Let me give some definitions to begin with.

Let $\Lambda$ be a compact orientable two-dimensional (over $\mathbb R$) manifold of genus $p>2$. Let $g$ be a conformal structure on $\Lambda$, and $f:\Lambda\to\Lambda$ a diffeomorphism, called a marking of $\Lambda$. Then we can form a triple $(\Lambda,g,f)$. We say two such triples $(\Lambda,g_1,f_1)$ and $(\Lambda,g_2,f_2)$ are Teichmuller equivalent if there exists a conformal map $$k:(\Lambda,g_1)\to(\Lambda,g_2)$$ for which $f_2\circ f_1^{-1}$ and $k$ are homotopic.

I am having some trouble understanding what role the marking $f$ plays in defining the Teichmuller equivalence.

(1) Before giving $\Lambda$ a conformal structure, we have already defined it to be a mainfold, i.e. it already has a differentiable structure. Also, as is known, a conformal structure gives rise to a differentiable structure, so I am wondering whether a conformal structure in our definition needs to be compatible with the original differentiable structure. Ideally if we can prove all conformal structures over $\Lambda$ have the same differentiable structure then I wouldn't have to ask this at all, but that doesn't seem to be the case.

(2) Suppose different conformal structures indeed determine different differentiable structures, then with respect to what structure is the marking $f$ a diffeomorphism? In other words, does the choice of $f$ depend on $g$?

(3) Is there any additional restriction on the homotopy between $f_2\circ f_1^{-1}$ and $k$? More specifically, do we need the intermediate mappings to be diffeomorphisms (or of some other nontrivial type) as well?

I am pretty new to these concepts so I am very confused about what role $f$ plays here. Any help is greatly appreciated!

  • $\begingroup$ I've provided an answer, but let me ask: What is your source for this definition? At first appearance this is a somewhat clunky definition; it seems to be an amalgam of two standard but distinct approaches to defining Teichmuller space. But on the other hand, perhaps this definition fits neatly between the two standard approaches and is useful for proving their equivalence. That's outside the scope of your question, though, hence this is only a comment. $\endgroup$
    – Lee Mosher
    Sep 3, 2019 at 16:25
  • $\begingroup$ @LeeMosher Thanks for your answer! I'll have to get back to it later when I have the time. This is taken from chapter 4 of the book Compact Riemann Surfaces by Jurgen Jost. The author mentions in the book that there indeed are two ways to define this equivalence, but he doesn't elaborate any further beyond this remark. $\endgroup$
    – trisct
    Sep 3, 2019 at 16:34

1 Answer 1


Regarding question (1), once the differentiable structure is specified then yes, the conformal structures are required to be compatible with the differentiable structure. There are in fact infinitely many inequivalent differentiable structures on $\Lambda$, and a conformal structure is compatible with exactly one of those structures. On the other hand, it sort of doesn't matter which differentiable structure you picked, because any two differentiable structures on $\Lambda$ are related by a homeomorphism $h : \Lambda \to \Lambda$ that is isotopic to the identity. As you trace through the definitions of the Teichmuller space of $\Lambda$, it follows that the Teichmuller space is (in some appropriate sense) independent of the choice of differentiable structure on $\Lambda$, because that homeomorphism $h$ can be used to canonically identify the Teichmuller spaces associated to the two given differentiable structures.

In short, feel free to pick the differential structure of your desire on $\Lambda$. In your context (i.e. whatever book or course or web site you are looking at), since the maps $f$ are required to be diffeomorphisms, it must be that the surface $\Lambda$ comes with a differentiable structure already given, either implicitly or explicitly. The maps $f$ occuring in this definition must be diffeomorphisms with respect to the given differentiable structure, and the formal structures $g$ must be compatible with that structure.

Regarding question (2), once you understand that there is an underlying differential structure on $\Lambda$ with respect to which each $g$ and $f$ are combatible, this answers the first subquestion of (2).

But your second subquestion in (2) does not quite make sense to me:

...does the choice of $f$ depend on $g$?

The wording of the definition you quote is slightly deficient, I would suggest that you refer to an ordered triple $(\Lambda,g,f)$ as a marked conformal structure on $\Lambda$. The fact is that the two components $g,f$ of a marked conformal structure may be independently chosen: any choice of conformal structure $g$ on $\Lambda$ and any choice of diffeomorphism $f$ of $\Lambda$ (subject to compatibility with the given underlying differentiable structure) defines a marked conformal structure $(\Lambda,g,f)$. So in that sense no, the choice of $f$ does not depend on $g$.

Regarding question (3), in some references you might indeed see the requirement strengthened to say that $f_2 \circ f_1^{-1}$ and $k$ are diffeotopic, which means isotopic so that the intermediate maps are diffeomorphisms. However, as it turns out, if I am given two diffeomorphisms $F,G : \Lambda \to \Lambda$, the following are equivalent:

  • $F,G$ are diffeotopic;
  • $F,G$ are isotopic;
  • $F,G$ are homotopic.

For the connection between isotopy and homotopy, a good reference is the paper by D.B.A. Epstein, "Curves on 2-manifolds and isotopies", Acta Math. 115 1966 83–107. I don't have a good reference for the connection between diffeotopy and isotopy, but the point is that every isotopy can be "smoothed"; there is probably some reference out there which proves this.

Final remark: You expressed confusion about the role of $f$. There's actually a good reason for your confusion, which is that $f$ is not necessary for formulating the definition of Teichmuller space:

Fact: Every marked conformal structure $(\lambda,g,f)$ is equivalent to another marked conformal structure $(\lambda,g',f')$ such that $f' = \text{Id}$.

To prove this, given $(\lambda,g,f)$, one defines $g' = (f^{-1})_*(g)$ (the "pullback" conformal structure). Now one checks the equivalence condition: does there exists a conformal map $k : (\Lambda,g,f) \to (\Lambda,g',f')$ such that $f' \circ f^{-1}$ is isotopic to $k$? Yes there does, namely $k = f' \circ f^{-1} = Id \circ f^{-1} = f^{-1}$. We just have to check that $k(\Lambda,g)=(\Lambda,g')$, which is equivalent to $k_*(g)=g'$ which is equivalent to $(f^{-1})_*(g)=g'$ which is how $g'$ was defined.

So, one could now write the following definition, which is more simply stated but is equivalent to the definition you quoted, where the equivalence proof applies the "Fact" written above:

Let $\Lambda$ be a compact orientable two-dimensional (over $\mathbb R$) manifold of genus $p>2$. Let $g$ be a conformal structure on $\Lambda$. Then we can form a pair $(\Lambda,g)$. We say two such pairs $(\Lambda,g_1)$ and $(\Lambda,g_2)$ are Teichmuller equivalent if there exists a conformal map $$k:(\Lambda,g_1)\to(\Lambda,g_2)$$ for which $k$ is homotopic to the identity.


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