# Signature of compact oriented 4-manifold

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?

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.
• Can one prove that the 4-dimensional fibration over $S^1$ is the boundary of a (singular) 5-dimensional fibration over the disc? That would of course imply that the signature is zero, since the 4-manifold would bound a 5-manifold. I was wondering if one could construct the 5-manifold explicitly. – Bruno Martelli Sep 14 '13 at 9:50