2
$\begingroup$

This is a more detailed question about my first question Representation theory and topology of Teichmüller space, I asked there how to understand:

$$T_{g}\hookrightarrow Hom(\pi_{1}({S}),PSL_{2}(\mathbb{R}))/PSL_{2}(\mathbb{R})$$

below is the definition of Teichmüller space I used.

Suppose S is a topological surface of genus $g\geq 2$. A marked Riemann surface $(X, f)$ is a Riemann surface X together with a homemorphism $f:S\rightarrow X$. Two marked surfaces $ (X,f)\sim (Y,g)$ are equivalent if $gf^{-1}: X\rightarrow Y$ is isotopic to an isomorphism(biholomorphism). The Teichmüller space $T_g$ is given by:

$$T_g=\{(X,f)\}/\sim$$

Thanks to @Andy Putman, where he gave some hints in the comments. Let me summarize here, below S is always a hyperbolic Riemann surface of genus at least 2:

The group of biholomorphisms of D is $PSL_{2}(\mathcal{ℝ})$. This shows that $𝑆\cong 𝐷/\Gamma$, where $\Gamma$ is a discrete subgroup of $PSL_{2}(\mathcal{ℝ})$. The marking on 𝑆 lets you identify $\pi_{1}(𝑆)$ with $\Gamma$(up to conj due to basepts), so we get a point in your rep space. Reversing this process gives you the inverse map from the rep space to Teichmuller space.

About how markings of S identify $\pi_{1}(𝑆)$ with $\Gamma$(up to conj due to basepts). If 𝑆 and 𝑆′ are two surfaces of genus $𝑔\geq 1$, then they're both $𝐾(\pi,1)$'s for their fundamental group, so the set of homotopy classes of homeomorphisms from 𝑆 to 𝑆′ is the same as the set of conjugacy classes of isomorphisms between their fundamental groups. Thus a marking of a surface of genus at least 1 is the same as an identification of its fundamental group with that of your fixed reference surface (up to conjugacy)

If I don't understand wrongly, he used the property of $K(G,1)$ here, namely Proposition 1B.9 in Hatcher's book "Algebraic topology".

Let X be a connected CW-complexes, and Y be a K(G,1). Then any homomorphism $\pi_{1}(X,x_{0})\rightarrow \pi_{1}(Y,y_{0}) $ is induced by a map $(X,x_{0})\rightarrow (Y,y_{0})$, which is unique up to homotopy fixing $x_{0}$

What I want to show is the marking, which is the homotopy classes of biholomorphism of hyperbolic Riemann Surfaces, identify with $Hom(\pi_{1}({S}),PSL_{2}(\mathbb{R}))/PSL_{2}(\mathbb{R})$. So my questions are:

  1. Are homotopy classes of homeomorphism between hyperbolic Riemann Surfaces the same thing as homotopy classes of biholomorphsim between hyperbolic Riemann Surfaces?

  2. I want to prove if (X, f)~(Y, g), then their corresponding representations are conjugate in $PSL_2(\mathbb{R})$. The representation corresponding to (X, f) is induced from f by:

$$\pi_{1}(S)\stackrel{f_{*}}\rightarrow \pi_{1}(X)\cong \Gamma_{X}\subset Aut(D)$$

If the answer to question 1 is right, how can I show "if (X, f)~(Y, g), then their corresponding representations are conjugate in $PSL_2(\mathbb{R})$" using "the set of homotopy classes of homeomorphisms from 𝑆 to 𝑆′ is the same as the set of conjugacy classes of isomorphisms between their fundamental groups"?

In other words, we have known that:

$$\pi_{1}(S)\stackrel{f_{*}}\rightarrow \pi_{1}(X)\cong \Gamma_{X}\subset Aut(D)$$ and:

$$\pi_{1}(S)\stackrel{g_{*}}\rightarrow \pi_{1}(Y)\cong \Gamma_{Y}\subset Aut(D)$$

are two representations of $\pi_{1}(S)$ on D, then the set $Hom(\pi_{1}({S}),PSL_{2}(\mathbb{R}))/PSL_{2}(\mathbb{R})$ is bijective to the set of all the group isomorphism:

$$h:\pi_{1}(X)\rightarrow \pi_{1}(Y)$$

up to conjugacy?

P.S: Maybe the second question is easy to answer because right now I'm stuck in these definitions and maybe I ignore something obvious.

$\endgroup$

1 Answer 1

6
$\begingroup$
  1. No, certainly not. Biholomorphic maps are very rare and very rigid. Generically, given Riemann surfaces $X$ and $Y$, there will be no biholomorphic maps between them.

  2. Yes. You will want to read up on the "developing map" derived from a hyperbolic structure, and its associated "holonomy representation". These are part of Thurston's more general notion of $(G, X)$ structures. You can find a discussion of the history and many references in this mathoverflow question.

$\endgroup$
1
  • $\begingroup$ Thank you for answering my question again! As to the first question I asked, your answer actually tells me that Prop 1B.9 in Hatcher's book is not enough right? According to your answer of my second question, even in this case, I must asked for Thurston's (G,X) structures to give a detailed proof...After all, thank you for providing another aspect to view this problem! $\endgroup$
    – Kenny S
    Commented Dec 20, 2023 at 1:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .