Triviality of line bundles

Question

Let $L$ be a smooth line bundle over $\mathbb{R}$ or $\mathbb{C}$ on a manifold $M$ such that $L^{\otimes m}$ is trivial for some $m \ge 1$. Can we find a smooth $m$-fold covering $(\tilde M,\pi)$ of $M$ such that $\pi^*(L)$ is trivial?

Answer 1

In this answer, I will prove that the assertion is true in the complex case.

We begin by investigating the Chern class, as there is a nice connection between the Chern class of the tensor power of a line bundle and the original.

Complex line bundles over $M$ are classified by elements of $H^2(M,\mathbb{Z})$, determined by the first Chern class $c_1(L)\in H^2(M,\mathbb{Z})$. One has that $c_1(L^{\otimes m})=mc_1(L)=0$ and in the case that $L^{\otimes m}\cong \underline{\mathbb{C}}$, we conclude that $c_1(L)$ is $m$-torsion in $H^{2}(M,\mathbb{Z})$. If $\tilde{M}$ is a cover of $M$ such that $\pi^*{L}\cong \underline{\mathbb{C}}$ we have by naturality of $c_1$, $c_1(\pi^*L)=\pi^*c_1(L)=0$. This allows us to rephrase the question as follows:

Let $c\in H^2(M,\mathbb{Z})$ be $m$ torsion. Does there exist a cover $\pi: \tilde{M}\to M$ such that $\pi^*(c)=0$.

Consider the case in which $M=\mathbb{R}P^2$. One has $H^2(\mathbb{R}P^2,\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}\langle c\rangle$ and hence $c$ is $2$-torsion. The only covering space of $\mathbb{R}P^2$ is $S^2$ and we see that the induced map $\pi^*: H^2(\mathbb{R}P^2,\mathbb{Z})\to H^2(S^2,\mathbb{Z})$ must be $0$ affirming the proposition.

Chern-Weil theory tells us that any complex line bundle for which $c_1(L)$ is torsion admits a curvature which is exact and hence can be given a connection which has vanishing curvature. This means that such a bundle is given by a $U(1)$ valued local system, i.e. (assuming that $M$ is connected) a representation $\rho_L: \pi_1(M)\to U(1)$. Here we see that since $L^{\otimes m}\cong \mathbb{C}$ we have $\rho_L^m:\pi_1(M)\to U(1)$ the trivial homomorphism. Given a map $f: N\to M$ (of connected, based spaces), the pullback local system corresponding to $f^*L$ is given by $\rho_L\circ f_*: \pi_1(N)\to \pi_1(M)\to U(1)$ and this composition must be trivial for $f^*L$ to be trivial over $N$.

If $\rho_L: \pi_1(M)\to U(1)$ has $\rho_L^m=1$ then $\mathrm{Im}\rho_L\leq \mu_{m}$, the $m$th roots of unity. If we assume that $\mathrm{Im}\rho_L=\mu_m$ then the $\ker\rho_L$, has a corresponding cover with fiber cardinality given by $|\mu_m|=m$. If $\pi: \tilde{M}\to M$ is the cover, we have $\pi^*L\leftrightarrow\rho_L\circ \pi_*: \pi_1(\tilde{M})\to \pi_1(M)\to U(1)=1$ and hence $\pi^*L$ is trivial. We see that $|\mathrm{Im}\rho_L|\mid m$. Writing $m=n|\mathrm{Im}\rho_L|$, if $\hat{M}$ is the cover corresponding to $\mathrm{Im}\rho_L$ then $\bigsqcup_{n} \hat{M}$ yields the desired cover.