# Pontryagin numbers on a fiber bundle over $S^1$

Let $F$ be a closed manifold. What are the Pontryagin numbers on $E=F\times S^1$? More generaly, let $E$ be a closed manifold which is a fiber bundle over $S^1$ (with fiber $F$). $E$ is also called mapping torus. What are the Pontryagin numbers on $E$?

When $E$ is 4-dimensional, the signature of such a fiber bundle over $S^1$ is zero, which implies that the corresponding Pontryagin number is zero. I wonder if all the Pontryagin numbers of closed orientable mapping tori in any dimensions are always zero.

The special case of $S^1 \times M$ is the boundary of $D^2 \times M$, so of course its Pontryagin numbers vanish.
The answer is 0. The proof of the first case is as follows. Consider a family of product Riemannian metrics $g_{t}=t^{2}ds^{2}+h$, $0<t<1$, on E, where h is a Riemannian metric on F. The straight forward calculation shows that the sectional curvature (or the norm of curvature operator) of each $g_{t}$ is bounded by a constant C independent of t, i.e. $|Rm(g_{t})|<C$. Now the Chern-Weil theory says that for any Pontryagin number P we have $$P= \int_{E} R_{t} dvol_{g_{t}} ,$$ where $R_{t}$ is a polynomial of curvature operators. When t tends to 0, we have $$|\int_{E} R_{t} dvol_{g_{t}} | < C Vol(E, g_{t}) \rightarrow 0.$$ However the Pontryagin number P is an integer independent of t, and thus it must be 0.