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 $r=2$.

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 $r=2$.

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 two special cases. In both of them, $g$ is large enough that we're in the stable range.

1. $H^k(\text{Mod}_g;H) = 0$ for all $k$.
2. $H^k(\text{Mod}_g;H^{\otimes 2}) = H^k(\text{Mod}_g;\mathbb{Q})$. The explanation for this is that there is a $1$-dimensional trivial subrepresentation of $H^{\otimes 2}$ coming from the symplectic form.

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 $r=2$.