Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $S$ be an orientable compact surface. A homeomorphism $f: S \to S$ induces an isomorphism $f_{*}: H_1(S) \to H_1(S)$.

How much can we say the converse? Namely, if we are given an element of $\alpha \in$ $\operatorname{Aut}(H_1(S))$, is there a self-homeomorphism $f$ of $S$ (unique up to isotopy or something) such that $f_*=\alpha$?

share|improve this question
add comment

3 Answers 3

up vote 4 down vote accepted

The set of self-homeomorphisms of a surface up to isotopy is called the mapping class group of the surface. For genus one, this is determined by the action on homology, but for higher genus there is a very large and interesting subgroup of the mapping class group, called the Torelli group, which consists of self-homeomorphisms inducing the trivial map on homology.

share|improve this answer
This is interesting. Can you give me an example of an element of the Torelli group for genous 2 surface? –  Primo Apr 8 '12 at 22:37
Yes. There is a map called a Dehn twist en.wikipedia.org/wiki/Dehn_twist which is supported in a neighborhood of a simple closed curve. A Dehn twist supported on a nontrivial null-homologous curve is in the Torelli group. (E.g. a curve that separates the surface into two genus one surfaces.) –  Grumpy Parsnip Apr 8 '12 at 23:49
@Primo Forgot to ping you. –  Grumpy Parsnip Apr 9 '12 at 0:03
Thank you. I understand. –  Primo Apr 9 '12 at 0:17
add comment

The answer depends on what you mean by $\mathrm{Aut}(H^1(S))$. If you mean general linear group, then the answer is no, but if you mean the symplectic group, then the answer is yes.

There is a cup product pairing $H^1(S) \times H^1(S) \to H^2(S) \cong \mathbf Z$ which is symplectic, and any automorphism of the surface preserves this pairing (up to a sign, if you don't require it to preserve the orientation of $S$). This is the only condition: any symplectic automorphism can be realized by a homeomorphism of the surface.

See also http://en.wikipedia.org/wiki/Mapping_class_group#Torelli_group

share|improve this answer
add comment

Here's an example without the assumption that $S$ is a surface:

Let $S = S^1 \vee A$ where $A$ is a closed annulus. Then $H_1(S) = \mathbb Z \oplus \mathbb Z$. Take the isomorphism $\varphi: (a,b) \mapsto (b,a)$.

A homeomorphism $f$ inducing $\varphi$ would have to map $x$ in $S^1$ to $f(x) \in A$. But $S^1$ and $A$ are not homeomorphic.

share|improve this answer
My assumption of $S$ is an orientable compact surface. Can you construct similar example with this assumption? –  Primo Apr 8 '12 at 21:06
@Primo Oh, sorry, I missed that somehow. Let me try to come up with a different example. –  Matt N. Apr 8 '12 at 21:24
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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