0
$\begingroup$

Let $\Sigma_n$ and $\Sigma_m$ be two closed oriented surfaces of genus $n$ and $m$, with $n \leq m$. We may think about these surfaces as connected sums of tori, so there is an canoical inclusion map $j: \Sigma_n \to \Sigma_m$.

Furthermore, we know that the fundamental group of a closed oriented surface $\Sigma_g$ of genus $g$ is given by \begin{equation} \pi_1(\Sigma_g)= \langle x_1,y_1,\ldots,x_g,y_g \mid \prod_{i=1}^g[x_i,y_i]=e \rangle \end{equation}

My question is: how does the induced map $\pi_1(j)$ look like on fundamental groups? I do have trouble visualizing how the fundamental polygon of $\Sigma_n$ lies in the fundamental polygon of $\Sigma_m$. Any help would be appreciated.

$\endgroup$
  • 5
    $\begingroup$ What's the canonical inclusion? $\endgroup$ – Dan Rust Jul 31 '14 at 13:32
  • 1
    $\begingroup$ To echo @DanielRust, it is not surprising that you are having trouble visualizing something that does not exist. $\endgroup$ – Lee Mosher Jul 31 '14 at 13:38
  • $\begingroup$ In fact I'm pretty sure the only maps $\Sigma_n\to\Sigma_m$ with $n<m$ are homotopic to the constant map. This can be shown using a cohomology argument and considering the cup product. $\endgroup$ – Dan Rust Jul 31 '14 at 13:38
  • $\begingroup$ It's not that easy to visualize maps between surfaces of genus 2 or more. One way of generating examples is to look at congruence subgroups in arithmetic groups in SL(2,R) but basically it's a world very different from tori. $\endgroup$ – Mikhail Katz Jul 31 '14 at 14:21
  • $\begingroup$ Thank you all. Looks like I was a bit confused. @Daniel Rust: Could you sketch the argument that the only maps $\Sigma_n \to \Sigma_m$ are homotopic to the constant map or give a reference? $\endgroup$ – Hierkommtdiemaus Jul 31 '14 at 15:35
1
$\begingroup$

There are no canonical inclusions among the $\Sigma_n$. On the other hand there are canonical projections from $\Sigma_n$ to $\Sigma_m$ when $n>m$. Here the induced map in $\pi_1$ simply sends the extra generators to the identity element.

$\endgroup$
  • $\begingroup$ Thanks! How do these canonical projections look like on the level of spaces? $\endgroup$ – Hierkommtdiemaus Jul 31 '14 at 14:51
  • 1
    $\begingroup$ You "pinch off" the superfluous handles. $\endgroup$ – Mikhail Katz Jul 31 '14 at 14:54
0
$\begingroup$

For at least "a little bit canonical" maps beween closed oriented surfaces, you need the contrary $n\geq m$. (you tried to imagine a map from a torus to $\Sigma_g$ which didn't work out)

(In the following, by component I mean the $i$-th part of the surface homeomorphic to a torus with $0,1$ or $2$ discs removed.)

A map $\Sigma_n \to \Sigma_m$ with $n \geq m$ could map some components canonically and some constant. In such a case you always get mapped the two generators of the $i$-th component canonically to the $j$-th component, and the two generators from the collapsed parts will be mapped to zero. This would correspond to a map $\mathbb Z^n \to \mathbb Z^m$ with corresponding matrix

$$(I_{m\times m} | 0 )$$

(note that this map has to look like this, in particular must have rank equal to $m$)

where you choose the basis of $\mathbb Z^n$ so that all canonical generators corresponding to a collapsed component of $\Sigma_n$ are appended in the end.

$\endgroup$
  • $\begingroup$ What do you mean by "This would correspond to a map $\mathbb{Z}^n\to\mathbb{Z}^m$"? $\endgroup$ – Dan Rust Jul 31 '14 at 14:08

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.