Computation of $H^*_c(\mathcal{M}_{1,[2]},\mathbb{V}_1)$

Question

As a term of a Serre spectral sequence, I would like to compute the cohomology group with compact support $H^*_c(\mathcal{M}_{1,[2]},\mathbb{V}_1)$ of the moduli space of genus $1$ curves with $2$ unordered marked points, with values in $\mathbb{V}_1$, the standard representation of $SL_2(\mathbb{Z})$, coming from the cohomology group $H^1(E,\mathbb{Q})$, where $E$ is an elliptic curve.

I know that $H^*_c(\mathcal{M}_{1,1},\mathbb{V}_1)=0$, and I would like to try to use this result to obtain a similar one for the case of $2$ unordered marked points.

My attempt so far consists in using mapping class groups, the Birman exact sequence

\begin{equation*} 1\to \pi_1(S_{1,1})\to \text{PMod}(S_{1,2})\to \text{PMod}(S_{1,1})\to 1 \end{equation*} and the Hochschild-Serre spectral sequence of group cohomology, to obtain $H^*(\mathcal{M}_{1,2},\mathbb{V}_1)$ (which I believe to be identically $0$), using the fact that the action of the fibre $\pi_1(S_{1,1})\cong \mathbb{Z}*\mathbb{Z}$ on the local system is trivial, as $\mathbb{V}_1$ is a representation of $\text{PMod}(S_{1,1})\cong SL_2(\mathbb{Z})$.

Since I am not used to mapping class groups and group cohomology, I have, now, two questions:

1. Is the reasoning so far correct?
2. Assuming it is, this takes care only of the moduli stack $\mathcal{M}_{1,2}$ with ordered points. How do I obtain the cohomology of the unordered one? I think that would consist in getting rid of the action of the elliptic involution, exchanging the marked points, but it is not clear to me how to effectively do it.

To be coherent with other results I have, I would hope this cohomology groups not to be identically $0$.

I would appreciate any comment on my problem, or any helpful reference for the computation of $H^*_c(\mathcal{M}_{1,[2]},\mathbb{V}_1)$, also with other methods.

Answer 1

Your approach is good but you must have messed something up with the Hochschild-Serre spectral sequence.

I prefer the sheafy approach over thinking about the Hochschild-Serre spectral sequence, so let me explain it in these terms. So we have the forgetful map $\pi : M_{1,2} \to M_{1,1}$, and we may write $$ H^\bullet_c(M_{1,2},V) = H^\bullet_c(M_{1,1},R\pi_!V ).$$ Now $V=\pi^\ast V$ is actually pulled back from $M_{1,1}$ and we have $$R\pi_! \pi^\ast V \cong (R\pi_!\mathbf Q) \otimes V \cong (V \otimes V)[-1] \oplus V(-1)[-2]$$ where in the first step we used the projection formula and in the second we used a section to split $R\pi_!\mathbf Q$ in the derived category. Representation theory of $\mathrm{SL}(2)$ shows that $V \otimes V \cong V_2 \oplus \mathbf Q(-1)$, where $V_2$ is the second symmetric power of $V$ and $\mathbf Q(-1)$ is a Tate twist of the trivial representation. From the Eichler-Shimura isomorphism we know $H^\bullet_c(M_{1,1},-)$ with coefficients taken in all of these summands: $H^2_c(M_{1,1},\mathbf Q(-1)) \cong \mathbf Q(-2)$, $H^1_c(M_{1,1},V_2)\cong \mathbf Q(0)$, all other cohomologies vanish. So $H^\bullet_c(M_{1,2},V)$ is $\mathbf Q(-2)$ in degree $3$, $\mathbf Q(0)$ in degree $2$, and vanishes otherwise.

To deal with unordered points we use that $H^\bullet_c(M_{1,[2]},V) = H^\bullet_c(M_{1,2},V)^{\Sigma_2}$, where $\Sigma_2$ acts by permuting the markings. Unfortunately the computation in the previous paragraph isn't immediately $\Sigma_2$-equivariant. But in fact $\Sigma_2$ acts on $M_{1,2}$ preserving the map $\pi$; it acts by multiplication by $-1$ on the universal elliptic curve. To see this we consider a curve with two marked points $(x,y)$. We can translate to make the first coordinate the origin, and our markings are then $(0,y-x)$. Clearly if we had swapped $x$ and $y$ we would have gotten $x-y$ instead. Inversion in $V$ acts as the identity on $V \otimes V$, so $H^\bullet_c(M_{1,[2]},V) = H^\bullet_c(M_{1,2},V)$.

Hope this is coherent with other results you have!