Signature of a finite covering space

Question

Suppose $\tilde{M}\rightarrow M$ is a finite (k-fold) covering of the smooth, oriented, compact 4-manifold $M$. Is there a relation between the signatures (http://en.wikipedia.org/wiki/Signature_(topology)) of $M$ and $\tilde{M}$?

I have a line of reasoning involving the Hirzebruch signature theorem that suggests $\sigma(\tilde{M})=k\cdot\sigma(M)$. If this is true, I would love to see independent lines of reasoning that support it.

Answer 1

Pontryagin numbers are multiplicative under coverings, hence, the same applies to the signature.

Answer 2

Let $M$ (and hence $\widetilde{M}$) be a closed, connected, oriented, smooth manifold of dimension $4n$, and $\pi : \widetilde{M} \to M$ a smooth $k$-sheeted covering.

As $\pi$ is a smooth covering map, it is a local diffeomorphism, so $\pi^*TM \cong T\hat{M}$. In particular,

$$\pi^*p_i(M) = \pi^*p_i(TM) = p_i(\pi^*TM) = p_i(T\hat{M}) = p_i(\hat{M}).$$

So if $p_{i_1}(M)\dots p_{i_l}(M) \in H^{4n}(M; \mathbb{Z})$, then $\pi^*(p_{i_1}(M)\dots p_{i_l}(M)) = p_{i_1}(\widetilde{M})\dots p_{i_l}(\widetilde{M})$. Therefore

\begin{align*} p_{i_1\dots i_l}(\widetilde{M}) &= \langle p_{i_1}(\widetilde{M})\dots p_{i_l}(\widetilde{M}), [\widetilde{M}]\rangle_{\widetilde{M}}\\ &= \langle \pi^*(p_{i_1}(M)\dots p_{i_l}(M)), [\widetilde{M}]\rangle_{\widetilde{M}}\\ &= \langle p_{i_1}(M)\dots p_{i_l}(M), \pi_*[\widetilde{M}]\rangle_{M}\\ &= \langle p_{i_1}(M)\dots p_{i_l}(M), k[M]\rangle_M\\ &= k\langle p_{i_1}(M)\dots p_{i_l}(M), [M]\rangle_M\\ &= k\,p_{i_1\dots i_l}(M). \end{align*}

The above calculation uses the equality $\pi_*[\widetilde{M}] = k[M]$ which follows from the fact that the degree of a finite covering is the number of sheets.

By the Hirzebruch signature theorem, there are rational numbers $a_{i_1\dots i_l}$ such that $\sigma(X) = \sum a_{i_1\dots i_l}p_{i_1\dots i_l}(X)$ for every closed, connected, oriented, smooth manifold of dimension $4n$. Therefore,

$$\sigma(\widetilde{M}) = \sum a_{i_1\dots i_l}p_{i_1\dots i_l}(\widetilde{M}) = \sum a_{i_1\dots i_l}k\, p_{i_1\dots i_l}(M) = k\sum a_{i_1\dots i_l}p_{i_1\dots i_l}(M) = k\sigma(M).$$