In the previous set of notes we introduced the notion of a complex diffeomorphism between two open subsets of the complex plane (or more generally, two Riemann surfaces): an invertible holomorphic map whose inverse was also holomorphic. (Actually, the last part is automatic, thanks to Exercise 40 of Notes 4.) Such maps are also known as biholomorphic maps or conformal maps (although in some literature the notion of “conformal map” is expanded to permit maps such as the complex conjugation map that are angle-preserving but not orientation-preserving, as well as maps such as the exponential map from to that are only locally injective rather than globally injective). Such complex diffeomorphisms can be used in complex analysis (or in the analysis of harmonic functions) to change the underlying domain to a domain that may be more convenient for calculations, thanks to the following basic lemma:
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:
- (i) (Elliptic model) The Riemann sphere ;
- (ii) (Parabolic model) The complex plane ; and
- (iii) (Hyperbolic model) The unit disk .
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 (besides the constant function ) 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 algebra 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 connected 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:
- Translation maps for some , with the convention that is mapped to ;
- Dilation maps for some , with the convention that is mapped to ; and
- The inversion map , with the convention that is mapped to .
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 Schwarz 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 connected 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 connected 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 (after first shrinking the neighbourhoods and if necessary). 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 (Schottky’s theorem) 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 3) 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. Schwarz-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 . In a similar vein, the arguments in the preceding section show that under the hypotheses of Theorem 49, there exists a surjective map from to , but one needs something like Theorem 51 to obtain the crucial additional property of injectivity (which was automatic in the preceding section, since one already started with an injection in hand).
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 one 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 with , such that and agree on a neighbourhood for each .
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 continuous Perron family on is a family of continuous 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 .
One can also consider more general Perron families of subharmonic functions that are merely upper semi-continuous rather than continuous, but for the current application continuous Perron families will suffice.
The fundamental theorem that powers the Perron method is then
Theorem 62 (Perron method) Let be a continuous 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 to find a continuous function that equals on the boundary of this disk (viewed as a subset of ) and is harmonic and at least as large as in the interior; 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.)
30 comments
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18 October, 2016 at 10:05 pm
maxbaroi
Int he proof of Schwarz’s lemma (Lemma 15), the statement should be replace with
18 October, 2016 at 10:33 pm
maxbaroi
I have a notational question. Why do you write instead of when the later seems more commonly used to describe the quotient spaces?
Sorry for the spam, I have midterms this week and of course trying to do anything but study for them.
[This is an artefact of the fact that my automorphisms area acting on the left. Of course in the case of additive actions, there is no distinction between and ; I’ll adjust the text accordingly. -T.]
19 October, 2016 at 8:41 am
Lior Silberman
The adjustment has a typo: to the text above you added instead of (wrong direction of slash).
[Corrected, thanks – T.]
19 October, 2016 at 12:36 am
Anonymous
In the proof of lemma 1(part ii), it should be “By part (i)”.
[Corrected, thanks – T.]
22 October, 2016 at 9:32 am
Anonymous
According to definition 58, if identically, it is (trivially!) subharmonic. Should this trivial example be excluded from the definition?
[Some authors adopt this convention, but it will not make a difference in the current notes. -T.]
22 October, 2016 at 1:10 pm
Anonymous
In theorem 51 (line 5), should be .
[Corrected, thanks – T.]
22 October, 2016 at 6:08 pm
Anonymous
Mathoverflow: A Generalization of Goldbach’s Conjecture : 2n=p+q,p≡q≡a(mod m)
http://mathoverflow.net/questions/252597/a-generalization-of-goldbach-s-conjecture-2n-pq-p-equiv-q-equiv-a-pmod
23 October, 2016 at 7:32 am
Anonymous
The Schwarz-Christoffel formulas in theorem 45 and exercise 46 are for bounded polygons. Are there similar formulas for unbounded polygons?
24 October, 2016 at 6:42 pm
Anonymous
Few typos
1. Second paragraph of section 1, “fundamental theorem of calculus” should be “fundamental theorem of algebra”.
2. Proof of Theorem 25, “concatenating it with the path “, the expression should be .
3. Proof of Proposition 40, Corollary 27 is from Notes 3, not 2 (the link is actually correct).
4. Same paragraph, “ converge locally uniformly” should be instead.
Thank you!
[Corrected, thanks – T.]
4 November, 2016 at 6:10 am
Anonymous
In proposition 40, is it possible to make the proof more constructive by using the explicit construction of the function from the function to give an explicit recursive construction of from ?
4 November, 2016 at 7:34 am
Terence Tao
In principle, yes, but the convergence of this algorithm is incredibly slow. Numerically, there are faster methods, such as the “Zipper” method of Kuhnau and Marshall, discussed for instance here.
4 November, 2016 at 8:11 pm
Joseph
Is it obvious somehow that the group of Moebius-transformations is a simple group? This is an exercise in Conway, preceded by an easy check that the kernel of a homomorphism from GL2(C) to the Moebius transformations under composition is all the multiples of the identity-matrix by the units in C, i.e. GL2(C) is not simple. So, identifying the group of Moebius trans. with GL2(C) seems to contradict this exercise (?)
5 November, 2016 at 6:20 am
Terence Tao
The Mobius transformations are identified with , not .
Saying a group is simple is equivalent to saying that every non-trivial conjugacy class generates the group, so one can establish the simpleness of the Mobius transforms through an understanding of the conjugacy classes. For instance, all parabolic Mobius transforms are conjugate to each other, and it is not difficult to see (e.g. by Gaussian elimination) that all Mobius transforms can be expressed as products of parabolic transforms. If instead one starts with a hyperbolic or elliptic element, one can find a conjugate of it which differs from it by a parabolic element, so from the previous fact we know that these conjugacy classes also generate the group. One can also proceed using the more general structural theory of Lie groups and Lie algebras, but this is probably overkill.
5 November, 2016 at 11:24 am
Joseph
Thanks. Is there a book out there that takes a more algebraic approach to complex analysis?
6 November, 2016 at 2:52 am
Heinrichs Gerhard
There are books about Riemannian Surfaces with more (commutative) algebra e.g. S. Lang Therory of algebraic and abelian Function’s
11 November, 2016 at 5:30 am
Anonymous
Can you add some information on more recent approaches (e.g. the circle packing approach) to conformal mapping ?
27 November, 2016 at 9:36 am
venkyclement
In Proposition 12, why doesn’t the strategy of taking local line segments and passing to a finite subcover of the Riemann sphere not work for an open curve? This proposition suggests that all space filling curves of a unit square must be open.
27 November, 2016 at 3:56 pm
Terence Tao
Both open and closed curves can be space-filling (in the unit square or in the Riemann sphere). What the proof of Proposition 12 shows is that either such curve can be deformed by continuous homotopy to a set which is locally the union of finitely many line segments, and is thus not space filling (but this still permits the original curve to be space-filling).
27 November, 2016 at 5:10 pm
nassoro mtandani
pro. tao i am so interested on how you solve mathematical problems i need to study math please helpme
12 February, 2017 at 7:04 am
Anonymous
One can tell from context though, it might be better to indicate explicitly that the unit disk in the hyperbolic model is an “open” disk?
13 February, 2017 at 11:10 am
Anonymous
This video http://www-users.math.umn.edu/~arnold/moebius/ seems to suggest that Möbius transformations are simplest when viewed on the sphere. But actually, that video only “defines” a Mobius transformation on the sphere and then project the image on the sphere back to the complex plane. Can one show (without using the Cartician coordinates tediously maybe) that it is actually the same as the linear fractional transformation on the complex plane?
13 February, 2017 at 4:54 pm
Terence Tao
This is perhaps simplest to see for special Mobius transformations such as translation, dilation, rotation, and inversion; general Mobius transformations are compositions of these (see discussion after Exercise 9), so once special Mobius transformations are identified with three-dimensional motions of the sphere, general Mobius transformations are also (because three-dimensional motions of Euclidean space form a group).
13 February, 2017 at 5:49 pm
Anonymous
In this note, what “inversion” really means is “reciprocation” while in geometry, inversion means ? (https://en.wikipedia.org/wiki/Inversive_geometry) It might be just a taste of terminology though. Or is there a higher point of view that they are actually the same thing?
17 February, 2017 at 9:20 am
Anonymous
Is this true that the Schwarz-Christoffel map in Theorem 45 maps the exterior of onto the complement of the polygon ?
17 February, 2017 at 10:01 pm
Terence Tao
I’m not sure there is any natural way to extend holomorphically to the entire complex plane (basically because of the branch cuts of that one would start encountering). Note that as the only automorphisms of the Riemann sphere are the Mobius transformations, one would only expect to conformally map both the interior and exterior of a domain in a Riemann sphere simultaneously to a disk and its exterior if the domain was already a disk, an exterior of a disk, or a half-plane.
18 February, 2017 at 12:59 am
Anonymous
And if we wish to construct a conformal map that maps onto the complement of the polygon , can we use the same Schwarz-Christoffel construction (replacing with the complement angles)?
[Yes – T.]
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8 May, 2018 at 5:40 am
Anonymous
In exercise 9, it seems that M should be connected. From the contradiction, I can just deduce that f is constant on f^{-1}(U’_β)∩U_α(this is defined in notes 4 definition 17) rather than on M. If I wanna expand the domain to M, I find that it’s necessary f^{-1}(U’_β)∩U_α intersects with another f^{-1}(U’_p)∩U_q, but without connectivity I can’t ensure f^{-1}(U’_β)∩U_α intersects with at least one another f^{-1}(U’_p)∩U_q. (if we mark f^{-1}(U’_i)∩U_j as E_ij， we can get M actually equals the union of all the E_ij, and M=E_pq ∪ (the union of the left E_ij), both of the right-hand is open, if M is connected, E_pq ∩ (the union of the left E_ij)!=Φ, E_pq can intersect with at least one another E_ij)
[Corrected, thanks – T.]
9 May, 2018 at 5:44 am
Anonymous
In exercise 22, M should be revised again(2 position in exercise 22, beginning and end).
[Corrected, thanks – T.]