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I study braid groups and algebraic geometry, and I learned about the mapping class group (MCG).

Specifically, I encountered the MCG of the punctured unit disk (assume n punctures). I understand that the MCG of the punctured disk is a subgroup of the bijective diffeomorphisms group from the punctured disk to itself, and that diffeomorphisms are homotopic to the identity diffeomorphism on the boundary. How can I show that this subgroup is a normal subgroup of the bijective diffeomorphisms group?

If I name the diffeomorphisms group K, and the normal subgroup S, is the MCG K/S? If this is true, how do I prove this?

Thanks!

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Every topological group $G$ fits into a short exact sequence

$$1 \to G_0 \to G \to \pi_0(G) \to 1$$

where $\pi_0(G)$ is the group of connected components and $G_0$ is the connected component of the identity. In particular, the fact that $G \to \pi_0(G)$ is a surjective homomorphism with kernel $G_0$ means that $G_0$ is normal.

If $G$ is taken to be a diffeomorphism group, then $\pi_0(G)$ is the corresponding mapping class group. In particular, it is a quotient, not a subgroup, of the diffeomorphism group.

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    $\begingroup$ It's probably worth noting that the mapping class group usually can't be realized by diffeomorphisms; that is, there is usually not a section $\text{MCG}(M) \to \text{Diff}(M)$. This is true for surfaces of genus at least 3, for instance. $\endgroup$ – user98602 Sep 21 '15 at 0:46

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