## Holomorphic Maps and Riemann Surfaces (Pt. IV)

**Point of Post: **This is a continuation of this post.

The last theorem we prove tells us that biholomorphisms are really just bijective holomorphic maps–in other words, if a holomorphic map has a set theoretic inverse, this inverse is necessarily holomorphic.

**Theorem(Holomorphic Inverse Theorem): ***Let be a bijective holomorphic map, then is a biholomorphism.*

**Proof: **Let be arbitrary and let be such that . Choose charts around and around such that . We need to check that is holomorphic on . That said, by definition we know that is holomorphic , and since since evidently we know from the Holomorphic Inverse Theorem for domains in that is holomorphic. Since was arbitrary the conclusion follows.

An interesting corollary of this is the following, very surprising, characterization of biholomorphisms between compact Riemann surfaces:

**Corollary: ***Let and be compact Riemann surfaces. Then, a map is a biholomorphism if and only if it’s holomorphic and injective.*

**Holomorphic Functions**

As promised, we now specialize the above discussion to mappings where the codomain is . In particular, we define a *holomorphic function* on a Riemann surface to be a holomorphic mapping . We use the notation (or if is a domain inside a Riemann surface with the induced structure) for .

As promised in the motivation we shall now show one of the indications that compact and non-compact Riemann surfaces are fundamentally different:

**Theorem: ***Let be a compact Riemann surface. Then, every holomorphic function on is constant.*

**Proof: **Let , and suppose that is non-constant, then by previous theorem we see that this implies that is surjective and thus compact, a contradiction.

While this theorem is easy to prove, an observation based off of it actually can lead one to very deep intuition about the differences between compact and non-compact Riemann surfaces (if one is willing to take a little bit of knowledge on faith!). Namely, we shall prove later that every Riemann surface admits a non-constant meromorphic (to be defined, but should be clear–a holomorphic function with poles) function. Now, don’t try to prove this on your own. Don’t scoff at it either! This is one of the deepest, most powerful, and most difficult theorems to prove in all of (basic) Riemann surface theory (it will occupy a vast amount of our time!). Most of the theorems we shall prove generalize to higher dimensions, but this is truly a theorem of dimension one! Regardless, this existence allows us to notice something strange.

Regardless of me telling you that the existence of non-constant meromorphic functions is amazing, it really does inspire a ‘so what’? Why does this tell us anything important? Well, we shall prove in a soon-to-come post that, much like in the case of domains in , is a ring for any Riemann surface, but more specifically, is an integral domain (this is where the term ‘domain’ in integral domain came from, by the way–integral still seems to be a mystery[at least to me]). Moreover, when we define meromorphic functions on a Riemann surface we shall be able to prove that, much like in the case of domains in , one has that the set of mermorphic functions on is a field. Now, since is an integral domain, we can consider the field of fractions . Well, then the above theorem tells us that if is compact then,

Once again, so what? Well, combining these two ‘so what’ statements gives us the following fact: if is compact, then

Well, depending on your point-of-view we may have just cashed in two ‘so what’ statements for another ‘so what’ statement. Ah, but here is where you are wrong! Remember that the statement is true for domains . The proof of this relied heavily on the Weierstrass product theorem, which allowed us to create entire functions with a specified (discrete!) zero set and specified multiplicities (we basically multiplied by an entire function whose zeros with multiplicity matched the poles with multiplicity, and then divided both sides by it). Thus, we see that for compact is an indication that there isn’t a Weierstrass product theorem for compact Riemann surfaces.

This tells us that we can’t just pick, willy-nilly, and set of zeros with specified multiplicties and find holomorphic functions on compact with them. This tells us that compact are less malleable, their structure more rigid. This intuition about compact Riemann surfaces is one that will hold true throughout all of our studies, and should be the first indication (before the big categorical justification of the intuition!) as to why compact Riemann surfaces are more algebraic–algebraic things are rigid. Conversely, we shall eventually prove that, spitting in the face of compacts, non-compact Riemann surfaces do have , which is because non-compacts do have a Weierstrass theorem. This allows us to intuit that non-compacts are more malleable, more analytic.

Ok, coming back from rants we can, as was alluded to in the motivation, actually find analogs in holomorphic maps on Riemann surfaces for every holomorphic mapping property between domains in . In particular, we can prove the big one that was left out of the discussion above: the maximum modulus principle:

**Theorem: ***Let be a Riemann surface, and . Then, if achieves its maximum on (i.e. for some ) then is contant.
*

**Proof: **Suppose that achieves its maximum at . Choose a coordinate disc centered at . We see then that is a holomorphic mapping on which achieves its maximum modulus at . Thus, by the Maximum Modulus Principle for domains this implies that is contant on and so is constant on . Since is open it follows from the Identity Theorem that is actually constant on all of .

This allows us to give another proof that consists only of constants for compact . Indeed, since is compact we know that achieves its maximum on for any , from where the Maximum Modulus Principle implies that is constant.

The last theorem we currently prove is the adaptation of Riemann’s Removable Singularity theorem to holomorphic functions on Riemann surfaces:

**Theorem(Riemann’s Removable Singularity): ***Let be a Riemann surface and suppose that and is bounded in some punctured neighborhood of . Then, then admits an extension .*

**Proof: **Let be the punctured neighborhood for which is bounded on. Now, choose a coordinate disc centered at such that . Then, we know that is bounded on and holomorphic, and thus by Riemann’s Removable Singularity theorem, admits a holomorphic extension . Define on by and . Clearly then is holomorphic at every point of and is holomorphic at since choosing the neighborhood one sees that and thus is holomorphic.

**References:**

[1] Varolin, Dror. *Riemann Surfaces by Way of Complex Analytic Geometry*. Providence, RI: American Mathematical Society, 2011. Print.

[2] Miranda, Rick. *Algebraic Curves and Riemann Surfaces*. Providence, RI: American Mathematical Society, 1995. Print.

[3] Forster, Otto. *Lectures on Riemann Surfaces*. New York: Springer-Verlag, 1981. Print.

[4] Conway, John B. *Functions of One Complex Variable*. New York: Springer-Verlag, 1978. Print.

[5] Gong, Sheng. *Concise Complex Analysis*. Singapore: World Scientific, 2001. Print

[6] Rudin, Walter. *Real and Complex Analysis.* New York: McGraw-Hill, 1966. Print.

[…] singularity' makes sense since evidently if then is bounded in a neighborhood of and so Riemann's Removable Singularity Theorem allows us to holomorphically extend to . In particular, we see that is holomorphic at if and […]

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