4
$\begingroup$

It is my understanding that Dennis Johnson defined a `relative weight filtration,' of the mapping class group of an oriented surface. My question is what is this filtration, and how does it relate to the lower central series of the mapping class group? In particular, why is the relative weight filtration the "right" filtration as opposed to the lower central series.

$\endgroup$
3
  • 1
    $\begingroup$ There's the Johnson filtration, which I think is defined as follows: $J_k$, the $k$th term of the Johnson filtration, is the kernel of the natural map $\mathrm{MCG}(\Sigma)\to\mathrm{Out}(\pi_1\Sigma/\gamma_k\pi_1\Sigma)$ (where $\gamma_k\pi_1\Sigma$ is the $k$th term of the lower central series). Is that what you're after? There are numerous experts on MO (a fortiori from the fact that there are numerous Farb students everywhere). No doubt one of them will be along shortly to correct my definition and answer your question. $\endgroup$
    – HJRW
    Oct 2, 2010 at 16:32
  • 2
    $\begingroup$ Your definition is correct, Henry. I gave a more complete answer below (thus verifying your conjecture that a former student of Farb would answer the question). $\endgroup$ Oct 2, 2010 at 16:45
  • 1
    $\begingroup$ I was very confident about that one (the conjecture, not the definition, which I was only fairly confident about). $\endgroup$
    – HJRW
    Oct 2, 2010 at 16:50

1 Answer 1

10
$\begingroup$

Let $S$ be a surface (for simplicity, assume that $S$ has exactly one boundary component) and let $Mod(S)$ be its mapping class group. Let's assume that the genus of $S$ is at least $3$. To begin with, $Mod(S)$ is perfect, so its lower central series is not interesting. Define $\mathcal{I}(S)$ to be the Torelli group, ie the kernel of the action of $Mod(S)$ on $H_1(S)$. Things work better here. Johnson proved that the intersection of the lower central series for $\mathcal{I}(S)$ is trivial. Let me describe exactly what he did.

For a group $G$, let $\gamma_k(G)$ be the kth term in the lower central series for $G$, indexed so that $\gamma_0(G) = G$. Since $S$ has a boundary component, we can stick a basepoint $p$ on that boundary component and get an honest action of $Mod(S)$ on $\Gamma:=\pi_1(S,p)$ (if $S$ had no boundary, then we would only get an outer action). Johnson defined $\mathcal{I}(S,k)$ to be the kernel of the action of $Mod(S)$ on $\Gamma / \gamma_k(\Gamma)$. This gives a filtration $$\mathcal{I}(S) = \mathcal{I}(S,1) \supset \mathcal{I}(S,2) \supset \cdots$$ I believe that this is the filtration you are referring to (it has become known as the Johnson filtration). Johnson proved that $\cap_{k=1}^{\infty} \mathcal{I}(S,k) = 1$.

It is not hard to show that $\mathcal{I}(S,k) / \mathcal{I}(S,k+1)$ is abelian. One might thus be led to conjecture that $\mathcal{I}(S,k) = \gamma_{k-1}(\mathcal{I},k)$. Johnson proved that this is false. More specifically, he calculated $\mathcal{I}(S) / \gamma_1(\mathcal{I}(S))$ and showed that it contains a lot of 2-torsion coming from the Rochlin invariants of homology 3-spheres (the appropriate quotients were originally constructed by Birman and Craggs).

It is true that $\mathcal{I}(S,2)$ is the kernel of the universal torsion-free abelian quotient of $\mathcal{I}(S,1)$. One might thus conjecture that $\mathcal{I}(S,k)$ is the "torsion-free lower central series" of $\mathcal{I}(S)$. This hope was dashed by Morita, who showed that $\mathcal{I}(S,2)$ has a $\mathcal{I}(S)$-invariant $\mathbb{Z}$-quotient coming from the Casson invariant of homology 3-spheres that doesn't vanish on $\mathcal{I}(S,3)$.

We thus have two filtrations of $\mathcal{I}(S)$, the Johnson filtration and the "torsion-free lower central series". I'm not sure which is "better", but it is certainly true that the "torsion-free lower central series" is better understood due to a lot of work by Hain. In particular, he calculated a presentation for the Malcev completion of $\mathcal{I}(S)$, which is the completion of the filtered Lie algebra associated to the "torsion-free lower central series" filtration.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.