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Dec 19, 2023 at 1:29 vote accept Kenny S
Dec 18, 2023 at 6:28 comment added Kenny S @AndyPutman, I try to write down the details of these, then I encounter some questions. "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. " I want to derive the homotopy classes of isometry and not just homeomorphism, when I want to use the Prop 1.B.9 in Hatcher's book, which says a proposition of K(G,1).
Dec 15, 2023 at 21:51 answer added Sam Nead timeline score: 3
Dec 15, 2023 at 3:36 comment added Kenny S @AndyPutman, Very good explanation! By the way, how can I compute the rep space? Why it's $char_{2}\pi_{1}(S)$?
Dec 15, 2023 at 3:08 comment added Andy Putman @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).
Dec 15, 2023 at 2:53 comment added Kenny S @AndyPutman, thank you for your comment, the first half of what you said I can fully understand, but why "$\pi_{1}(S)$ is identified with $\Gamma$" can be related to the markings? This fact seems established anytime when we have a universal covering and identify the fundamental group with deck transformation group. I can't see where the marking appears. May you explain in details when you have time? As an answer I'd prefer
Dec 15, 2023 at 2:14 comment added Andy Putman 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.
Dec 15, 2023 at 2:05 history edited Kenny S CC BY-SA 4.0
added 685 characters in body
Dec 15, 2023 at 1:45 comment added Kenny S @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!
Dec 15, 2023 at 1:43 comment added Kenny S @LeeMosher, sorry, I will add the definition in the question later. There are many equivalent definitions, I know what you said "In some treatments...", the definition you gave is indeed what higher Teichmüller theory comes from. But as a beginner, I wonder how the definition you gave(discrete faithful representations) agrees with the definitions that uses "marking Riemannian surface" (as I added later in the question)
Dec 14, 2023 at 14:56 comment added LSpice 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.
Dec 14, 2023 at 14:56 history edited LSpice CC BY-SA 4.0
Inlining image
Dec 14, 2023 at 14:16 comment added Lee Mosher 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.
Dec 14, 2023 at 10:17 history asked Kenny S CC BY-SA 4.0