Let $M\_{g,1}$ be the mapping class group of surfaces of genus g $\geq 1$ with one boundary component. By $S\_g$ we denote a closed surface of genus $g$.


In the paper "Families of jacobian manifolds and characteristic classes of surface bundles.I" S.Morita proved that the twisted cohomology $H^1(M\_{g,1};H^1(S\_g;\mathbb{Z}))$ is $\mathbb{Z}$ if $g \geq 2$.


My question is if there is the some result known for $g=1$ that is $H^1(M\_{1,1};H^1(S\_1;\mathbb{Z}))$ ?



Twisted cohomology of the mapping class group
This can be calculated from


1) the extension $\mathbb{Z} \to M\_{1,1} \to SL\_2(\mathbb{Z})$


and


2) the decomposition $SL\_2(\mathbb{Z}) \cong \mathbb{Z}/4 \*\_{\mathbb{Z}/2} \mathbb{Z}/6$.


The extension shows that the group you want is isomoprhic to $H^1(SL\_2(\mathbb{Z});\mathbb{Z}^2)$. You can the use the Mayer--Vietoris sequence for the amalgamated product decomposition, and the standard way of computing group cohomology with cyclic groups, to get the answer.


I don't know what the answer is, though.


