Index of the mapping class group $\Gamma_{g,n}$ inside $\text{Out}(\Pi_{g,n})$

Question

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 $g\ge 2, \, n = 1$.

Answer 1

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 $g \geq 2$.

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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 $(g,n)=1$.

Answer 2

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.