# Cohomolgoy of $S^1$-bundle over genus $g$ surface

Suppose $\Sigma_g$ is the closed, orientable surface of genus $g\ge 1$. Isomorphism classes of principal $S^1$-bundle on $\Sigma_g$ is then classified by $H^2(\Sigma_g,\mathbb{Z})=\mathbb{Z}$. Suppose $S^1\hookrightarrow{} P_n\to \Sigma_g$ is the bundle corresponding to the class represented by the integer $n$. I'd like to compute the cohomology groups of $P_n$.

The obvious tool was to use Serre spectral sequence. But since $\pi_1(\Sigma_g)\ne 0$, the coefficient system is not necessarily simple. Is the action of $\pi_1(\Sigma_g,b)$ on $H^1(F_b,\mathbb{Z})=\mathbb{Z}$ is always trivial? I can see that this action should somehow depend on the Euler class of this bundle, but I'm not sure how to proceed.

If the coefficient system is indeed non-trivial, is there any way to compute, say, $E_2^{2,1}=\mathrm{H}^2(\Sigma_g, \mathcal{H}^1(S^1,\mathbb{Z}))$? Here $\mathcal{H}^1(S^1,\mathbb{Z})$ denotes the local coefficient system.

Any help is appreciated!

• The local system is simple. For $[\gamma]\in \pi_1(\Sigma)$, the action of $[\gamma]$ on $H^1(F;\mathbb{Z})$ is induced by the monodromy of $[\gamma]$. But the restriction of the principal $S^1$-bundle on (or to be more precise, pullback along) $\gamma$ is always trivial (as can be seen from $H^2(S^1;\mathbb{Z})=0$), so the monodromy is homotopic to the identity. – cjackal Apr 20 '18 at 6:33
• @cjackal So for any $S^1$-bundle, I can pullback the bundle along the loop and get an $S^1$-bundle over $S^1$, which is forced to be the torus. So the monodromy is always trivial and we always have a simple coefficient system for any $S^1$-bundle. Am I correct? – ChesterX Apr 20 '18 at 7:23
• That's my understanding. – cjackal Apr 20 '18 at 7:36
• I am confused. The frame bundle of a topologically non-trivial complex line bundle over $\Sigma_g$ cannot have a section or else the line bundle would be trivial. This seems to contradict what @cjackal is saying. – PVAL-inactive Apr 20 '18 at 8:07
• @PVAL-inactive I don't think that's what cjackal is saying. The $S^1$-bundle over the surface can be non-trivial, but over $S^1$ it is always the product. – ChesterX Apr 20 '18 at 8:12

Here is a geometric topological approach. Since $P_n$ is the frame bundle for the complex line bundle with Chern number n, $P_n$ is the boundary of the disk bundle over $\Sigma_g$ with Euler number $n$. Call this 4-manifold $D_n$. One way to think about this 4-manifold is to take $(\Sigma_g-D^2) \times D^2$ and attach a n-framed 2-handle along the $S^1$ boundary of $\Sigma_g-D^2$ (times a constant in the other factor). Now there is a Morse function on the 4 manifold which has 1 0-handle and 2g 1-handles (all coming from the standard Morse function on $\Sigma_g-D^2$ and one n-framed 2-handle attached along a curve representing a product of commutators of the generators of $\pi_1(\natural^{2g}S^1 \times D^3)$. In particular the boundary of this 4-manifold $P_n$, is $n$-surgery on a null homologous knot in $\#^{2g}S^1 \times S^2$ hence we have $H_1(P_n)=\Bbb Z^{2g}\oplus \Bbb Z/n$.