value obtained by the last good function at the endpoint of the curve . From analytic continuation and a continuity argument we see that if we perform a homotopy of with fixed endpoints, this final value does not change (even if the number of good functions may vary). Thus we can define a function by setting whenever is a path from to and is the final good function constructed by the above procedure. From construction we see that is locally equal to a good function at every point in , and is thus itself a good function, as required.
Exercise 54 (Monodromy theorem) Let be a simply connected Riemann surface, let be another Riemann surface, let be a point in , let be an open neighbourhood, and let be holomorphic. Prove that the following statements are equivalent.
- (i) has a holomorphic extension to ; that is to say, there is a holomorphic function whose restriction to is equal to .
- (ii) For every curve starting at , we can find and holomorphic functions for such that and agree on a neighbourhood for each , and and also agree on a neighbourhood of .
Furthermore, if (i) holds, show that the holomorphic extension is unique. Give a counterexample that shows that the monodromy theorem fails if is only assumed to be connected rather than simply connected.
We remark that while the condition (ii) in the monodromy theorem looks somewhat complicated, it becomes more geometrically natural if one adopts the language of sheaves, which we will not do here.
In view of Lemma 53, we may reduce the task of establishing Theorem 51 to that of establishing the existence of a special type of harmonic function on (with one point removed), namely a Green’s function:
Definition 55 (Green’s function) Let be a connected Riemann surface, and let be a point in . A Green’s function for at is a function with the following properties:
- (i) is harmonic on .
- (ii) is non-negative on .
- (iii) has a logarithmic singularity at in the sense that is bounded near for some coordinate chart that maps to .
- (iv) is minimal with respect to the properties (i)-(iii), in the sense that for any other obeying (i)-(iii), we have pointwise in .
Clearly if a Green’s function for at exists, it is unique by property (iv), so we can talk about the Green’s function for at , if it exists. In the case of the disk , a Greens’ function may be explicitly computed:
Exercise 56 If , show that the function defined by is a Green’s function for at .
Theorem 51 may now be deduced from the following claim.
Proposition 57 (Existence of Green’s function) Let be a connected Riemann surface, let be a point in , and suppose that the collection of holomorphic maps that map to contains at least one non-constant map. Then the Green’s function for at exists. Furthermore, for any , one has for any .
(Note that in this proposition we no longer need to be simply connected.) Indeed, suppose that Proposition 57 held. Let be a simply connected Riemann surface, and let with containing a non-constant map. By hypothesis, the Green’s function is non-negative on . Noting that remains connected if we remove a small disk around , and from (iii) that will be strictly positive on the boundary of that disk, we observe from the maximum principle (Exercise 18) and (ii) that is in fact strictly positive on . By Lemma 53 we can find a holomorphic function with a simple zero at and no other zeroes, such that on . As is strictly positive, takes values in and is thus in . From Proposition 57 we see that for all . If equality occurs anywhere, then the quotient (after removing the singularity) is a function taking values in the closed unit disk , which has magnitude at ; by the maximum principle we then have for some real . Thus obeys all the properties required for Theorem 51.
It remains to obtain the existence of the Green’s function . To do this, we use a powerful technique for constructing harmonic functions, known as Perron’s method of subharmonic functions. The basic idea is to build a harmonic function by taking a suitable large family of subharmonic functions and then forming their supremum. We first give a definition of subharmonic function.
Definition 58 (Subharmonic function) Let be a Riemann surface. A subharmonic function on is an upper semi-continuous function obeying the following upper maximum principle: for any compact set in and any function that is continuous on and harmonic on the interior of , if for all , then for all .
A superharmonic function is similarly defined as a lower semi-continuous function such that for any compact and any function continuous on and harmonic on the interior of , the bound for implies that for all .
Clearly subharmonicity and superharmonicity are conformal invariants in the sense that the analogue of Lemma 1 holds for these concepts. We have the following elementary properties of subharmonic functions and superharmonic functions:
Exercise 59 Let be a Riemann surface.
- (i) Show that a function is subharmonic if and only if is superharmonic.
- (ii) Show that a function is harmonic if and only if it is both subharmonic and superharmonic.
- (iii) If are subharmonic, show that is also.
- (iv) Let , and let be an open subset of . Show that the restriction of to is subharmonic.
- (v) (Subharmonicity is a local property) Conversely, let , and suppose that for each there is a neighbourhood of such that the restriction of to is subharmonic. Show that is itself subharmonic. (Hint: If is continuous on a compact set and harmonic on the interior, and attains a maximum at an interior point of , show that is constant in some neighbourhood of that point.)
- (vi) (Maximum principle) Let be subharmonic, let be superharmonic, and let be a compact subset of such that for all . Show that for all . (This is a similar argument to (v).)
- (vii) Show that the sum of two subharmonic functions is again subharmonic (using the usual conventions on adding to itself or to another real number).
- (viii) (Harmonic patching) Let be subharmonic, let be compact, and let be a continuous function on that is harmonic on the interior of and agrees with on the boundary of . Show that the function , defined to equal on and on , is subharmonic.
- (ix) Let be a holomorphic function. Show that is subharmonic, with the convention that . (Hint: first use the maximum principle and harmonic conjugates to show that if contains a copy of a closed disk , and on the boundary of this disk for some continuous that is harmonic in the interior of the disk, then in the interior of the disk also.)
For smooth functions on an open subset of , one can express the property of subharmonicity quite explicitly:
Exercise 60 Let be an open subset of , and let be continuously twice (Fréchet) differentiable. Show that the following are equivalent:
- (i) is subharmonic.
- (ii) For all closed disks in , one has
- (iii) One has for all .
Show that the equivalence of (i) and (ii) in fact holds even if is only assumed to be continuous rather than continuously twice differentiable.
However, we will not use the above exercise in our analysis here as it will not be convenient to impose a hypothesis of continuous twice differentiability on our subharmonic functions.
The Perron method is based on the observation that under certain conditions, the supremum of a family of subharmonic functions is not just subharmonic (as per Exercise 59(iii)), but is in fact harmonic. A key concept here is that of a Perron family:
Definition 61 Let be a Riemann surface. A Perron family on is a family of subharmonic functions with the following properties:
- (i) If , then .
- (ii) (Harmonic patching) If , is a compact subset of , and is a continuous function that is harmonic in the interior of and equals on the boundary of , then the function defined to equal on and outside of also lies in .
- (iii) For every , there exists with .
The fundamental theorem that powers the Perron method is then
Theorem 62 (Perron method) Let be a Perron family on a connected Riemann surface , and set to be the function (note that cannot equal thanks to axiom (iii) of a Perron family). Then one of the following two statements hold:
- (i) for all .
- (ii) is a harmonic function on .
Proof: Let us first work locally in some open subset of that is complex diffeomorphic to a disk ; to simplify the discussion we abuse notation by identifying with in the following discussion.
Assume for the moment that is not identically equal to on . Let be an arbitrary point in (viewed as a subset of ). Then we can find a sequence such that as .
We can use Exercise 42 find a harmonic function that equals on the boundary of this disk (viewed as a subset of ); if we then let be the function defined to equal on and outside of this disk, then is larger than and also lies in thanks to axiom (ii). Thus, by replacing with , we may assume that is harmonic on . Next, by replacing with and using axiom (i), we may assume that pointwise; replacing with a harmonic function on as before we may assume that is harmonic on . Continuing in this fashion we may assume that and that are harmonic on . Form the function , then we have pointwise with . By the Harnack principle (Exercise 58 of Notes 4), we thus see that is either harmonic on , or equal to on . The latter cannot occur since we are assuming not identically equal to , thus is harmonic.
Now let be another point in . We can find another sequence with . As before we may assume that the are increasing and are harmonic on ; we may also assume that pointwise. Setting , we conclude that is harmonic with on . In particular . The harmonic function is non-negative on and vanishes at , hence is identically zero on by the maximum principle. Since , we conclude that and agree at . Since was an arbitrary point on , we conclude that is harmonic at .
Putting all this together, we see that for any point in there is a neighbourhood (corresponding to the disk in the above arguments) with the property that is either equal to on , or is harmonic on . By a continuity argument we conclude that one of the two options (i), (ii) of the theorem must hold.
Now we can conclude the proof of Proposition 57, and hence the hyperbolic case of the uniformisation theorem, by applying the above theorem to a well-chosen Perron family. Let be a simply connected Riemann surface, and let be the collection of all continuous subharmonic functions that vanishes outside of a compact subset of , and which have a logarithmic singularity at in the sense that is bounded near for some coordinate chart that takes to (note that the precise choice of chart here is irrelevant). This collection is non-empty, for it contains the function that equals (say) on , and zero elsewhere (this follows from the observation that is harmonic away from the origin, and is harmonic everywhere, as well as the various properties in Exercise 59). From Exercise 59 we see that is a Perron family; thus, by Theorem 62, the function is either harmonic on , or is infinite everywhere. Using the element of used above we see that is non-negative.
Let be an arbitrary element of . By Exercise 59(ix), is subharmonic, hence is superharmonic and also non-negative since takes values in ; as vanishes at , has at least a logarithmic singularity at in the sense that is bounded from below near . If , then vanishes outside of a compact set , hence outside of for any . As has a logarithmic singularity at we also have in a sufficiently small neighbourhood of . Appying the maximum principle (Exercise 59(vi)) we conclude that on all of ; sending to zero and then taking suprema in we conclude that
or equivalently
pointwise on . In particular, since contains at least one non-constant map, cannot be infinite everywhere and must therefore be harmonic.
Similarly, if is a function obeying the properties (i)-(iii) of a Green’s function, and , then another application of the maximum principle shows that on for any ; sending and taking suprema in we see that pointwise.
The only remaining task to show is that has a logarithmic singularity at . Certainly it has at least this much of a singularity, in that is bounded from below near , as can be seen by comparing to any element of . To get the upper bound, observe that for any and , the function is subharmonic on and diverges to at , and is hence in fact subharmonic on all of . In particular, for in the disk , we have from the maximum principle that
and hence on taking suprema in and limits in
The right-hand side is finite, and this gives the required upper bound to complete the proof that has a logarithmic singularity at . This concludes the proof of Proposition 57 and hence Theorem 49.
Before we turn to the non-hyperbolic case of the uniformisation theorem, we record a symmetry property of the Green’s functions that is used to establish that case:
Proposition 63 (Symmetry of Green’s functions) Let be a connected Riemann surface, and suppose that the Green’s functions exist for all . Then for all distinct , we have .
When is simply connected, this symmetry can be deduced from (7). For that are not simply connected, the argument is trickier, requiring one to pass to a universal cover of , establish the existence of Green’s functions on , and find an identity relating the Green’s functions on with the Green’s functions on . For details see Marshall’s notes.
Now we can discuss to the non-hyperbolic case of the uniformisation theorem, Theorem 50. Now we do not have any Green’s functions, or any non-constant bounded holomorphic functions. However, note that all three of the model Riemann surfaces , and still have plenty of meromorphic functions: in particular, for any two distinct points in , one can find a holomorphic function that has a simple zero at , a simple pole at , and no other zeroes and poles, namely ; one can think of this function with a zero-pole pair as a “dipole“. Similarly if one works on the domain or rather than . From this we see that Theorem 50 would imply the following claim:
Theorem 64 (Existence of dipoles) Let be a simply connected Riemann surface. Let be distinct points in . Then there exists a holomorphic map that has a simple zero at , a simple pole at , and no other zeroes and poles. Furthermore, outside of a compact set containing , the function can be chosen to be bounded away from both and (that is, there exists such that for all ).
In the converse direction, we can use Theorem 64 to recover Theorem 50 in a manner analogous to how Theorem 51 implies Theorem 49. Indeed, let be a simply connected Riemann surface without non-constant holomorphic maps from to . Given any three distinct points in , we consider the dipoles and . The function
has removable singularities at and at , no poles, and is also bounded away from a compact set. Thus this function extends to a bounded holomorphic function on . Since does not have any non-constant bounded holomorphic functions, the function (8) must be constant, thus for some complex numbers ; as is non-constant, must be non-zero. Since vanishes only at , we conclude that for any . Since also has its only zero at and its only pole at , we conclude that is injective. By Exercise 40 of Notes 4, is thus a complex diffeomorphism from to an open subset of , which of course is simply connected since is. If is all of then we are in the elliptic case and we are done. If omits at least one point in then by applying a Möbius transform is complex diffeomorphic to a simply connected open subset of ; by the Riemann mapping theorem, we conclude that is either complex diffeomorphic to or to . The latter case cannot occur by hypothesis, and we are done.
It remains to prove Theorem 64. As before, we convert the problem to one of finding a specific harmonic function. More precisely, one can derive Theorem 64 from
Theorem 65 (Existence of dipole Green’s functions) Let be a connected Riemann surface. Let be distinct points in , and let and be coordinate charts on disjoint neighbourhoods of respectively, which map and respectively to . Then there exists a harmonic function such that is bounded near , and is bounded near . Furthermore, is bounded outside of a compact subset of .
In the case , one can take the dipole Green’s function to be the function for an arbitrary constant .
Exercise 66 Adapt the proof of Lemma 53 to show that Theorem 65 implies Theorem 64 (and hence Theorem 50).
We still need to prove Theorem 65. If admitted Green’s functions for every point , we could simply take to be the difference . Unfortunately, as we are in the non-hyperbolic case, is not expected to have Green’s functions, and it does not appear possible to construct the dipole Green’s functions directly from Perron’s method due to the indefinite sign of these functions. However, it turns out that if one removes a small disk from of some small radius in a given coordinate chart, then the resulting Riemann surface will admit Green’s functions , and by considering limits of the sequence as using a version of Montel’s theorem one will be able to obtain the required dipole Green’s function, after first making heavy use of the maximum principle (and an important variant of that principle known as Harnack’s inequality, see Exercise 68 below) to obtain some locally uniform control on the difference in . To obtain this locally uniform control, the symmetry property in (63) is key, as it allows one to write
so that the main challenge is to show that the differences and are bounded uniformly in , which can be done from the maximum principle and the Harnack inequality. The details are unfortunately a little complicated, and we refer the reader to Marshall’s notes for the complete argument.
To close this section we give a quick corollary to the uniformisation theorem, namely Rado’s theorem on the topology of Riemann surfaces:
Corollary 67 (Rado’s theorem) Every connected Riemann surface is second countable and separable.
Proof: By passing to the universal cover, it suffices to verify this claim for simply connected Riemann surfaces. But the three model surfaces , , are clearly second countable and separable, so the claim follows from the uniformisation theorem.
It is remarkably difficult to prove this theorem directly, without going through the uniformisation theorem. (As just one indication of the difficulty of this theorem, the analogue of Rado’s theorem for complex manifolds in two and higher dimensions is known to be false.)
Exercise 68 (Harnack inequality) Let be a non-negative continuous function on a closed disk that is harmonic on the interior of the disk. Show that for every and , one has
(Hint: use Exercise 42.)
has very rapidly decaying coefficients (of order ), leading to an infinite radius of convergence; also, as the series converges to , the series decays very rapidly as approaches . The problem is whether this is essentially the only example of this type. More precisely:
Problem 1 Let be a bounded sequence of real numbers, and suppose that the power series
(which has an infinite radius of convergence) decays like as , in the sense that the function remains bounded as . Must the sequence be of the form for some constant ?
As it turns out, the problem has a very nice solution using complex analysis methods, which by coincidence I happen to be teaching right now. I am therefore posing as a challenge to my complex analysis students and to other readers of this blog to answer the above problem by complex methods; feel free to post solutions in the comments below (and in particular, if you don’t want to be spoiled, you should probably refrain from reading the comments). In fact, the only way I know how to solve this problem currently is by complex methods; I would be interested in seeing a purely real-variable solution that is not simply a thinly disguised version of a complex-variable argument.
(To be fair to my students, the complex variable argument does require one additional tool that is not directly covered in my notes. That tool can be found here.)
Singularities come in varying levels of “badness” in complex analysis. The least harmful type of singularity is the removable singularity – a point which is an isolated singularity (i.e., an isolated point of the singular set ) where the function is undefined, but for which one can extend the function across the singularity in such a fashion that the function becomes holomorphic in a neighbourhood of the singularity. A typical example is that of the complex sinc function , which has a removable singularity at the origin , which can be removed by declaring the sinc function to equal at . The detection of isolated removable singularities can be accomplished by Riemann’s theorem on removable singularities (Exercise 35 from Notes 3): if a holomorphic function is bounded near an isolated singularity , then the singularity at may be removed.
After removable singularities, the mildest form of singularity one can encounter is that of a pole – an isolated singularity such that can be factored as for some (known as the order of the pole), where has a removable singularity at (and is non-zero at once the singularity is removed). Such functions have already made a frequent appearance in previous notes, particularly the case of simple poles when . The behaviour near of function with a pole of order is well understood: for instance, goes to infinity as approaches (at a rate comparable to ). These singularities are not, strictly speaking, removable; but if one compactifies the range of the holomorphic function to a slightly larger space known as the Riemann sphere, then the singularity can be removed. In particular, functions which only have isolated singularities that are either poles or removable can be extended to holomorphic functions to the Riemann sphere. Such functions are known as meromorphic functions, and are nearly as well-behaved as holomorphic functions in many ways. In fact, in one key respect, the family of meromorphic functions is better: the meromorphic functions on turn out to form a field, in particular the quotient of two meromorphic functions is again meromorphic (if the denominator is not identically zero).
Unfortunately, there are isolated singularities that are neither removable or poles, and are known as essential singularities. A typical example is the function , which turns out to have an essential singularity at . The behaviour of such essential singularities is quite wild; we will show here the Casorati-Weierstrass theorem, which shows that the image of near the essential singularity is dense in the complex plane, as well as the more difficult great Picard theorem which asserts that in fact the image can omit at most one point in the complex plane. Nevertheless, around any isolated singularity (even the essential ones) , it is possible to expand as a variant of a Taylor series known as a Laurent series . The coefficient of this series is particularly important for contour integration purposes, and is known as the residue of at the isolated singularity . These residues play a central role in a common generalisation of Cauchy’s theorem and the Cauchy integral formula known as the residue theorem, which is a particularly useful tool for computing (or at least transforming) contour integrals of meromorphic functions, and has proven to be a particularly popular technique to use in analytic number theory. Within complex analysis, one important consequence of the residue theorem is the argument principle, which gives a topological (and analytical) way to control the zeroes and poles of a meromorphic function.
Finally, there are the non-isolated singularities. Little can be said about these singularities in general (for instance, the residue theorem does not directly apply in the presence of such singularities), but certain types of non-isolated singularities are still relatively easy to understand. One particularly common example of such non-isolated singularity arises when trying to invert a non-injective function, such as the complex exponential or a power function , leading to branches of multivalued functions such as the complex logarithm or the root function respectively. Such branches will typically have a non-isolated singularity along a branch cut; this branch cut can be moved around the complex domain by switching from one branch to another, but usually cannot be eliminated entirely, unless one is willing to lift up the domain to a more general type of domain known as a Riemann surface. As such, one can view branch cuts as being an “artificial” form of singularity, being an artefact of a choice of local coordinates of a Riemann surface, rather than reflecting any intrinsic singularity of the function itself. The further study of Riemann surfaces is an important topic in complex analysis (as well as the related fields of complex geometry and algebraic geometry), but unfortunately this topic will probably be postponed to the next course in this sequence (which I will not be teaching).
— 1. Laurent series —
Suppose we are given a holomorphic function and a point in . For a sufficiently small radius , the circle and its interior both lie in , and the Cauchy integral formula tells us that
in the interior of this circle. In Corollary 18 of Notes 3, this was used to form a convergent Taylor series expansion
in the interior of this circle, where the coefficients could be reconstructed from the values of on the circle by the formula
Now suppose that is only known to be holomorphic outside of . Then the Cauchy integral formula no longer directly applies, because the interiors of contours such as are no longer contained in the region where is holomorphic. To deal with this issue, we use the following convenient decomposition.
Lemma 1 (Cauchy integral formula decomposition in annular regions) Let be a holomorphic function. Let , be simple closed anticlockwise contours in such that is contained in the interior of (or equivalently, by Exercise 49 of Notes 3, that is contained in the exterior of ). Suppose also that the “annular region” is contained in . Then there exists a decomposition
on , where is holomorphic on the union of and the interior of , and is holomorphic on the union of and the exterior of , with as . Furthermore, if is connected, then this decomposition is unique.
In addition, we have the Cauchy integral type formulae
for in the interior of , and
Proof: We begin with uniqueness. Suppose we have two decompositions
on , where and holomorphic, and both going to zero at infinity. Then the holomorphic functions and agree on the common domain , and are hence restrictions of a single entire function . But goes to zero at infinity and is hence bounded; applying Liouville’s theorem (Theorem 28 of Notes 3) we see that vanishes entirely. This gives and on the non-empty open set , and then we have and on and by analytic continuation (Corollary 23 of Notes 3).
Now for existence. Suppose that we can establish the identity (1) for in . Then we can define on by
and on by
noting from (1) that this consistently defines on
From Exercise 36 of Notes 3 we see that is holomorphic. Similarly if we define on by
and on by
One can then verify that obey all the required properties.
Thus it remains to establish (1). This follows from the homology form of the Cauchy integral formula (Exercise 63(v) of Notes 3), but we can also avoid explicit use of homology by the following “keyhole contour” argument. For , we have
and
and so to prove (1), it suffices to show that
By the factor theorem (Corollary 22 of Notes 3) it thus suffices to show that
By perturbing using Cauchy’s theorem we may assume that these curves are simple closed polygonal paths (if one wishes, one can also restrict the edges to be horizontal and vertical, although this is not strictly necessary for the argument). By connecting a point in to a point in by a polygonal path in the interior of , and removing loops, self-intersections, or excursions into the interior (or image) of , we can find a simple polygonal path from a point in to a point in that lies entirely in except at the endpoints. By rearranging and we may assume that is the initial and terminal point of , and is the initial and terminal point of . Then the closed polygonal path has vanishing winding number in the interior of or exterior of , thus contains all the points where the winding number is non-zero. This path is not simple, but we can approximate it to arbitrary accuracy by a simple closed polygonal path by shifting the simple polygonal paths and slightly; for small enough, the interior of will then lie in . Applying Cauchy’s theorem (Theorem 52 of Notes 3) we conclude that
taking limits as we obtain (2) as claimed.
Exercise 2 Let be a simple closed anticlockwise contour, and let be simple closed anticlockwise contours in the interior of whose images are disjoint, and such that the interiors are also disjoint. Let be an open set containing and the region
Show that for any , one has
(Hint: induct on using Lemma 1.)
Exercise 3 (Painlevé’s theorem on removable singularities) Let be an open subset of . Let be a compact subset of which has zero length in the following sense: for any , one can cover by a countable number of disks such that . Let be a bounded holomorphic function. Show that the singularities in are removable in the sense that there is an extension of to which remains holomorphic. (Hint: one can work locally in some disk in that contains a portion of . Cover this portion by a finite number of small disks, group them into connected components, use the previous exercise, and take an appropriate limit.) Note that this result generalises Riemann’s theorem on removable singularities, see Exercise 35 from Notes 3. The situation when has positive length is considerably more subtle, and leads to the theory of analytic capacity, which we will not discuss further here.
Now suppose that is holomorphic for some open set that contains an annulus of the form
for some and . From Lemma 1, we can split , where is holomorphic in , and is holomorphic in the exterior region , with going to zero as . From Corollary 18 of Notes 3, one has a Taylor expansion
for some coefficients that is absolutely convergent in the disk . One cannot directly apply this Taylor expansion to . However, observe that the function is holomorphic in the punctured disk , and goes to zero as one approaches zero. By Riemann’s theorem (Exercise 35 from Notes 3), this function may be extended to to a holomorphic function that vanishes at the origin. Applying Corollary 18 of Notes 3 again, we conclude that there is a Taylor expansion
for some coefficients that is absolutely convergent in the punctured disk . Changing variables, we conclude that
for all in (3), with the doubly infinite series on the right-hand side being absolutely convergent. This series is known as the Laurent series in the annulus (3). The coefficients may be explicitly computed in terms of :
Exercise 4 (Fourier inversion formula) Let be holomorphic on some open set that contains an annulus of the form (3), and let be the coefficients of the Laurent expansion (4) in this annulus. Show that the coefficients are uniquely determined by and , and are given by the formula
for all integers , whenever is a simple closed curve in the annulus with . Also establish the bounds
The following modification of the above exercise may help explain the terminology “Fourier inversion formula”.
Exercise 5 (Fourier inversion formula, again) Let .
- (i) Show that if is holomorphic on the annulus , then we have the Fourier expansion
for all , where the Fourier coefficients are given by the formula
Furthermore, show that the Fourier series in (7) is absolutely convergent, and the coefficients obey the asymptotic bounds (5), (6).
- (ii) Conversely, if are complex numbers obeying the asymptotic bounds (5), (6), show that there exists a function holomorphic on the annulus obeying the Fourier expansion (7) and the inversion formula (8).
The Laurent series for a given function can vary as one varies the annulus. Consider for instance the function . In the annulus , the Laurent expansion coincides with the Taylor expansion:
On the other hand, in the exterior region , the Taylor expansion is no longer convergent. Instead, if one writes and uses the geometric series formula, one instead has the Laurent expansion
in this region.
Exercise 6 Find the Laurent expansions for the function in the regions , . (Hint: use partial fractions.)
We can use Laurent series to analyse an isolated singularity. Suppose that is holomorphic on a punctured disk . By the above discussion, we have a Laurent series expansion (4) in this punctured disk. If the singularity is removable, then the Laurent series must coincide with the Taylor series (by the uniqueness component of Exercise 4), so in partcular for all negative ; conversely, if vanishes for all negative , then the Laurent series matches up with a convergent Taylor series and so the singularity is removable. We then adopt the following classification:
It is clear that any holomorphic function will be of exactly one of the above three categories. Also, from the uniqueness of Laurent series, shrinking does not affect which of the three categories will lie in (or what order of pole will have, in the second category). Thus, we can classify any isolated singularity of a holomorphic function with singularities as being either removable, a pole of some finite order, or an essential singularity by restricting to a small punctured disk and inspecting the Laurent coefficients for negative .
Example 7 The function has a Laurent expansion
and thus has an essential singularity at .
It is clear from the definition (and the holomorphicity of Taylor series) that (as discussed in the introduction), a holomorphic function has a pole of order at an isolated singularity if and only if it is of the form for some holomorphic with . Similarly, a holomorphic function would have a zero of order at if and only if for some with .
We can now define a class of functions that only have “nice” singularities:
Definition 8 (Meromorphic functions) Let be an open subset of . A function defined on outside of a singular set is said to be meromorphic on if
- (i) is closed and discrete (i.e., all points in are isolated); and
- (ii) Every is either a removable singularity or a pole of finite order.
Two meromorphic functions , are said to be equivalent if they agree on their common domain of definition . It is easy to see that this is an equivalence relation. It is common to identify meromorphic functions up to equivalence, similarly to how in measure theory it is common to identify functions which agree almost everywhere.
Exercise 9 (Meromorphic functions form a field) Let denote the space of meromorphic functions on a connected open set , up to equivalence. Show that is a field (with the obvious field operations). What happens if is not connected?
Exercise 10 (Order is a valuation) If is a meromorphic function, and , define the order of at as follows:
- (a) If has a removable singularity at , and has a zero of order at once the singularity is removed, then .
- (b) If is holomorphic at , and has a zero of order at , then .
- (c) If has a pole of order at , then .
- (d) If is identically zero, then .
Establish the following facts:
- (i) If and are equivalent meromorphic functions, then for all . In particular, one can meaningfully define the order of an element of at any point in , where is as in the preceding exercise.
- (ii) If and , show that . If is not zero, show that .
- (iii) If and , show that . Furthermore, show if , then the above inequality is in fact an equality.
In the language of abstract algebra, the above facts are asserting that is a valuation on the field .
The behaviour of a holomorphic function near an isolated singularity depends on the type of singularity.
Theorem 11 Let be holomorphic on an open set outside of a singular set , and let be an isolated singularity in .
- (i) If is a removable singularity of , then converges to a finite limit as .
- (ii) If is a pole of , then as .
- (iii) (Casorati-Weierstrass theorem) If is an essential singularity of , then every point of is a limit point of as , that is to say there exists a sequence converging to such that converges to (where we adopt the convention that converges to if converges to ).
Proof: Part (i) is obvious. Part (ii) is immediate from the factorisation and noting that converges to the non-zero value as . The case of (iii) follows from Riemann’s theorem on removable singularities (Exercise 35 from Notes 3). Now suppose is finite. If (iii) failed, then there exist such that avoids the disk on the domain . In particular, the function is bounded and holomorphic on , and thus extends holomorphically to by Riemann’s theorem. This function cannot vanish identically, so we must have on for some and some holomorphic that does not vanish at . Rearranging this as , we see that has a pole or removable singularity at , a contradiction.
In Theorem 56 below we will establish a significant strengthening of the Casorati-Weierstrass theorem known as the Great Picard Theorem.
Exercise 12 Let be holomorphic in outside of a discrete set of singularities. Let . Show that the radius of convergence of the Taylor series of around is equal to the distance from to the nearest non-removable singularity in , or if no such non-removable singularity exists. (This fact provides a neat way to understand the rate of growth of a sequence : form its generating function , locate the singularities of that function, and find out how close they get to the origin. This is a simple example of the methods of analytic combinatorics in action.)
A curious feature of the singularities in complex analysis is that the order of singularity is “quantised”: one can have a pole of order , , or (for instance), but not a pole of order or . This quantisation can be exploited: if for instance one somehow knows that the order of the pole is less than for some integer and real number , then the singularity must be removable or a pole of order at most . The following exercise formalises this assertion:
Exercise 13 Let be holomorphic on a disk except for a singularity at . Let be an integer, and suppose that there exist , such that one has the upper bound
for all . Show that the singularity of at is either removable, or a pole of order at most (the latter option is only possible when is positive). (Hint: use Lemma 4 and a limiting argument to evaluate the Laurent coefficients for .) In particular, if one has
for all , then the singularity is removable.
As mentioned in the introduction, the theory of meromorphic functions becomes cleaner if one replaces the complex plane with the Riemann sphere. This sphere is a model example of a Riemann surface, and we will now digress to briefly introduce this more general concept (though we will not develop the general theory of Riemann surfaces in any depth here). To motivate the definition, let us first recall from differential geometry the notion of a smooth -dimensional manifold (over the reals).
Definition 14 (Smooth manifold) Let , and let be a topological space. An (-dimensional real) atlas for is an open cover of together with a family of homeomorphisms (known as coordinate charts) from each to an open subset of . Furthermore, the atlas is said to be smooth if for any , the transition map , which maps one open subset of to another, is required to be smooth (i.e., infinitely differentiable). A map from one topological space (equipped with a smooth atlas of coordinate charts for ) to another (equipped with a smooth atlas of coordinate charts for some ) is said to be smooth if, for any and , the maps are smooth; if is invertible and and are both smooth, we say that is a diffeomorphism, and that and are diffeomorphic. Two smooth atlases on are said to be equivalent if the identity map from (equipped with one of the two atlases) to (equipped with the other atlas) is a diffeomorphism; this is easily seen to be an equivalence relation, and an equivalence class of such atlases is called a smooth structure on . A smooth -dimensional real manifold is a Hausdorff topological space equipped with a smooth structure. (In some texts the mild additional condition of second countability on is also imposed.) A map between two smooth manifolds is said to be smooth, if the map from (equipped with one of the atlases in the smooth structure on ) to (equipped with one of the atlases in the smooth structure on ) is smooth; it is easy to see that this definition is independent of the choices of atlas. We may similarly define the notion of a diffeomorphism between two smooth manifolds.
This definition may seem excessively complicated, but it captures the modern geometric philosophy that one should strive as much as possible to work with objects that are coordinate-independent in that they do not depend on which atlas of coordinate charts one picks within the equivalence class of the given smooth structure in order to perform computations or to define foundational concepts. One can also define smooth manifolds more abstractly, without explicit reference to atlases, by working instead with the structure sheaf of the rings of smooth real-valued functions on open subsets of the manifold , but we will not need to do so here.
Example 15 A simple example of a smooth -dimensional manifold is the unit circle ; there are many equivalent atlases one could place on this circle to define the smooth structure, but one example would be the atlas consisting of the two charts , , defined by setting , , , , for , and for . Another smooth manifold, which turns out to be diffeomorphic to the unit circle , is the one-point compactification of the real numbers, with the two charts , defined by setting , , , to be the identity map, and defined by setting for and .
Exercise 16 Verify that the unit circle is indeed diffeomorphic to the one-point compactification .
A Riemann surface is defined similarly to a smooth manifold, except that the dimension is restricted to be one, the reals are replaced with the complex numbers, and the requirement of smoothness is replaced with holomorphicity (thus Riemann surfaces are to the complex numbers as smooth curves are to the real numbers). More precisely:
Definition 17 (Riemann surface) Let be a Hausdorff topological space. A holomorphic atlas on is an open cover of together with a family of homeomorphisms (known as coordinate charts) from each to an open subset of , such that, for any , the transition map , which maps one open subset of to another, is required to be holomorphic. A map from one space (equipped with coordinate charts for ) to another (equipped with coordinate charts for some ) is said to be holomorphic if, for any and , the maps are holomorphic; if is invertible and and are both holomorphic, we say that is a complex diffeomorphism, and that and are complex diffeomorphic. Two holomorphic atlases on are said to be equivalent if the identity map from (equipped with one of the atlases) to (equipped with the other atlas) is a complex diffeomorphism; this is easily seem to be an equivalence relation, and we refer to such an equivalence class as a (one-dimensional) complex structure on . A Riemann surface is a Hausdorff topological space , equipped with a one-dimensional complex structure. (Again, in some texts the hypothesis of second countability is imposed. This makes little difference in practice, as most Riemann surfaces one actually encounters will be second countable.)
By considering dimensions greater than one, one can arrive at the more general notion of a complex manifold, the study of which is the focus of complex geometry (and also plays a central role in the closely related fields of several complex variables and complex algebraic geometry). However, we will not need to deal with higher-dimensional complex manifolds in this course. The notion of a Riemann surface should not be confused with that of a Riemannian manifold, which is the topic of study of Riemannian geometry rather than complex geometry.
Clearly any open subset of the complex numbers is a Riemann surface, in which one can use the atlas that only consists of one “tautological” chart, the identity map . More generally, any open subset of a Riemann surface is again a Riemann surface. If are open subsets of the complex numbers, and is a map, then by unpacking all the definitions we see that is holomorphic in the sense of Definition 17 if and only if it is holomorphic in the usual sense.
Now we come to the Riemann sphere , which is to the complex numbers as is to the real numbers. As a set, this is the complex numbers with one additional point (the point at infinity) attached. Topologically, this is the one-point compactification of the complex numbers : the open sets of are either subsets of that were already open, or complements of compact subsets of . As a Riemann surface, the complex structure can be described by the atlas of coordinate charts , , where , , , is the identity map, and equals for with . It is not difficult to verify that this is indeed a Riemann surface (basically because the map is holomorphic on ). One can identify the Riemann sphere with a geometric sphere, and specifically the sphere , through the device of stereographic projection through the north pole , identifying a point in with the point on collinear with that point, and the point at infinity identified with the north pole . This geometric perspective is especially helpful when thinking about Möbius transformations, as is for instance exemplified by this excellent video. (We may cover Möbius transformations in a subsequent set of notes.)
By unpacking the definitions, we can now work out what it means for a function to be holomorphic to or from the Riemann sphere. For instance, if is a map from an open subset of to the Riemann sphere , then is holomorphic if and only if
Similarly, if a function is a map from an open subset of the Riemann sphere to the Riemann sphere, then is holomorphic if and only if
We can then identify meromorphic functions with holomorphic functions on the Riemann sphere:
Exercise 18 Let be open, let be a discrete subset of , and let be a function. Show that the following are equivalent:
- (i) is meromorphic on .
- (ii) is the restriction of a holomorphic function to the Riemann sphere.
Furthermore, if (ii) holds, show that is uniquely determined by , and is unaffected if one replaces with an equivalent meromorphic function.
Among other things, this exercise implies that the composition of two meromorphic functions is again meromorphic (outside of where the composition is undefined, of course).
Exercise 19 Let be a holomorphic map from the Riemann sphere to itself. Show that is a rational function in the sense that there exist polynomials of one complex variable, with not identically zero, such that for all with . (Hint: show that has finitely many poles, and eliminate them by multiplying by appropriate linear factors. Then use Exercise 29 from Notes 3.)
Exercise 20 (Partial fractions) Let be a polynomial of one complex variable, which by the fundamental theorem of algebra we may write as
for some distinct roots , some non-zero , and some positive integers . Let be another polynomial of one complex polynomial. Show that there exist unique polynomials , with each having degree less than for , such that one has the partial fraction decomposition
for all . Furthermore, show that vanishes if the degree of is less than the degree of , and has degree otherwise.
— 2. The residue theorem —
Now we can prove a significant generalisation of the Cauchy theorem and Cauchy integral formula, known as the residue theorem.
Suppose one has a function holomorphic on an open set outside of a singular set . If is an isolated singularity of , then we have a Laurent expansion
which is convergent in some punctured disk . The coefficient plays a privileged role and is known as the residue of at ; we denote it by . Clearly this is quantity is local in the sense that it only depends on the behaviour of in a neighbourhood of ; in particular, it does not depend on the domain so long as remains inside of that domain. By convention, we also set if is holomorphic at (i.e., if ).
We then have
Theorem 21 (Residue theorem) Let be a simply connected open set, and let be holomorphic outside of a closed discrete singular set (thus all singularities in are isolated singularities). Let be a closed curve in . Then
where only finitely many of the terms on the right-hand side are non-zero.
Proof: The image of is contained in some large ball; restricting and to this ball, we may assume without loss of generality that is both discrete and compact, and thus finite (by the Bolzano-Weierstrass theorem).
Next, we reduce to the case where all the residues vanish. We introduce the rational function defined by
From Laurent expansion around each singularity we see that for all , thus . Also, from the definition of winding number (see Definition 38 of Notes 3) we have
Setting , it thus suffices to show that
As is simply connected, is homotopic in (as closed contours) to a point. Let denote the homotopy. We would like to mimic the proof of Cauchy’s theorem (Theorem 4 of Notes 3) to conclude (9). The difficulty is that the homotopy may pass through points in . However, note from the vanishing of the residue that one has a Laurent expansion of the form
for some coefficients , in some punctured disk , with both series being absolutely convergent in this punctured disk. From term by term differentiation (see Theorem 15 of Notes 1) we see that has an antiderivative in this punctured disk, namely
(note how crucial it is that the term is absent in order to form this antiderivative). The absolute convergence of the series on the right-hand side in can be seen from the comparison test. From the fundamental theorem of calculus, we thus conclude that is conservative on . Also, for any that is not in , we see from Cauchy’s theorem that is conservative on for some radius . Putting this together using a compactness argument, we conclude that there exists a radius , such that for all in the image of the homotopy , the function is conservative in .
Now we repeat the proof of Cauchy’s theorem (Theorem 4 of Notes 3), discretising the homotopy into short closed polygonal paths (each of diameter less than ) around which the integral of is zero, to conclude (9). The argument is completely analogous, save for the technicality that the paths may occasionally pass through one of the points in . But this can be easily rectified by perturbing each of the paths by adding a short detour around any point of that is passed through; we leave the details to the interested reader.
Combining the residue theorem with the Jordan curve theorem, we obtain the following special case, which is already enough for many applications:
Corollary 22 (Residue theorem for simple closed contours) Let be a simple closed anticlockwise contour in . Suppose that is holomorphic on an open set containing the image and interior of , outside of a closed discrete that does not intersect the image of . Then we have
If is oriented clockwise instead of anticlockwise, then we instead have
Exercise 23 (Homology version of residue theorem) Show that the residue theorem continues to hold when the closed curve is replaced by a -cycle (as in Exercise 63 of Notes 3) that avoids all the singularities in , and the requirement that be simply connected is replaced by the requirement that contains all the points outside of the image of where .
Exercise 24 (Exterior version of residue theorem) Let be a simple closed anticlockwise contour in . Suppose that is holomorphic on an open set containing the image and exterior of , outside of a finite that does not intersect the image of . Suppose also that converges to a finite limit in the limit . Show that
If is oriented clockwise instead of anticlockwise, show instead that
In order to use the residue theorem effectively, one of course needs some tools to compute the residue at a given point. The Fourier inversion formula (4) expresses such residues as a contour integral, but this is not so useful in practice as often the best way to compute such integrals is via the residue theorem, leaving one back where one started! But if the singularity is not an essential one, we have some useful formulae:
Exercise 25 Let be holomorphic on an open set outside of a singular set , and let be an isolated point of .
- (i) If has a removable singularity at , show that .
- (ii) If has a simple pole at , show that .
- (iii) If has a pole of order at most at for some , show that
In particular, if near for some that is holomorphic at , then
Using these facts, show that Cauchy’s theorem (Theorem 14 from Notes 3), the Cauchy integral formula (Theorem 39 from Notes 3), and the higher order Cauchy integral formula (Exercise 40 from Notes 3) can be derived from the residue theorem. (Of course, this is not an independent proof of these theorems, as they were used in the proof of the residue theorem!)
The residue theorem can be applied in countless ways; we give only a small sample of them below.
Exercise 26 Use the residue theorem to give an alternate proof of the fundamental theorem of algebra, by considering the integral for a polynomial of degree and some large radius .
Exercise 27 Let be a Dirichlet polynomial of the form
for some sequence of complex numbers, with only finitely many of the non-zero. Establish Perron’s formula
for any real numbers with not an integer. What happens if is an integer? Generalisations and variants of this formula, particularly with the Dirichlet polynomial replaced by more general Dirichlet series in which infinitely many of the are allowed to be non-zero, are of particular use in analytic number theory; see for instance this previous blog post.
Exercise 28 (Spectral theorem for matrices) This exercise presumes some familiarity with linear algebra. Let be a positive integer, and let denote the ring of complex matrices. Let be a matrix in . The characteristic polynomial , where is the identity matrix, is a polynomial of degree in with leading coefficient ; we let be the distinct zeroes of this polynomial, and let be the multiplicities; thus by the fundamental theorem of algebra we have
We refer to the set as the spectrum of . Let be any closed anticlockwise curve that contains the spectrum of in its interior, and let be an open subset of that contains and its interior.
- (i) Show that the resolvent is a meromorphic function on with poles at the spectrum of , where we call a matrix-valued function meromorphic if each of its components are meromorphic. (Hint: use the adjugate matrix.)
- (ii) For any holomorphic , we define the matrix by the formula
(cf. the Cauchy integral formula). We refer to as the holomorphic functional calculus for applied to . Show that the matrix does not depend on the choice of , depends linearly on , and equals the identity matrix when is the constant function . Furthermore, if is the function , show that
Conclude in particular that if is a polynomial
with complex coefficients , then the function (as defined by the holomorphic functional calculus) matches how one would define algebraically, in the sense that
- (iii) Prove the Cayley-Hamilton theorem . (Note from (ii) that it does not matter whether one interprets algebraically, or via the holomorphic functional calculus.)
- (iv) If is holomorphic, show that the matrix-valued function has only removable singularities in .
- (v) If are holomorphic, establish the identity
- (vi) Show that there exist matrices that are idempotent (thus for all ), commute with each other and with , sum to the identity (thus ), annihilate each other (thus for all distinct ) and are such that for each , one has the nilpotency property
In particular, we have the spectral decomposition
where each is a nilpotent matrix with . Finally, show that the range of (viewed as a linear operator from to itself) has dimension . Find a way to interpret each as the (negative of the) “residue” of the resolvent operator at .
Under some additional hypotheses, it is possible to extend the analysis in the above exercise to infinite-dimensional matrices or other linear operators, but we will not do so here.
— 3. The argument principle —
We have not yet defined the complex logarithm of a complex number , but one of the properties we would expect of this logarithm is that its derivative should be the reciprocal function: . In particular, by the chain rule we would expect the formula
for a holomorphic function , at least away from the zeroes of . Inspired by this formal calculation, we refer to the function as the log-derivative of . Observe the product rule and quotient rule, when applied to complex differentiable functions that are non-zero at some point , gives the formulae
which are of course consistent with the formal calculation (10), given how we expect the logarithm to act on products and quotients. Thus, for instance, if , are polynomials that are factored as
and
for some non-zero complex numbers , distinct complex numbers , and positive integers , then the log-derivative of the rational function is given by
In particular, the log-derivative of is meromorphic with poles at , with a residue of at each zero of , and a residue of at each pole of .
A general rule of thumb in complex analysis is that holomorphic functions behave like generalisations of polynomials, and meromorphic functions behave like generalisations of rational functions. In view of this rule of thumb and the above calculation, the following lemma should thus not be surprising:
Lemma 29 Let be a holomorphic function on an open set outside of a singular set , and let be either an element of or an isolated point of .
- (i) If is holomorphic and non-zero at , then the log-derivative is also holomorphic at .
- (ii) If is holomorphic at with a zero of order , then the log-derivative has a simple pole at with residue .
- (iii) If has a removable singularity at , and is non-zero once the singularity is removed, then the log-derivative has a removable singularity at .
- (iv) If has a removable singularity at , and has a zero of order once the singularity is removed, then the log-derivative has a simple pole at with residue .
- (v) If has a pole of order at , then the log-derivative has a simple pole at with residue .
Proof: The claim (i) is obvious. For (ii), we use Taylor expansion to factor for some holomorphic and non-zero near , and then from (11) we have
Since is holomorphic at , the claim (ii) follows. The claim (v) is proven similarly using a factorisation , and using (12) in place of (11). The claims (iii), (iv) then follow from (i), (ii) respectively after removing the singularity.
Remark 30 Note that the lemma does not cover all possible singularity and zero scenarios. For instance, could be identically zero, in which case the log-derivative is nowhere defined. If has an essential singularity then the log-derivative can be a pole (as seen for instance by the example for some ) or another essential singularity (as can be seen for instance by the example ). Finally, if has a non-isolated singularity, then the log-derivative could exhibit a wide range of behaviour (but probably will be quite wild as one approaches the singular set).
By combining the above lemma with the residue theorem, we obtain the argument principle:
Theorem 31 (Argument principle) Let be a simple closed anticlockwise contour. Let be an open set containing and its interior. Let be a meromorphic function on that is holomorphic and non-zero on the image of . Suppose that after removing all the removable singularities of , has zeroes in the interior of (of orders respectively), and poles in the interior of (of orders respectively). ( is also allowed to have zeroes and poles in the exterior of .) Then we have
where is the closed contour .
Proof: The first equality of (13) follows from the residue theorem and Lemma 29. From the change of variables formula (Exercise 16(ix) of Notes 2) we have
and the second identity also follows.
We isolate the special case of the argument principle when there are no poles for special mention:
Corollary 32 (Special case of argument principle) Let be a simple closed anticlockwise contour, let be an open set containing the image of and its interior, and let be holomorphic. Suppose that has no zeroes on the image of . Then the number of zeroes of (counting multiplicity) in the interior of is equal to the winding number of around the origin.
Recalling that the winding number is a homotopy invariant (Lemma 41 of Notes 3), we conclude that the number of zeroes of a holomorphic function in the interior of a simple closed anticlockwise contour is also invariant with respect to continuous perturbations, so long as zeroes never cross the contour itself. More precisely:
Corollary 33 (Stability of number of zeroes) Let be an open set. Let , be simple closed anticlockwise contours that are homotopic as closed curves via some homotopy ; suppose also that contains the interiors of and . Let be holomorphic, and let be a continuous function such that and for all . Suppose that for all and (i.e., at time , the curve never encounters any zeroes of ). Then the number of zeroes (counting multiplicity) of in the interior of equals the number of zeroes of in the interior of (counting multiplicity).
Proof: By Corollary 32, it suffices to show that
But the curves and are homotopic as closed curves in , using the homotopy defined by
(note that this avoids the origin by hypothesis). The claim then follows from Lemma 41 of Notes 3.
Informally, the above corollary asserts that zeroes of holomorphic functions cannot be created or destroyed, as long as they are confined within a closed contour.
Example 34 Let be the unit circle . The polynomial has a double zero at , so (counting multiplicity) has two zeroes in the interior of . If we consider instead the perturbation for some , this has simple zeroes at and respectively, so as long as , the holomorphic function also has two zeroes in the interior of ; but as crosses , the zeroes of pass through , and one no longer has any zeroes of in the interior of . The situation can be contrasted with the real case: the function has a double zero at the origin when , but as soon as becomes positive, the zeroes immediately disappear from the real line. Note that the stability of zeroes fails if we do not count zeroes with multiplicity; thus, as a general rule of thumb, one should always try to count zeroes with multiplicity when doing complex analysis. (Heuristically, one can think of a zero of order as simple zeroes that are “infinitesimally close together".)
Example 35 When one considers meromorphic functions instead of holomorphic ones, then the number of zeroes inside a region need not be stable any more, but the number of zeroes minus the number of poles will be stable. Consider for instance the meromorphic function , which has a removable singularity at but no zeroes or poles. If we perturb it to for some , then we suddenly have a double pole at , but this is balanced by two simple zeroes at and ; in the limit as we see that the two zeroes “collide” with the double pole, annihilating both the zeroes and the poles.
A particularly useful special case of the stability of zeroes is Rouche’s theorem:
Theorem 36 (Rouche’s theorem) Let be a simple closed contour, and let be an open set containing the image of and its interior. Let be holomorphic. If one has for all in the image of , then and have the same number of zeroes (counting multiplicity) in the interior of .
Proof: We may assume without loss of generality that is anticlockwise. By hypothesis, and cannot have zeroes on the image of . The claim then follows from Corollary 33 with , , , , and .
Rouche’s theorem has many consequences for complex analysis. One basic consequence is the open mapping theorem:
Theorem 37 (Open mapping theorem) Let be an open connected non-empty subset of , and let be holomorphic and not constant. Then is also open.
Proof: Let . As is not constant, the zeroes of are isolated (Corollary 24 of Notes 3). Thus, for sufficiently small, is nonvanishing on the image of the circle . Clearly has at least one zero in the interior of this circle. Thus, by Rouche’s theorem, if is sufficiently close to , then will also have at least one zero in the interior of this circle. In particular, contains a neighbourhood of , and the claim follows.
Exercise 38 Use Rouche’s theorem to obtain another proof of the fundamental theorem of algebra, by showing that a polynomial with and has exactly zeroes (counting multiplicity) in the complex plane. (Hint: compare with inside some large circle .)
Exercise 39 (Inverse function theorem) Let be an open subset of , let , and let be a holomorphic function such that . Show that there exists a neighbourhood of in such that the map is a complex diffeomorphism; that is to say, it is holomorphic, invertible, and the inverse is also holomorphic. Finally, show that
for all . (Hint: one can either mimic the real-variable proof of the inverse function theorem using the contraction mapping theorem, or one can use Rouche’s theorem and the open mapping theorem to construct the inverse.)
Exercise 40 Let be an open subset of , and be a map. Show that the following are equivalent:
- (i) is a local complex diffeomorphism. That is to say, for every there is a neighbourhood of in such that is open and is a complex diffeomorphism (as defined in the preceding exercise).
- (ii) is holomorphic on and is a local homeomorphism. That is to say, for every there is a neighbourhood of in such that is open and is a homeomorphism.
- (iii) is holomorphic on and is a local injection. That is to say, for every there is a neighbourhood of in such that is injective.
- (iv) is holomorphic on , and the derivative is nowhere vanishing.
Exercise 41 (Hurwitz’s theorem) Let be an open connected non-empty subset of , and let be a sequence of holomorphic functions that converge uniformly on compact sets to a limit (which is then necessarily also holomorphic, thanks to Theorem 34 of Notes 3). Prove the following two versions of Hurwitz’s theorem:
- (i) If none of the have any zeroes in , show that either also has no zeroes in , or is identically zero.
- (ii) If all of the are univalent (that is to say, they are injective holomorphic functions), show that either is also univalent, or is constant.
Exercise 42 (Bloch’s theorem) The purpose of this exercise is to establish a more quantitative variant of the open mapping theorem, due to Bloch; this will be useful later in this notes for proving the Picard and Montel theorems. Let be a holomorphic function on a disk , and suppose that is non-zero
- (i) Suppose that for all . Show that there is an absolute constant such that contains the disk . (Hint: one can normalise , , . Use the higher order Cauchy integral formula to get some bound on for near the origin, and use this to approximate by near the origin. Then apply Rouche’s theorem.)
- (ii) Without the hypothesis in (i), show that there is an absolute constant such that contains a disk of radius . (Hint: if one has for all , then we can apply (i) with replaced by . If not, pick with , and start over with replaced by and replaced by . One cannot iterate this process indefinitely as it will create a singularity of in .)
— 4. Branches of the complex logarithm —
We have refrained until now from discussing one of the most basic transcendental functions in complex analysis, the complex logarithm. In real analysis, the real logarithm can be defined as the inverse of the exponential function ; it can also be equivalently defined as the antiderivative of the function , with the initial condition . (We use here for the real logarithm in order to distinguish it from the complex logarithm below.)
Let’s see what happens when one tries to extend these definitions to the complex domain. We begin with the inversion of the complex exponential. From Euler’s formula we have that ; more generally, we have whenever for some integer . In particular, the exponential function is not injective. Indeed, for any non-zero , we have a multi-valued logarithm
which, by Euler’s formula, can be written as
where
denotes all the possible arguments of in polar form. These arguments are a coset of the group , and so the complex logarithm is a coset of the group . For instance, if , then
and
The complex exponential never vanishes, so by our definitions we see that is the empty set. As such, we will usually omit the origin from the domain when discussing the complex exponential.
Of course, one also encounters multi-valued functions in real analysis, starting when one tries to invert the squaring function , as any given positive number has two square roots. In the real case, one can eliminate this multi-valuedness by picking a branch of the square root function – a function which selects one of the multiple choices for that function at each point in the domain. In particular, we have the positive branch of the square root function on , as well as the negative branch . One could also create more discontinuous branches of the square root function, for instance the function that sends to for , and to for .
Suppose now that we have a branch of the logarithm function, thus
for any . If is complex differentiable at some point , then by differentiating (14) at using the chain rule, we see that
and hence by (14) again we have
(which is of course consistent with the real-variable formula ). If now is a closed contour in , and is differentiable on the entire image of , then the fundamental theorem of calculus then tells us that
On the other hand, is equal to . We thus conclude that for any branch of the complex logarithm, the set on which is complex differentiable cannot contain any closed curve that winds non-trivially around the origin. Thus for instance one cannot find a branch of that is holomorphic on all of , or even on a neighbourhood of the unit circle (or any other curve going around the origin).
On the other hand, if is a simply connected open subset of , then from Cauchy’s theorem the function is conservative on . If we pick a point in and arbitrarily select a logarithm of , we can then use the fundamental theorem of calculus to find an antiderivative of on with . By definition, is holomorphic, and from the chain rule we have for all that
and hence by the quotient rule
As is connected, must therefore be constant; by construction we have , and thus
for all . In other words, is a branch of the complex logarithm.
Thus, for instance, the region formed by excluding the negative real axis from the complex plane is simply connected (it is star-shaped around ), and so must admit a holomorphic branch of the complex logarithm. One such branch is the standard branch of the complex logarithm, defined as
where is the standard branch of the argument, defined as the unique argument in in the interval . This branch of the logarithm is continuous on , and hence (by the exercise below) is holomorphic on this region, and is thus an antiderivative of here. Similarly if one replaces the negative real axis by other rays emenating from the origin (or indeed from arbitrary simple curves from zero to infinity, see Exercise 44 below.)
Exercise 43 Let be a connected non-empty open subset of .
- (i) If and are continuous branches of the complex logarithm, show that there exists a natural number such that for all .
- (ii) Show that any continuous branch of the complex logarithm is holomorphic.
- (iii) Show that there is a continuous branch of the logarithm if and only if and lie in the same connected component of . (Hint: for the “if” direction, use a continuity argument to show that the winding number of any closed curve in around vanishes. For the “only if”, encircle the connected component of in (which is a compact subset of by hypothesis) by a simple polygonal path in .)
Exercise 44 Let be a continuous injective map with and as .
- (i) Show that is not all of . (Hint: modify the construction in Section 4 of Notes 3 that showed that a simple closed curve admitted at least one point with non-zero winding number.)
- (ii) Show that the complement is simply connected. (Hint: modify the remaining arguments in Section 4 of Notes 3). In particular, by the preceding discussion, there is a branch of the complex logarithm that is holomorphic outside of .
It is instructive to view the identity
, through the lens of branches of the complex logarithm such as the standard branch . From the fundamental theorem of calculus, one has
for any curve that avoids the negative real axis. Of course, the contour does not avoid this negative axis, but it can be approximated by (non-closed) contours that do. More precisely, one has
where is the map . As each avoids the negative real axis, we thus have
We observe that has a jump discontinuity of on the negative real axis, and specifically
and
which gives an alternate derivation of the identity (15). More generally, the identity
for any closed curve avoiding the origin can be interpreted using the standard branch of the logarithm as a version of the Alexander numbering rule (Exercise 55 of Notes 3): each crossing of across the branch cut triggers a jump up or down in the count towards the winding number, depending on whether the crossing was in the anticlockwise or clockwise direction.
One can use branches of the complex logarithm to create branches of the root functions for natural numbers . As with the complex exponential, the function is not injective, and so is multivalued (see Exercise 15 of Notes 0). One cannot form a continuous branch of this function on for any , as the corresponding branch of would then contradict the quantisation of order of singularities (Exercise 13). However, on any domain where there is a holomorphic branch of the complex logarithm, one can define a holomorphic branch of the function by the formula
It is easy to see that is indeed holomorphic with for all . Thus for instance we have the standard branch of the root function, which is holomorphic away from the negative real axis. More generally, one can define a “standard branch of ” for any complex by the formula , for instance the standard branch of can be computed to be .
The presence of branch cuts can prevent one from directly applying the residue theorem to calculate integrals involving branches of multi-valued functions. But in some cases, the presence of the branch cut can actually be exploited to compute an integral. The following exercise provides an example:
Exercise 45 Compute the improper integral
by applying the residue theorem to the function for some branch of with branch cut on the positive real axis, and using a “keyhole” contour that is a perturbation of
the key point is that the branch cut makes the contribution of (the perturbations) of and fail to cancel each other.
The construction of holomorphic branches of can be extended to other logarithms:
Exercise 46 Let be a simply connected subset of , and let be a holomorphic function with no zeroes on .
- (i) Show that there exists a holomorphic branch of the complex logarithm , thus .
- (ii) Show that for any natural number , there exists a holomorphic branch of the root function , thus .
Actually, one can invert other non-injective holomorphic functions than the complex exponential, provided that these functions are a covering map. We recall this topological concept:
Definition 47 (Covering map) Let be a continuous map between two connected topological spaces . We say that is a covering map if, for each , there exists an open neighbourhood of in such that the preimage is the disjoint union of open subsets of , such that for each , the map is a homeomorphism. In this situation, we call a covering space of .
In complex analysis, one specialises to the situation in which are Riemann surfaces (e.g. they could be open subsets of ), and is a holomorphic map. In that case, the homeomorphisms are in fact complex diffeomorphisms, thanks to Exercise 40.
Example 48 The exponential map is a covering map, because for any element of written in polar form as , one can pick (say) the neighbourhood
of , and observe that the preimage of is the disjoint union of the open sets
for , and that the exponential map is a diffeomorphism. A similar calculation shows that for any natural number , the map is a covering map from to . However, the map is not a covering map from to , because it fails to be a local diffeomorphism at zero due to the vanishing derivative (here we use Exercise 40). One final (non-)example: the map is not a covering map from the upper half-plane to , because the preimage of any small disk around splits into two disconnected regions, and only one of them is homeomorphic to via the map .
From topology we have the following lifting property:
Lemma 49 (Lifting lemma) Let be a continuous covering map between two path-connected and locally path-connected topological spaces . Let be a simply connected and path connected topological space, and let be continuous. Let , and let be such that . Then there exists a unique continuous map such that and , which we call a lift of by .
Proof: We first verify uniqueness. If we have two continuous functions with and , then the set is clearly closed in and contains . From the covering map property we also see that is open, and hence by connectedness we have on all of , giving the claim.
To verify existence of the lift, we first prove the existence of monodromy. More precisely, given any curve with we show that there exists a unique curve such that and (the reader is encouraged to draw a picture to describe this situation). Uniqueness follows from the connectedness argument used to prove uniqueness of the lift , so we turn to existence. As in previous notes, we rely on a continuity argument. Let be the set of all for which there exists a curve such that , where is the restriction of to . Clearly is closed in and contains ; using the covering map property it is not difficult to show that is also open in . Thus is all of , giving the claim.
Now let , be homotopic curves with fixed endpoints, with initial point and some terminal point , and let be a homotopy. For each , we have a curve given by , and by the preceding paragraph we can associate a curve such that and . Another application of the continuity method shows that for all , the map is continuous; in particular, the map . On the other hand, lies in , which is a discrete set thanks to the covering map property. We conclude that is constant in , and in particular that .
Since is simply connected, any two curves with fixed endpoints are homotopic. We can thus define a function by declaring for any to be the point , where is any curve from to , and is constructed as before. By construction we have , and from the local path connectedness of and the covering map property of we can check that is continuous. The claim follows.
We can specialise this to the complex case and obtain
Corollary 50 (Holomorphic lifting lemma) Let be a holomorphic covering map between two path-connected Riemann surfaces . Let be a simply connected and path connected Riemann surface, and let be holomorphic. Let , and let be such that . Then there exists a unique holomorphic map such that and , which we call a lift of by .
Proof: A Riemann surface is automatically locally path-connected, and a connected Riemann surface is automatically path connected (observe that the set of all points on the surface that can be path-connected to a reference point is open, closed, and non-empty). Applying Lemma 49, we obtain all the required claims, except that the lift produced is only known to be continuous rather than holomorphic. But then we can locally express as the composition of one of the local inverses of with . Applying Exercise 40, these local inverses are holomorphic, and so is holomorphic also.
Remark 51 It is also possible to establish the above corollary using the monodromy theorem and analytic continuation.
Exercise 53 Let be simply connected, and let be holomorphic and avoid taking the values . Show that there exists a holomorphic function such that . (This can be proven either through Corollary 50, or by using the quadratic formula to solve for and then applying Exercise 46.)
In some cases it is also possible to obtain lifts in non-simply connected domains:
Exercise 54 Show that there exists a holomorphic function such that for all . (Hint: use the Schwartz reflection principle, see Exercise 37 of Notes 3.)
As an illustration of what one can do with all this machinery, let us now prove the Picard theorems. We begin with the easier “little” Picard theorem.
Theorem 55 (Little Picard theorem) Let be entire and non-constant. Then omits at most one point of .
The example of the exponential function , whose range omits the origin, shows that one cannot make any stronger conclusion about .
Proof: Suppose for contradiction that we have an entire non-constant function such that omits at least two points. After applying a linear transformation, we may assume that avoids and , thus takes values in .
At this point, the most natural thing to do from a Riemann surface point of view would be to cover by a bounded region, so that Liouville’s theorem may be applied. This can be done easily once one has the machinery of elliptic functions; but as we do not have this machinery yet, we will instead use a more ad hoc covering of using the exponential and trigonometric functions to achieve a passable substitute for this strategy.
We turn to the details. Since avoids , we may apply Exercise 46 to write for some entire . As avoids , must avoid the integers .
Next, we apply Exercise 53 to write for some entire . The set must now avoid all complex numbers of the form for natural numbers and integers . In particular, if is large enough, we see that does not contain any disk of the form . Applying Bloch’s theorem (Exercise 42(ii)) in the contrapositive, we conclude that for any disk in , one has for some absolute constant . Sending to infinity and using the fundamental theorem of calculus, we conclude that is constant, hence and are also constant, a contradiction.
Now we prove the more difficult “great” Picard theorem.
Theorem 56 (Great Picard theorem) Let be holomorphic on a disk outside of a singularity at . If this singularity is essential, then omits at most one point of .
Note that if one only has a pole at , e.g. if for some natural number , then the conclusion of the great Picard theorem fails. This result easily implies both the little Picard theorem (because if is entire and non-polynomial, then has an essential singularity at the origin) and the Casorati-Weierstrass theorem (Theorem 11(iii)). By repeatedly passing to smaller neighbourhoods, one in fact sees that with at most one exception, every complex number is attained infinitely often by a function holomorphic in a punctured disk around an essential singularity.
Proof: This will be a variant of the proof of the little Picard theorem; it would again be more natural to use elliptic functions, but we will use some passable substitutes for such functions concocted in an ad hoc fashion out of exponential and trigonometric functions.
Assume for contradiction that has an essential singularity at and avoids at least two points in . Applying linear transformations to both the domain and range of , we may normalise , , and assume that avoids and , thus we have a holomorphic map with an essential singularity at .
The domain is not simply connected, so we work instead with the function
defined by
Clearly is holomorphic on the right-half plane and avoids . We also observe that obeys the periodicity property
As the right-half plane is simply connected, we may (as before) express for some holomorphic function , and then write for some holomorphic function that avoids all numbers of the form for natural numbers and integers . Using Bloch’s theorem as before, we see that for any disk in the right-half plane , we have for some absolute constant . We cannot set to infinity any more, but we can make as large as the real part of , giving the bound
In particular, on integrating along a line segment from to , and using the boundedness of on the compact set , we obtain a bound of the form
for some , and all and . Taking cosines using the formula , we obtain a polynomial type bound
On the other hand, from (16) one has
and hence
for all in the right half-plane. The set is discrete, the function is continuous, and the right half-plane is connected, so this function must in fact be constant. That is to say, there exists an integer such that
for all in the upper half plane. Equivalently, the function is periodic with period . From (17) and the triangle inequality we conclude that
We now upgrade this bound on (18) by exploiting the quantisation of pole orders (Exercise 13). As the function is periodic with period on the right half-plane, we may write
for some function , which is holomorphic thanks to the chain rule. From (18) we have
when . Applying Exercise 13 (with, say, and ), we conclude that has a removable singularity and is thus in particular bounded on (say) the disk . From (19) we conclude that is bounded on the region ; taking exponentials, we conclude that is also bounded on this region. Since , we conclude that is bounded on , and thus by Riemann’s theorem (Exercise 35 from Notes 3) has a removable singularity at the origin. But by taking Laurent series, this implies that has a pole of order at most at the origin, contradicting the hypothesis that the singularity of at the origin was essential.
Exercise 57 (Montel’s theorem) Let be an open subset of the complex plane. Define a holomorphic normal family on to be a collection of holomorphic functions with the following property: given any sequence in , there exists a subsequence which is uniformly convergent on compact sets (i.e., for every compact subset of , the sequence converges uniformly on to some limit). Similarly, define a meromorphic normal family to be a collection of meromorphic functions such that for any sequence in , there exists a subsequence that are uniformly convergent on compact sets, using the metric on the Riemann sphere induced by the identification with the geometric sphere . (More succinctly, normal families are those families of holomorphic or meromorphic functions that are precompact in the locally uniform topology.)
- (i) (Little Montel theorem) Suppose that is a collection of holomorphic functions that are uniformly bounded on compact sets (i.e., for each compact there exists a constant such that for all and ). Show that is a holomorphic normal family. (Hint: use the higher order Cauchy integral formula to establish some equicontinuity on this family on compact sets, then use the Arzelá-Ascoli theorem.
- (ii) (Great Montel theorem) Let be three distinct elements of the Riemann sphere, and suppose that is a family of meromorphic functions which avoid the three points . Show that is a meromorphic normal family. (Hint: use some elementary transformations to reduce to the case . Then, as in the proof of the Picard theorems, express each element of locally in the form and use Bloch’s theorem to get some uniform bounds on .)
Exercise 58 (Harnack principle) Let be an open connected subset of , and let be a sequence of harmonic functions which is pointwise nondecreasing (thus for all and ). Show that is either infinite everywhere on , or is harmonic. (Hint: work locally in a disk. Write each on this disk as the real part of a holomorphic function , and apply Montel’s theorem followed by the Hurwitz theorem to .) This result is known as Harnack’s principle.
Exercise 59
- (i) Show that the function is harmonic on but has no harmonic conjugate.
- (ii) Let , and let be a harmonic function obeying the bounds
for all and some constants . Show that there exists a real number and a harmonic function such that
for all . (Hint: one can find a conjugate of outside of some branch cut, say the negative real axis restricted to . Adjust by a multiple of until the conjugate becomes continuous on this branch cut.)
Exercise 60 (Local description of holomorphic maps) Let be a holomorphic function on an open subset of , let be a point in , and suppose that has a zero of order at for some . Show that there exists a neighbourhood of in on which one has the factorisation , where is holomorphic with a simple zero at (and hence a complex diffeomorphism from a sufficiently small neighbourhood of to a neighbourhood of ). Use this to give an alternate proof of the open mapping theorem (Theorem 37).
Exercise 61 (Winding number and lifting) Let , let be a closed curve avoiding , and let be an integer. Show that the following are equivalent:
- (i) .
- (ii) There exists a complex number and a curve from to such that for all .
- (iii) is homotopic up to reparameterisation as closed curves in to the curve that maps to for some .
Definition 1 (Homotopy) Let be an open subset of , and let , be two curves in .
- (i) If have the same initial point and final point , we say that and are homotopic with fixed endpoints in if there exists a continuous map such that and for all , and such that and for all .
- (ii) If are closed (but possibly with different initial points), we say that and are homotopic as closed curves in if there exists a continuous map such that and for all , and such that for all .
- (iii) If and are curves with the same initial point and same final point, we say that and are homotopic with fixed endpoints up to reparameterisation in if there is a reparameterisation of which is homotopic with fixed endpoints in to a reparameterisation of .
- (iv) If and are closed curves, we say that and are homotopic as closed curves up to reparameterisation in if there is a reparameterisation of which is homotopic as closed curves in to a reparameterisation of .
In the first two cases, the map will be referred to as a homotopy from to , and we will also say that can be continously deformed to (either with fixed endpoints, or as closed curves).
Example 2 If is a convex set, that is to say that whenever and , then any two curves from one point to another are homotopic, by using the homotopy
For a similar reason, in a convex open set , any two closed curves will be homotopic to each other as closed curves.
Exercise 3 Let be an open subset of .
- (i) Prove that the property of being homotopic with fixed endpoints in is an equivalence relation.
- (ii) Prove that the property of being homotopic as closed curves in is an equivalence relation.
- (iii) If are closed curves with the same initial point, show that is homotopic to with fixed endpoints if and only if is homotopic to as closed curves.
- (iv) Define a point in to be a curve of the form for some and all . Let be a closed curve in . Show that is homotopic with fixed endpoints to a point in if and only if is homotopic as a closed curve to a point in . (In either case, we will call homotopic to a point, null-homotopic, or contractible to a point in .)
- (v) If are curves with the same initial point and the same terminal point, show that is homotopic to with fixed endpoints in if and only if is homotopic to a point in .
- (vi) If is connected, and are any two curves in , show that there exists a continuous map such that and for all . Thus the notion of homotopy becomes rather trivial if one does not fix the endpoints or require the curve to be closed.
- (vii) Show that if is a reparameterisation of , then and are homotopic with fixed endpoints in U.
- (viii) Prove that the property of being homotopic with fixed endpoints in up to reparameterisation is an equivalence relation.
- (ix) Prove that the property of being homotopic as closed curves in up to reparameterisation is an equivalence relation.
We can then phrase Cauchy’s theorem as an assertion that contour integration on holomorphic functions is a homotopy invariant. More precisely:
Theorem 4 (Cauchy’s theorem) Let be an open subset of , and let be holomorphic.
- (i) If and are rectifiable curves that are homotopic in with fixed endpoints up to reparameterisation, then
- (ii) If and are closed rectifiable curves that are homotopic in as closed curves up to reparameterisation, then
This version of Cauchy’s theorem is particularly useful for applications, as it explicitly brings into play the powerful technique of contour shifting, which allows one to compute a contour integral by replacing the contour with a homotopic contour on which the integral is easier to either compute or integrate. This formulation of Cauchy’s theorem also highlights the close relationship between contour integrals and the algebraic topology of the complex plane (and open subsets thereof). Setting to be a point, we obtain an important special case of Cauchy’s theorem (which is in fact equivalent to the full theorem):
Corollary 5 (Cauchy’s theorem, again) Let be an open subset of , and let be holomorphic. Then for any closed rectifiable curve in that is contractible in to a point, one has .
Exercise 6 Show that Theorem 4 and Corollary 5 are logically equivalent.
An important feature to note about Cauchy’s theorem is the global nature of its hypothesis on . The conclusion of Cauchy’s theorem only involves the values of a function on the images of the two curves . However, in order for the hypotheses of Cauchy’s theorem to apply, the function must be holomorphic not only on the images on , but on an open set that is large enough (and sufficiently free of “holes”) to support a homotopy between the two curves. This point can be emphasised through the following fundamental near-counterexample to Cauchy’s theorem:
Example 7 Let , and let be the holomorphic function . Let be the closed unit circle contour . Direct calculation shows that
As a consequence of this and Cauchy’s theorem, we conclude that the contour is not contractible to a point in ; note that this does not contradict Example 2 because is not convex. Thus we see that the lack of holomorphicity (or singularity) of at the origin can be “blamed” for the non-vanishing of the integral of on the closed contour , even though this contour does not come anywhere near the origin. Thus we see that the global behaviour of , not just the behaviour in the local neighbourhood of , has an impact on the contour integral.
One can of course rewrite this example to involve non-closed contours instead of closed ones. For instance, if we let denote the half-circle contours and , then are both contours in from to , but one has
whereas
In order for this to be consistent with Cauchy’s theorem, we conclude that and are not homotopic in (even after reparameterisation).
In the specific case of functions of the form , or more generally for some point and some that is holomorphic in some neighbourhood of , we can quantify the precise failure of Cauchy’s theorem through the Cauchy integral formula, and through the concept of a winding number. These turn out to be extremely powerful tools for understanding both the nature of holomorphic functions and the topology of open subsets of the complex plane, as we shall see in this and later notes.
— 1. Proof of Cauchy’s theorem —
The underlying reason for the truth of Cauchy’s theorem can be explained in one sentence: complex differentiable functions behave locally like complex linear functions, which are conservative thanks to the fundamental theorem of calculus. More precisely, if is any complex linear function of , then has an antiderivative , and hence
for any rectifiable closed curve in the complex plane.
Perhaps the slickest way to make this intuition rigorous is through the following special case of Cauchy’s theorem.
Theorem 8 (Goursat’s theorem) Let be an open subset of , and be complex numbers such that the solid (and closed) triangle spanned by (or more precisely, the convex hull of ) is contained in . (We allow the triangle to degenerate in that we allow the to be collinear, or even coincident.) Then for any holomorphic function , one has
where is the closed polygonal path that traverses the vertices of the solid triangle in order.
Proof: Let us denote the triangular contour as . It is convenient (though odd-looking at first sight) to prove this theorem by contradiction. That is to say, suppose for contradiction that we had
for some . We now run the following “divide and conquer” strategy. We let , , be the midpoints of . Then from the basic properties of contour integration (see Exercise 16 of Notes 2) we can split the triangular integral as the sum of four integrals on smaller triangles, namely
(The reader is encouraged to draw a picture to visualise this decomposition.) By (2) and the triangle inequality (or, if one prefers, the pigeonhole principle), we must therefore have
where is one of the four triangular contours , , , or . Regardless of which of the four contours is, observe that the triangular region enclosed by is contained in that of . Furthermore, the diameter of is precisely half that of , where the diameter of a curve is defined by the formula
similarly, the perimeter of is precisely half that of . If we iterate the above process, we can find a nested sequence of triangular contours, each of which is contained in the previous one with half the diameter and perimeter, such that
for all . If we let be any point enclosed by , then from the decreasing diameters it is clear that the are a Cauchy sequence and thus converge to some limit , which is then contained in all of the closed triangles enclosed by any of the .
In particular, lies in and so is differentiable at . This implies, for any , that there exists a such that
whenever . We can rearrange this as
on . In particular, for large enough, this bound holds on the image on . In this case we can bound by , and hence by Exercise 16(v) of Notes 2,
From (1), the second integral vanishes. As each has half the diameter and perimeter of the previous, we thus have
But if one chooses small enough depending on and , we contradict (3).
Remark 9 This is a rare example of an argument in which a hypothesis of differentiability, rather than continuous differentiability, is used, because one can localise any failure of the conclusion all the way down to a single point. Another instance of such an argument is the standard proof of Rolle’s theorem.
Exercise 10 Find a proof of Goursat’s theorem that avoids explicit use of proof by contradiction. (Hint: use the fact that a solid triangle is compact, in the sense that every open cover has a finite subcover. For the purposes of this question, ignore the possibility that the proof of this latter fact might also use proof by contradiction.)
Goursat’s theorem only directly handles triangular contours, but as long as one works “locally”, or more precisely in a convex domain, we can quickly generalise:
Corollary 11 (Local Cauchy’s theorem for polygonal paths) Let be a convex open subset of , and let be a holomorphic function. Then for any closed polygonal path in , we have .
Proof: We induct on the number of vertices . The cases are trivial, and the case follows directly from Goursat’s theorem (using the convexity of to ensure that the interior of the polygon lies in ). If , we can split
The second integral on the right-hand side vanishes by Goursat’s theorem. The claim then follows from induction.
Exercise 12 By using the (real-variable) fundamental theorem of calculus and Fubini’s theorem in place of Goursat’s theorem, give an alternate proof of Corollary 11 in the case that is a rectangle and the derivative of is continuous. (One can also use Stokes’ theorem in place of the fundamental theorem of calculus and Fubini’s theorem.)
We can amplify Corollary 11 using the fundamental theorem of calculus again:
Corollary 13 (Local Cauchy’s theorem) Let be a convex open subset of , and let be a holomorphic function. Then has an antiderivative . Also, for any closed rectifiable curve in , and whenever are two rectifiable curves in with the same initial point and same terminal point. In other words, is conservative on .
Proof: The first claim follows from Corollary 11 and the second fundamental theorem of calculus (Theorem 30 from Notes 2). The remaining claims then follow from the first fundamental theorem of calculus (Theorem 27 from Notes 2).
We can now prove Cauchy’s theorem in the form of Theorem 4.
Proof: We will just prove part (i), as part (ii) is similar (and in any event it follows from part (i)). Since reparameterisation does not affect the integral, we may assume without loss of generality that and are homotopic with fixed endpoints, and not merely homotopic with fixed endpoints up to reparameterisation.
Let be a homotopy from to . Note that for any and , lies in the open set . From compactness, there must exist a radius such that for all and . Next, as is continuous on a compact set, it is uniformly continuous. In particular, there exists such that
whenever and are such that and .
Now partition and as and in such a way that and for all and . For each such and , let denote the closed polygonal contour
(the reader is encouraged here to draw a picture of the situation; we are using polygonal contours here rather than the homotopy because we did not require any rectifiability properties on the homotopy). By construction, the diameter of this contour is at most , so the contour is contained entirely in the disk . This disk is convex and contained in . Applying Corollary 11 or Corollary 13, we conclude that
for all and . If we sum this over all and , and noting that the homotopy fixes the endpoints, we conclude after a lot of cancelling that
(again, the reader is encouraged to draw a picture to see this cancellation). However, from a further application of Corollary 11 we have
for , where is the restriction of to , and similarly for . Putting all this together we conclude that
as required.
One nice feature of Cauchy’s theorem is that it allows one to integrate holomorphic functions on curves that are not necessarily rectifiable. Indeed, if is a curve in , then for a sufficiently fine partition , the polygonal (and hence rectifiable) path will be contained in , and furthermore be homotopic to with fixed endpoints. One can then define when is holomorphic in and is non-rectifiable by declaring
where is any rectifiable curve that is homotopic (with fixed endpoints) to . This is a well defined definition thanks to the above discussion as well as Cauchy’s theorem; also observe that the exact open set in which the homotopy lives is not relevant, since given any two open sets containing the image of one can find a rectifiable curve which is homotopic to with fixed endpoints in , and hence in and separately. With this extended notion of the contour integral, one can then remove the hypothesis of rectifiability from many theorems involving integration of holomorphic functions. In particular, Cauchy’s theorem itself now holds for non-rectifiable curves. This reflects some duality in the integration concept ; if one assumes more regularity on the function , one can get away with worse regularity on the curve , and vice versa.
A special case of Cauchy’s theorem is worth recording explicitly. We say that an open set in the complex plane is simply connected if it is non-empty, connected, and if every closed curve in is contractible in to a point. For instance, from Example 2 we see that any convex non-empty open set is simply connected. From Theorem 4 we then have
Theorem 14 (Cauchy’s theorem, simply connected case) Let be a simply connected subset of , and let be holomorphic. Then for any closed curve in . In particular (by Exercise 31 of Notes 2), is conservative and has an antiderivative.
— 2. Consequences of Cauchy’s theorem —
Now that we have Cauchy’s theorem, we use it to quickly give a large number of striking consequences. We begin with a special case of the