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There are certain embeddings which are impossible. Let

Revision: For $S_{g,n}$ denote the surface n>6$, there is no embedding of genus $g$with$n$punctures to keep with conventional notation\mathcal{S}_n \hookrightarrow \mathcal{S}_{n+1}$.

First, consider homomorphisms between recall that there is an extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$, where $Mod(S_{0,n})$ is the (orientation preserving) mapping class groups. Consider group of the $n$-punctured sphere.

Inside $Mod(S_{0,n})$, there is a subgroup of isomorphic to $Mod(S_{0,12})$ \mathbb{Z}/(n-2)\mathbb{Z}$, which is a rotation of order$n-2$of$S^2$, and fixes the north and south poles. The$n$punctures include the north and south poles and one orbit of size$n-2$. The preimage of this group in $\mathcal{S}_n$ is isomorphic to $A_5$, by the orientation preserving symmetries \mathbb{Z}/(n-2)\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ or to $\mathbb{Z}/2(n-2)\mathbb{Z}$ (of an icosahedroncourse if $n$ is odd, where these are isomorphic). Then we get a corresponding subgroup of $\mathcal{S}_{n+1}$ from the injection. Project this group to $12$ vertices are permuted by Mod(S_{0,n+1})$via the map $A_5$. There is no such subgroup in \mathcal{S}_{n+1}\to Mod(S_{0,n+1})$. The kernel of this projection will have size at most $Mod(S_{0,13})$. 2$. By the Nielsen realizationRealization Theorem, any finite subgroup of$Mod(S_{0,n+1})$must preserve a mapping class group is realized by isometries complete hyperbolic metric of finite area on$S_{0,n+1}$, and in particular by uniformization extends to a surfacefinite group of conformal automorphisms of$S^2$which permutes$n+1$marked points. Thus, an embedding The finite subgroups of$Mod(S_{0,12})\hookrightarrow Mod(S_{0,13})$would give PSL_2(\mathbb{C})$ lie inside a conjugate of $SO(3)$, so the image is an isometric action abelian subgroup of $A_5$ on SO(3)$. The only abelian subgroups of$S_{0,13}$, and therefore SO(3)$ are cyclic (or $\mathbb{Z}/2\mathbb{Z}^2$), so the image is either $\mathbb{Z}/(n-2)\mathbb{Z}$ or $\mathbb{Z}/2(n-2)\mathbb{Z}$ (in which case we may take an embedding into index 2 subgroup isomorphic to $SO(3)$ permuting \mathbb{Z}/(n-2)\mathbb{Z}$; here we need$13$points. n>6$ to conclude that the image is not $\mathbb{Z}/2^2$). However, for $n>5$, there is only one embedding no subgroup of $A_5$ into $SO(3)$ (up isomorphic to outer automorphisms)$\mathbb{Z}/(n-2)\mathbb{Z}$ which permutes $n+1$ points, and the orbits therefore there is no such subgroup of $Mod(S_{0,n+1})$, a contradiction. To see thisaction have cardinalities , note that a cyclic group of rotations of $12, 20S^2$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ has two fixed points, 30and every other orbit of size $n-2$. Thus, 60$. So there is no action must be some$k$and$e$such that there are$k$orbits of size$A_5$preserving n-2$, and $13$ points.

Now consider the corresponding braid groups e$orbits of size$\mathcal{S}_{12}, \mathcal{S}_{13}$following your notation. There is a central extension 1$, where $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$e\leq 2$. If we had an embedding$\mathcal{S}_{12}\hookrightarrow \mathcal{S}_{13}$k\leq 1$, then we would get a homomorphism $\mathcal{S}_{12}\to Mod(S_{0,13})$. There is n+1=k(n-2)+e \leq n$, a$\mathbb{Z}/2$extension of$A_5$in contradiction. If$\mathcal{S}_{12}$which is the binary icosahedral group, and again by Nielsen realizationk\geq 2$, one can show that the image of this group in then $Mod(S_{0,13})$ must be n+1=k(n-2)+e \geq 2(n-2)$, so$A_5$, giving the same n\leq 5$, a contradictionas above.

There ought to be obvious generalizations of this argument to other $n$, but I haven't thought enough yet about arguments that might work for all $n$.

1

There are certain embeddings which are impossible. Let $S_{g,n}$ denote the surface of genus $g$ with $n$ punctures to keep with conventional notation.

First, consider homomorphisms between mapping class groups. Consider the subgroup of $Mod(S_{0,12})$ which is isomorphic to $A_5$, by the orientation preserving symmetries of an icosahedron, where the $12$ vertices are permuted by $A_5$. There is no such subgroup in $Mod(S_{0,13})$. By Nielsen realization, any finite subgroup of a mapping class group is realized by isometries of a surface. Thus, an embedding $Mod(S_{0,12})\hookrightarrow Mod(S_{0,13})$ would give an isometric action of $A_5$ on $S_{0,13}$, and therefore an embedding into $SO(3)$ permuting $13$ points. However, there is only one embedding of $A_5$ into $SO(3)$ (up to outer automorphisms), and the orbits of this action have cardinalities $12, 20, 30, 60$. So there is no action of $A_5$ preserving $13$ points.

Now consider the corresponding braid groups $\mathcal{S}_{12}, \mathcal{S}_{13}$ following your notation. There is a central extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}_n \to Mod(S_{0,n})$. If we had an embedding $\mathcal{S}_{12}\hookrightarrow \mathcal{S}_{13}$, then we would get a homomorphism $\mathcal{S}_{12}\to Mod(S_{0,13})$. There is a $\mathbb{Z}/2$ extension of $A_5$ in $\mathcal{S}_{12}$ which is the binary icosahedral group, and again by Nielsen realization, one can show that the image of this group in $Mod(S_{0,13})$ must be $A_5$, giving the same contradiction as above.

There ought to be obvious generalizations of this argument to other $n$, but I haven't thought enough yet about arguments that might work for all $n$.