Write Mod(g,n) for the mapping class group of a genus-$g$ surface $\Sigma$ with $n$ boundary components. When $n=0,1$ we define the Torelli group $T$ to be the subgroup of Mod(g,n) which acts trivially on the homology $H = H_1(\Sigma,\mathbf{Z})$.

The Johnson homomorphism is a much-studied homomorphism from the Torelli group to $\mathrm{Hom}(H,\wedge^2 H)$ (when n=1) or a quotient of this (when n=0) whose kernel turns out to be the commutator subgroup of Torelli.

Morita showed in 1993 that the Johnson homomorphism extends to the whole group Mod(g,1), not as a homomorphism, but as a 1-cocycle in

$H^1(\mathrm{Mod}(g,1), \mathrm{Hom}(H,\wedge^2 H))$

where the action is given by the action of Mod(g,n) on $H$. (Thus the Morita cocycle restricts to a homomorphism on Torelli, as claimed.) It can be thought of as keeping track of the action of the mapping class on the quotient of $\pi_1(\Sigma)$ by the third term of its lower central series.

All of the above is well-known, or at least well-known to the people who know this kind of thing well. Now here's my question: is there a Morita cocycle on Mod(g,n) when n > 1?

Of course, such a cocycle would restrict to a Johnson homomorphism from the Torelli subgroup of Mod(g,n), and even this is subtle; but Church's paper "Orbits of curves under the Johnson kernel," gives a way to define a Torelli group and a Johnson homomorphism for Mod(g,n) which behaves well with respect to inclusion of subsurfaces. So a more specific version of my question would be: when n > 1, does Church's "Johnson homomorphism" extend to a "Morita cocycle" on all of Mod(g,n) which behaves well with respect to inclusion of subsurfaces?


1 Answer 1


The answer is "yes" -- in fact one can do better and get a class in

$$H^1(\text{Aut}(F_m), \text{Hom}(H, \wedge^2 H)),$$

where $F_m$ is the free group on $m$ generators and $H$ is the Abelianization of $F_m$, if I'm not mistaken. This gives the cocycle you want since there is an obvious map $\text{Mod}_{g, n}\to \text{Aut}(\pi_1(\Sigma_{g, {n-1}}))\simeq F_{2g+n-2}$ for $n\geq 2$, given by the conjugation action of $\text{Mod}_{g,n}$ on the point-pushing subgroup, namely $\pi_1(\Sigma_{g, {n-1}})$.

A construction goes as follows. Let $\mathbb{Z}[F_m]$ be the group ring of $F_m$, and let $\mathscr{I}$ be the augmentation ideal. Then $H \simeq \mathscr{I}/\mathscr{I}^2$ canonically (via the map sending $g$ to $g-1$) and $\mathscr{I}^2/\mathscr{I}^3\simeq H^{\otimes 2}$ canonically (via the multiplication map). There is a short exact sequence of $\text{Aut}(F_m)$ modules $$0\to \mathscr{I}^2/\mathscr{I}^3\to \mathscr{I}/\mathscr{I}^3\to \mathscr{I}/\mathscr{I}^2\to 0,$$ which we can think of as an extension of $H$ by $H^{\otimes 2}$, and hence gives a class in

$$H^1(\text{Aut}(F_m), \text{Hom}(H, H^{\otimes 2})).$$

But in fact a direct computation of a crossed homomorphism representing this class shows that it lands in $\text{Hom}(H, \text{Alt}^2(H)).$

EDIT: This is now worked out in detail here: https://arxiv.org/abs/2004.06146

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    $\begingroup$ Actually this will all be in a paper I'm writing with one of your students, among others! $\endgroup$ Commented Aug 13, 2019 at 0:00
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    $\begingroup$ To fix the conjugation action, I am worried you need to fix a base point which is $\operatorname{Mod}(g,n)$-invariant, which could be a point on one of the boundary components. It seems, though, that this cocycle could depend on the choice of boundary component. This is not so bad except it might cause trouble for the compatibility with inclusions of surfaces, if the chosen boundary component disappears or something. $\endgroup$
    – Will Sawin
    Commented Aug 13, 2019 at 0:57
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    $\begingroup$ @WillSawin: This is a good point. One can fix the base-point dependence by taking the Inn-coinvariants of $Hom(H, \wedge^2 H)$, but this does lose a bit of information (and in fact I think the class one gets if one does this is pulled back from $\text{Mod}(g, n-1)$...) $\endgroup$ Commented Aug 13, 2019 at 1:36
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    $\begingroup$ @JSE Don't you also have a paper on this stuff with one of your students? $\endgroup$
    – Will Sawin
    Commented Aug 13, 2019 at 14:11
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    $\begingroup$ We're working on it but for one of the theorems we want to prove I need this setup to work! And it's the same student, Wanlin Li! (Also Daniel Corey.) $\endgroup$
    – JSE
    Commented Aug 13, 2019 at 15:24

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