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Let $\text{Mod}_g$ be the mapping class group of a closed oriented genus-$g$ surface $\Sigma_g$ and let $H = H_1(\Sigma_g;\mathbb{Q})$. Fix some $r \geq 0$. It is known that the cohomology group $H^k(\text{Mod}_g;H^{\otimes r})$ is independent of $g$ once $g$ is sufficiently large relative to $g$ and $r$. Does anyone know a concrete description of it?

For $r=0$, this is just the Madsen-Weiss theorem. For $r \geq 1$, this could be extracted from the paper

E. Looijenga, Stable cohomology of the mapping class group with symplectic coefficients and of the universal Abel-Jacobi map, J. Algebraic Geom. 5 (1996), no. 1, 135-150.

However, this paper really answers a much more complicated question where you look at the cohomology in an irreducible algebraic representation of $\text{Sp}(2g,\mathbb{Q})$. You could assemble this to get information about $H^{\otimes r}$, but given how complicated Looijenga's answer is this would lead to something terrible. I'm hoping there is a reasonable closed-form answer for these specific representations.

I've worked through Looijenga's argument and extracted the following special case: if $k$ is even, then $H^k(\text{Mod}_g;H) = 0$, while if $k$ is odd of the form $k = 2n-1$, then $$H^k(\text{Mod}_g;H) = \bigoplus_{i=0}^{n-2} H^{2i}(\text{Mod}_g;\mathbb{Q}).$$ Thanks to Dan Petersen in the comments for pointing out that I had originally screwed this up, as well as a related calculation for $r=2$.

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  • $\begingroup$ Your calculations by hand are definitely wrong. The cohomology with coefficients in $H$ does not vanish stably. Neither does the cohomology with coefficients in the two nontrivial representations of weight two. $\endgroup$
    – Dan Petersen
    Commented Sep 20, 2022 at 19:08
  • $\begingroup$ One can give a kind of generating series for these cohomology groups in terms of plethysm of symmetric functions, would you be interested in this? $\endgroup$
    – Dan Petersen
    Commented Sep 20, 2022 at 19:09
  • $\begingroup$ @DanPetersen: Are you sure? Remember that we're just working rationally. The proof for $H$ uses the fact that the universal curve $\mathcal{M}_{g,1} \rightarrow \mathcal{M}_g$ is an algebraic fiber bundle with projective fiber, so the spectral sequence for it degenerates. I think this means that $H^k(\mathcal{M}_{g,1})$ is the direct sum of $H^k(\mathcal{M}_g)$ and $H^{k-1}(\mathcal{M}_g;H)$ and $H^{k-2}(\mathcal{M}_g)$. But it's easy to see that we also have $H^k(\mathcal{M}_{g,1})$ equal to the direct sum of $H^k(\mathcal{M}_g)$ and $H^{k-2}(\mathcal{M}_g)$ -- the marked (continued) $\endgroup$
    – Fred
    Commented Sep 20, 2022 at 19:13
  • $\begingroup$ point just introduces a new stable generator in $H^2$ corresponding to the first Chern class of the vertical tangent bundle. $\endgroup$
    – Fred
    Commented Sep 20, 2022 at 19:13
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    $\begingroup$ I recommend to have a look at Appendix B of Randal-Williams' "Cohomology of automorphism groups of free groups with twisted coefficients". The stable description of the graded $\Sigma_q$-module $H^*(\Gamma_g;H^{\otimes q})$ appears in the middle of page 1471. $\endgroup$
    – archipelago
    Commented Sep 20, 2022 at 20:37

1 Answer 1

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Appendix B of Randal-Williams' "Cohomology of automorphism groups of free groups with twisted coefficients" gives a stable description of the graded $\mathbb{Q}[\Sigma_q]$-module $H^*(\Gamma_g;H^{\otimes q})$.

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  • $\begingroup$ I assume that you intend to take $k=q$, right? My understanding of RW's description is that the symmetric group $\Sigma_q$ is acting by permuting the tensor factors in $H^{\otimes q}$. $\endgroup$
    – Fred
    Commented Sep 21, 2022 at 14:34
  • $\begingroup$ Yes, this was a typo. $\endgroup$
    – archipelago
    Commented Sep 21, 2022 at 17:14

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