7
$\begingroup$

Let $\Sigma_{g,n}$ denote an $n$-punctured surface of genus $g$, with $2g+n-2 > 0$. Let $\Pi_{g,n}$ be its fundamental group (for some choice of base point), and let $\Gamma_{g,n}$ denote its pure, orientation preserving mapping class group.

Then, by the Dehn-Nielsen-Baer theorem, the outer action of $\Gamma_{g,n}$ on $\Pi_{g,n}$ is faithful and induces an isomorphism between $\Gamma_{g,n}$ and the subgroup $\text{Out}^*(\Pi_{g,n})$ of $\text{Out}(\Pi_{g,n})$ which fixes the conjugacy class of each simple closed curve surrounding a puncture.

Question. When does $\text{Out}^*(\Pi_{g,n})$ have finite index inside $\text{Out}(\Pi_{g,n})$?

If $(g,\,n)=(1,\,1)$, then the index is $2$. Since $\Pi_{1,1} = \Pi_{0,3}$ and $\Gamma_{0,3}$ is trivial, one gets infinite index for $(g, \, n)=(0,\,3)$. What happens in the other cases? I'm particularly interested in the case where $g\ge 2, \, n = 1$.

$\endgroup$

2 Answers 2

6
$\begingroup$

This is surely not the most direct answer. But $\mathrm{Out}(\Pi_{g,1}) \cong \mathrm{Out}(F_{2g})$ surjects onto $\mathrm{GL}(2g,\mathbf Z)$, and the image of $\Gamma_{g,1}$ lands in $\mathrm{Sp}(2g,\mathbf Z)$. So the index is infinite for $g \geq 2$.


In fact a more careful version of this argument shows that $(g,n)=(1,1)$ is the unique case where you get a finite index subgroup. Let me flesh out the argument. For all $n>0$ we have similarly that $\mathrm{Out}(\Pi_{g,n}) \cong \mathrm{Out}(F_{2g+n-1})$ surjects to $\mathrm{GL}(2g+n-1,\mathbf Z)$. The corresponding representation $H$ of $\Gamma_{g,n}$ is just the action of the mapping class group on the first homology of your favorite genus $g$ surface $\Sigma$ with $n$ punctures. But the action must be compatible with a lot of extra structure coming from geometry: there is the short exact sequence $$ 0 \to \mathbf Z^{n-1} \to H_1(\Sigma,\mathbf Z) \to H_1(\overline \Sigma,\mathbf Z) \to 0 $$ and $\Gamma_{g,n}$ preserves it, where $\overline \Sigma$ is the compact surface obtained by filling in the punctures. Moreover, $H_1(\overline \Sigma,\mathbf Z)$ is of rank $2g$ and carries a symplectic form preserved by $\Gamma_{g,n}$; the action of $\Gamma_{g,n}$ on $\mathbf Z^{n-1}$ is trivial. These conditions define an infinite index subgroup of $\mathrm{GL}(2g+n-1,\mathbf Z)$ unless $(g,n)=1$.

$\endgroup$
2
  • 1
    $\begingroup$ You really don't need mixed Hodge structures to see this!!! $\endgroup$
    – HJRW
    Apr 28, 2021 at 14:52
  • 1
    $\begingroup$ @HJRW A very fair point, edited. $\endgroup$ Apr 28, 2021 at 14:57
7
$\begingroup$

It only has finite index in very low-complexity degenerate cases.

Here's a proof that it always has infinite index for $\Sigma_{g,1}$ with $g \geq 2$. This proof generalizes in an obvious way to deal with all the other cases too, but the notation gets worse.

Set $\pi = \pi_1(\Sigma_{g,1})$. Let $\{a_1,b_1,\ldots,a_g,b_g\}$ be the usual generating set for $\pi$ and let $$\omega = [a_1,b_1] \cdots [a_g,b_g] \in \pi$$ be the surface relation. It is enough to construct a sequence of elements $f_1,f_2,\ldots$ in $\text{Aut}(\pi)$ such that the conjugacy classes of the $f_k(\omega)$ are all distinct. This will immediately imply that the outer automorphisms associated to the $f_k$ are all in different mapping class group cosets.

In fact, you can take the $f_k$ to be powers of almost any random automorphism you write down. For instance, let $f_k$ be the automorphism defined by the formula $$f_k(a_1) = a_1 b_1^k \quad \text{and $f_k$ fixes all other generators},$$ so $f_k$ is the $k$th power of $f_1$. Letting exponentiation denote conjugation, we then have \begin{align*} f_k([a_1,b_1]) &= [a_1 b_1^k, b_1]\\ &= b_1^{-k} a_1^{-1} b_1^{-1} a_1 b_1^k b_1\\ &= b_1^{-k} a_1^{-1} b_1^{-1} a_1 b_1 b_1^k\\ &= [a_1,b_1]^{b_1^k}. \end{align*} and thus $$f_k(\omega) = [a_1,b_1]^{b_1^k} [a_2,b_2] \cdots [a_g,b_g].$$ As long as $g \geq 2$, these are all distinct conjugacy classes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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