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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 $H^1(M_{1,1};H^1(S_1;\mathbb{Z}))$ ?

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up vote 5 down vote accepted

This can be calculated from

1) the extension $\mathbb{Z} \to M_{1,1} \to SL_2(\mathbb{Z})$


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.

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Thanks, but it seems to me that you do not consider the twisting of the coefficients but just use that $H^1(S_1;\mathbb{Z})=\mathbb{Z}^2$. The action of $M_{1,1}$ on $H^1(S_1;\mathbb{Z})$ is not trivial but fixed point free. – berl13 Nov 22 '11 at 15:30
Everything I said works with twisted coefficients. – Oscar Randal-Williams Nov 22 '11 at 16:42

I think the answer is 0.

Indeed, $G=SL_2(\mathbb{Z})$ contains a normal $H=\mathbb{Z}/2\cong\{\pm I\}$. So if $V$ is set to be $\mathbb{Z}^2$ with the standard $G$-action, then the restriction of $H$ to $V$ is the direct sum of two non-trivial $H$-modules of rank 1. So $H^0(H,V)=0$ and $H^1(H,V)\cong\mathbb{Z}/2\oplus\mathbb{Z}/2$ with $G/H=PSL_2(\mathbb{Z})$ acting via the homomorphism $G/H\to SL_2(\mathbb{Z}/2)$. So $H^0(G/H,H^1(H,V))=0$ and from the Hochschild-Serre spectral sequence we get $H^1(G,V)=0$.

Now, depending on what one allows the elements of $M_{1,1}$ to do on the boundary, we have $M_{1,1}=G$ or a $\mathbb{Z}$-extension of $G$. In the second case we have using Hochschild-Serre again $H^1(M_{1,1},H^1(S_{1,1}))=H^1(G,V)=0$ as $H^0(G,V)=0$ and the action of $\mathbb{Z}$ on $H^1(S_{1,1})$ is trivial.

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In the second case you also get 0: you have to take the invariants, giving $H^1(\mathbb{Z}; \mathbb{Z}^2)^{SL_2(\mathbb{Z})}=0$. – Oscar Randal-Williams Nov 22 '11 at 16:54
Oscar -- yes, of course. – algori Nov 22 '11 at 17:01

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