3
$\begingroup$

$\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\char{char}$I am reading a note on Teichmüller space, and I come across a somewhat algebraic problem in the picture below, which may be easy to experts.

Teichmüller space $T_g$ is naturally a manifold homeomorphic to $\mathbb R^{6g - 6}$, and $\Gamma_g$ acts properly discontinuously on $T_g$. Thus, $M_g$ is an orbifold with $\pi_1^\text{orb}(M_g) = \Gamma_g$.

We are able to see the topology in two ways:

By Representation theory:

$$T_g \hookrightarrow \Hom(\pi_1(S), \PSL_2(\mathbb R))/\PSL_2(\mathbb R) = \char_2(\pi_1(S)),$$

where the image of $T_g$ is the open subset of discrete and faithful representations. A simple counting argument shows

$$\dim(\Gamma_g) = \dim \char_2(G) = (2g - 1)*3 - 3 = 6g - 6.$$

By Fenchel–Nielson Coordinates:

The definition of the Teichmüller space $T_g$ I used here is the following (the "isotopic" can be replaced by "homotopic" and the "homeomorphism" can be replaced by "orientation-preserving diffeomorphism" since any homeomorphism is homotopic to an diffeomorphsim between surfaces and the resulting space is the same ):

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:

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

I wonder how to understand this injective map:

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

explicitly?

(All I can understand is just an injective map from the fundamental group to $\operatorname{Aut}(\mathbb{H})$ just like what the author did here: Teichmuller space as Discrete Faithful Representations up to Conjugation.)

I'm also wonder how to compute $\Hom(\pi_1(S), \PSL_2(\mathbb R))/\PSL_2(\mathbb R)$, why it's $\char_{2}(\pi_{1}(S))$?

And how can I compute $\char_{2}(\pi_{1}(S))$?

I find something that looks just like the right hand of the mapping, which is called the representation of surface group, but I can't find a formal explanation or proof about how can it be related with Teichmüller space. Can anybody help me? Any advice or comment is welcome.

Improve this question
$\endgroup$
9
  • 5
    $\begingroup$ Your question is hard to answer without knowing how you want to define Teichmüller space, because your question is very close to the level of the very definition of $T_g$. In some treatments, $T_g$ is actually defined to be the subset of $Hom(\pi_{1}({S_g}),PSL_{2}(\mathbb{R}))/PSL_{2}(\mathbb{R})$ represented by faithful representations $\pi_{1}({S_g}) \mapsto PSL_{2}(\mathbb{R})$ with discrete image. $\endgroup$
    – Lee Mosher
    Commented Dec 14, 2023 at 14:16
  • 1
    $\begingroup$ What is the note? I also took the time to inline the image, and correct a typo ($6g - g$ was supposed to be $6g - 6$). Whenever possible, it is best to make your posts self-contained, not relying on an image (which, among other things, is not searchable) for textual content. $\endgroup$
    – LSpice
    Commented Dec 14, 2023 at 14:56
  • 2
    $\begingroup$ @LSpice, thank you so much for your edition of my question, I will remember next time to inline the image and make the question more self-contained! The note is written by Jing Tao and the name of the note is "Introduction to Teichmuller spaces", I somehow downloaded it but right now I can't find the original links, I will add the link of note later when I find it! $\endgroup$
    – Kenny S
    Commented Dec 15, 2023 at 1:45
  • 3
    $\begingroup$ If you have a marked Riemann surface $S$ of genus $g \geq 2$, the uniformization theorem says that its universal cover is biholomorphic to the unit disc $D$. It's a fun exercise in complex analysis that the group of biholomorphisms of $D$ is $\text{PSL}_2(\mathbb{R})$. This shows that $S \cong D / \Lambda$ where $\Lambda$ is a discrete subgroup of $\text{PSL}_2(\mathbb{R})$. The marking on $S$ lets you identify $\pi_1(S)$ with $\Lambda$ (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. $\endgroup$
    – Andy Putman
    Commented Dec 15, 2023 at 2:14
  • 2
    $\begingroup$ @KennyS: 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). $\endgroup$
    – Andy Putman
    Commented Dec 15, 2023 at 3:08

1 Answer 1

Reset to default
3
$\begingroup$

The notation $\mathrm{char_2(\pi_1(S))}$ is pronounced "the variety of characters of representations of $\pi_1(S)$ into $\mathrm{PSL}(2, \mathbb{C})$". This is also called the character variety.

You ask about the equality between $\mathrm{Hom}/\mathrm{PSL}$ and $\mathrm{char_2(\pi_1(S))}$. This is "just" the definition of the character variety - the left hand side defines the right hand side.

The injection you are asking about (namely from "the set of marked Riemann surfaces, up to biholomorphism making a certain square commute" into the character variety) follows from the uniformisation theorem.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Thank you for your answer! Forgive me that I'm not familiar with "character variety", can you tell me how to calculate the dimension of it for arbitrary fundamental groups of Riemann Surfaces of genus g? why it's 6g-6? I search this on wiki and the calculation is given only in special cases. Maybe a reference includes this calculation is enough if you think the computation is too complex to type here. $\endgroup$
    – Kenny S
    Commented Dec 17, 2023 at 0:54
  • $\begingroup$ Just a few points of etiquette: if my answer answers your original question, then you should accept it (by clicking on the "check" mark). If you have another question (for example, how to compute the dimension of the character variety), then you should ask that in a new question. $\endgroup$
    – Sam Nead
    Commented Dec 18, 2023 at 9:37
  • 1
    $\begingroup$ But the short answer to the question "why is the dimension 6g - 6?" is as follows. The surface has 3g - 3 curves in any pants decomposition. Every curve in a pants decomposition contributes two real parameters: its length and its twist. See the so-called Fenchel–Nielsen coordinates for Teichmuller space (which is homeomorphic to a component of the character variety). $\endgroup$
    – Sam Nead
    Commented Dec 18, 2023 at 9:42
  • $\begingroup$ I should have said “one short answer to the question”, as there are other computations of the dimension. $\endgroup$
    – Sam Nead
    Commented Dec 18, 2023 at 10:46
  • $\begingroup$ Okay I have cheaked your answer. I know the FN coordinates, but I think it's another way to derive the dimension, as you saw at the very beginning of my question, I assumed that the representation way to express the Teichmuller space can compute the dimension purely algebraically, without use of FN coordinates. Now I feel the representation definition is more like a "formal" or "abstract" definition with no practical use. $\endgroup$
    – Kenny S
    Commented Dec 19, 2023 at 1:35

You must log in to answer this question.

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