Pure mapping classes

Question

I am a new guy in mapping class groups and I am very confused about pure mapping classes tonight. From the Primer by Farb and Margalit, I learned the definition of pure mapping class groups PMod$(S_{g,n})$ of $S_{g,n}$ as ker(Mod$(S_{g,n})\to Sym(n))$ where $S_{g,n}$ denotes compact orientable surface with $g$ genus and $n$ punctures. However, when I read Johanna's paper on uniform uniform exponential growth, she defined pure mapping classes $\Gamma(S)$ as the kernel of the $Mod(S)$-action on homology with $\mathbb{Z}_3$ coefficients. From the book by Nikolai V. Ivanov, I learned a pure mapping class for a closed $1$-dimensional submanifold $C$ of a surface $S$ is defined by a diffeomorphism $F:S\to S$ satisfying:

The components of $C$ are nontrivial and are pairwise nonisotopic; $F$ is fixed on $C$, it does not rearrange the components of $S\backslash C$, and it induces on each component of the cut surface $S_C$ a diffeomorphism isotopic to either a pseudo-Anosov(pA) or the identity diffeomorphism.

and then the author proves a theorem

Theorem 3. If $m\geq 3$, then $\Gamma_S(m)$ consists of pure elements where $\Gamma_S(m)$ denotes the kernel of $Mod(S)\to Aut(H_1(S,\mathbb{Z}_m))$.

The following are my questions:

1. The idea of pure mapping classes come from pure braid groups. But what is the motivation for the condition that the induced diffeomorphism isotopic to either a pA or identity?

2. PMod$(S_{g,n})$ is defined by the kernel totally makes sense since the kernel fixed the punctures. However, I had trouble understanding $\Gamma_S(m)$. Even before that, why do we care about homology with coefficients $\mathbb{Z}_m$ (e.g. $\mathbb{Z}_3$ iin Johanna's paper)?

Answer 1

To answer question 1, since you are reading the Farb/Margalit primer, perhaps you know Thurston's classification theorem for mapping classes. One version of that theorem says the following: every mapping class of the surface $S$ is represented by a diffeomorphism $F : S \to S$ such that, for some closed 1-dimensional submanifold $C$ of $S$, the following hold (and these conditions are clearly more general than the Ivanov definition of pure):

• Components of $C$ are nontrivial and pairwise isotopic,
• $F(C)=C$ (in general $F$ need not be fixed on $C$, in fact it may not even take an individual component of $C$ to itself)
• For each component $S_i$ of the cut surface $S_C$, the first return map $F^k : S_i \to S_i$ is isotopic to either a pseudo-Anosov or a periodic diffeomorphism of $S_i$ (in general $F$ itself might not even take $S_i$ to itself, but of course $F^k$ does take $S_i$ to itself for some minimal value of $k \ge 1$, which is the meaning of the phrase "first return map").

For many applications, one needs the above three conditions to be strengthened in the manner of Ivanov's definition of "pure". The comment of @MoisheKahan alludes to one such application: the mapping class group of $S$ has a finite index torsion free subgroup, namely $\Gamma_m(S)$ for $m \ge 3$. For a more topological application, if $S_i$ is a component of $S_C$ then would like to be able to "restrict" the mapping class to $S_i$, obtaining an element of the mapping class group of the subsurface $S_i$ represented by $F | S_i : S_i \to S_i$. But this is clearly nonsense unless $F(S_i)=S_i$.

Anyway, although the strengthened versions of those three conditions are not generally true for all mapping classes, Ivanov's theorem says that the strengthened conditions are true if one assumes that the mapping class is contained in $\Gamma_S(m)$ for $m \ge 3$.

As for question 2, the reason why we care about coefficients $\mathbb Z_m$ for $m \ge 3$ is because they make Ivanov's theorem true. And the reason we focus on $m=3$ is because we're too lazy to keep saying "for all $m \ge 3$" (I've never seen an application for the $m \ge 3$ theorem which does not already follow from the $m=3$ theorem).

There is one special case of that theorem which I have always felt is quite instructive, and I'll leave it as an exercise for you to prove:

• If $F : S \to S$ is a finite order homeomorphism, and if the induced homology homomorphism $F_* : H_1(S;\mathbb Z_m) \to H_1(S;\mathbb Z_m)$ is the identity for some $m \ge 3$, then $F$ is the identity.

And if you get stuck on that exercise, here's an even easier warmup exercise, which captures the spirit of the previous one:

• If $G$ is a finite, 1-dimensional simplicial complex (i.e. a finite graph), if $F : G \to G$ is a simplicial isomorphism, if the induced homology homomorphism $F_* : H_1(G;\mathbb Z_m) \to H_1(G;\mathbb Z_m)$ is the identity for some $m \ge 3$, AND if $G$ has no vertex of valence 1, then $F$ is the identity.