Disclaimer - I don't have much experience in topology/complex geometry, so I apologize if what I'm asking is too elementary for this site.

Let $S$ be some orientable surface obtained by removing finitely many points from a closed surface.

The Riemann moduli space for $S$ is just the set of isomorphism classes of complex structures. In other words, it's the quotient of the set of all complex structures on $S$, by the equivalence relation where two complex structures $\mu,\mu'$ on $S$ are equivalent if there is an homeomorphism $f : S\rightarrow S$ such that $\mu\circ f = \mu'$.

According to Wikipedia, the Teichmuller space for $S$ can be described as also the quotient of the set of all complex structures on $S$ by the same equivalence relation, except where the $f$ is required to be isotopic to the identity.

In Lochak's paper here: http://www.math.jussieu.fr/~lochak/textes/curves.pdf

he says that points of the Teichmuller space can be described by pairs $(X,f)$ where $X$ is a Riemann surface (ie, a complex structure on $S$), and $f : S\rightarrow X$ is a diffeomorphism, and that $(X,f)\sim(X',f')$ if $f'\circ f^{-1}$ is homotopic to an isomorphism (ie, a biholomorphic map) from $X\rightarrow X'$.

My first question is - In this case, is $f'\circ f^{-1}$ being homotopic to an isomorphism equivalent to it being isotopic to one?

My second question is - Why are these two definitions equivalent? I can't seem to see why having an isomorphism $X\rightarrow X'$ that is isotopic to the identity is equivalent to $f'\circ f^{-1}$ being homotopic to an isomorphism.

In fact this just seems false. Ie, consider the case where $X = X'$ as Riemann surfaces, and pick markings $f,f'$ such that $f'\circ f^{-1}$ is not isotopic to the identity, then the equivalence of the two definitions would seem to imply that there are isomorphisms between $X$ and $X'$ that are isotopic to the identity, and ones that are not, which seems strange to me.

  • 3
    $\begingroup$ A small quibble: in defining the modular space and the Teichmuller space for $S$ one considers only those complex structures on $S$ for which the missing points are removable singularities, i.e. those induced by restriction of some complex structure on the closed surface from which $S$ was obtained by removing finitely many points. This is needed, for example, in the proof that the Teichmuller space of $S$ is homeomorphic to a ball of dimension $6g-6+2p$ where $g=$genus and $p=$number of points removed. $\endgroup$
    – Lee Mosher
    Jun 20, 2013 at 14:22

2 Answers 2

  1. D.B.A. Epstein proved in his paper "Curves on 2-manifolds and isotopies", Acta Math. 1966, that homotopy implies isotopy for homeomorphisms of surfaces of finite type.

  2. There are conformal automorphisms of Riemann surfaces which are not isotopic to the identity and there is the trivial automorphism, which is the identity (you can aso use rotations of tori as trivial examples). It is a corollary of Hurwitz theorem that a conformal automorphism of a hyperbolic surface is homotopic to identity iff it is the identity. The easiest examples of nontrivial automorphisms are hyperelliptic involutions, but there are many more. Just think about covering transformations of a finite cover $S'\to S$.


For completeness I'll record here the answer to my question of why the two definitions are equivalent.

Let $S$ be the underlying topological space of some Riemann surface of finite type. Let $\varphi,\varphi'$ denote two complex atlases on $S$, and let $X = (S,\varphi), X' = (S,\varphi')$ denote the associated Riemann surfaces. First consider the case where the markings are given by identity maps - that is,

$f : S\rightarrow X$ and $f' : S\rightarrow X'$ are both the identity map on underlying topological spaces. Then, it's a tautology that $f'\circ f^{-1}$ is homotopic/isotopic to a biholomorphism $X\rightarrow X'$ if and only if there is a biholomorphism $X\rightarrow X'$ homotopic/isotopic to the identity.

On the other hand, any marking $f : S\rightarrow X$ is equivalent to one where the map is the identity on underlying topological spaces. To see this, suppose $f : S\rightarrow X = (S,\varphi)$ is any marking, then we may pull back the atlas $\varphi$ on $X$ to an atlas $\varphi\circ f$ on $S$, and compare the marking $f$ with the marking $\text{id} : S\rightarrow (S,\varphi\circ f)$. These are clearly equivalent since $f\circ\text{id}^{-1} : (S,\varphi\circ f)\rightarrow (S,\varphi)$ respects the complex structures.

Note: I found this proof in section 2.1.4 of the book "Handbook of Teichmuller Theory, Volume 1".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.