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It is known that the mapping class group of the torus $\mathbb{T}^2$ is $\text{Mod}(\mathbb{T}^2) \cong \text{SL}_2(\mathbb{Z})$. We also know that for a pair of pants $P$ (a sphere with three boundary components), $\text{Mod}(P) \cong \mathbb{Z}^3$. Let $S$ be a pair of pants with a hole in it, i.e. a torus with three boundary components. Then what is $\text{Mod}(S)$? I think it should be something involving the above groups but I'm not sure how to go about finding it.

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  • $\begingroup$ What kind of description do you want? There isn't really something quite as concrete as those two groups. $\endgroup$
    – user29123
    Nov 20, 2019 at 21:06
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    $\begingroup$ It is a certain central extension of $PSL(2,Z)$ via $Z^4$. Is this what you are asking for? $\endgroup$ Nov 20, 2019 at 21:10
  • $\begingroup$ Just to verify, when $S$ is a compact surface with nonempty boundary, you are using $\text{Mod}(S)$ to denote the group of isotopy classes of homeomorphisms that restrict to the identity on the boundary, modulo isotopies which leave the boundary points stationary. Is that correct? $\endgroup$
    – Lee Mosher
    Nov 20, 2019 at 21:15

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I'll assume you are defining mapping class groups of compact surfaces with nonempty boundary as in my comment.

The answer is that it's complicated, but, when $S$ is the torus with three boundary components then $\text{Mod}(S)$ can be described as a certain multiple extension group of $\text{SL}_2(\mathbb Z)$.

The first step is to write $\text{Mod}(S)$ as a certain central extension as follows. By collapsing each component of $\partial S$ to a point we obtain the torus $\mathbb T^2$ back again, together with a 3 point subset $P=\{p_1,p_2,p_3\} \subset \mathbb T^2$ in one-to-one correspondence with the components of $\partial S$. Let $\text{Mod}(\mathbb T^2;P)$ denote the mapping class group of homeomorphisms of $\mathbb T^2$ that fix each point of $P$, modulo isotopies that leave each point of $P$ stationary. Then we get a certain central extension $$1 \mapsto \mathbb Z^3 \mapsto \text{Mod}(S) \mapsto \text{Mod}(\mathbb T^2;P) \mapsto 1 $$ That's not a complete description, because to completely determine the central extension requires more information. But I'll shove that under the rug and continue.

For the second step, one uses the Birman short exact sequence to remove the special nature of the point $p_3$, obtaining a certain short exact sequence $$1 \mapsto \underbrace{\pi_1(\mathbb T^2 - \{p_1,p_2\})}_{\text{free of rank 3}} \mapsto \text{Mod}(\mathbb T^2;\underbrace{\{p_1,p_2,p_2\}}_{P}) \mapsto \text{Mod}(T^2;\{p_1,p_2\}) \to 1 $$ Again........ this is not a complete description without giving further information to determine the extension.........

For the third step, one again uses the Birman short exact sequence to remove the special nature of the point $p_2$, obtaining another short exact sequence $$1 \mapsto \underbrace{\pi_1(\mathbb T^2 - \{p_1\})}_{\text{free of rank 2}} \mapsto \text{Mod}(\mathbb T^2;\{p_1,p_2\}) \mapsto \text{Mod}(\mathbb T^2;\{p_1\}) \mapsto 1 $$ Again........ (and it's getting crowded under that rug) ........

For the final step, one uses the fact that $\mathbb T^2$ is a Lie group to obtain an isomorphism $$\text{Mod}(\mathbb T^2;\{p_1\}) \approx \text{Mod}(\mathbb T^2) \approx \text{SL}_2(\mathbb Z) $$

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