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?
 A: 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.
A: I guess, the tangent bundle has a non-zero section, which is the pull back of
the section from the circle. It means that the first Pontryagin class is zero.
By the Hirzebruch signature theorem it follows that the signature is zero. Am I right?
