Is there a Morita cocycle for the mapping class group Mod(g,n) when n > 1?

Question

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?

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