I was told that the signature of $S\_1\times F\_3$ is zero, where $F\_3$ is a compact oriented 3-manifold. Let $M\_4$ be a fibre bundle with $S\_1$ as a the base manifold and $F\_3$ as the fibre. Assume $M\_4$ is oriented. Can one show that the signature of $M\_4$ is zero?



Signature of compact oriented 4-manifold
Yes, in several ways. It can be proven by cohomological methods (Meyer, ''Die Signatur von Faserbündeln'', PhD thesis in Bonn, early 1970s), L-theory (Lück-Ranicki ''Surgery obstructions if fibre bundles'') or index theory (a footnote in Atiyah ''The signature of fibre-bundles'', the details worked out by myself (arXiv:0902.4719)).
The result is that the signature is multiplicative in oriented fibre bundles of odd fibre dimension. When the base is $S^1$, there is a shorter argument:


Consider the rational Leray-Serre spectral sequence, which collapses for degree reasons. The terms that contribute to the middle dimensional cohomology of $M\_4$ are $E\_{2}^{0,2}=H^0 (S^1; H^2 (F))$ and $E\_{2}^{1,1}=H^1 (S^1 ;H^1 (F))$ (with local coefficients, of course). Now observe that both are dual to each other by Poincare-duality, so in particular have the same dimension. The term $E\_{2}^{1,1}=H^1 (S^1 ;H^1 (F))$ is a subgroup of $H^2 (M\_4)$. The point now is that the cup product is trivial on this subspace, by the multiplicativity of the spectral sequence.
By the above argument, $dim (E\_{2}^{1,1})=1/2 dim H^2 (M\_4)$, and so we have found a Lagrangian subspace, in particular, the signature is null.


