I was told that the signature of $S_1\times F_3$ is zero, where $F_3$ is a compact oriented 3manifold. 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), Ltheory (LückRanicki ''Surgery obstructions if fibre bundles'') or index theory (a footnote in Atiyah ''The signature of fibrebundles'', 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 LeraySerre 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 Poincareduality, 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.

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

$\begingroup$ @Johannes Ebert: Thank you very much for the answer. $\endgroup$ – XiaoGang Wen Sep 14 '13 at 10:08

$\begingroup$ @Bruno Martelli: Thank you very much for the question. I asked a related question here: mathoverflow.net/questions/142132/… $\endgroup$ – XiaoGang Wen Sep 14 '13 at 12:05
I guess, the tangent bundle has a nonzero 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?

2$\begingroup$ No, you are not; the Pontrjagin class does not have to vanish if there is a section. $\endgroup$ – Johannes Ebert Sep 14 '13 at 9:08