# The first Betti number of a finite covering space of a closed 3-manifold

Let $\tilde{M}\to M$ be a finite covering of closed 3-manifolds.

Is it possible that $\beta_1(\tilde{M})-1> [\pi_1(M):\pi_1(\tilde{M})](\beta_1(M)-1)>0$?

Here, $\beta_1(X)=\dim_{\mathbb{Q}}H_1(X;\mathbb{Q})$.

From the Reidemeister-Schreier method, $\operatorname{rank}(\pi_1(\tilde{M}))-1\leq [\pi_1(M):\pi_1(\tilde{M})](\operatorname{rank}\pi_1(M)-1)$.

Here, $\operatorname{rank}(G)$ is the least possible number of generators of $G$. By definition, $\operatorname{rank}(\pi_1(X))\geq \beta_1(X)$.

I'm mainly interested in the 3-manifold case though the question does make sense for general $n$-dimensional manifolds.

Edit: I know that there is a homology 3-sphere which has a finite cover which is not a homology 3-sphere. So I changed the inequality.

Take a closed (oriented) surface $\Sigma$ of genus $g$ with an orientation-preserving involution $\sigma$ with $2g-2$ fixed points (this is easy to realize with Riemann surfaces). The quotient $T=\Sigma /\sigma$ is a torus. Now take for $\tilde{M}$ $\mathbb{S}^1\times \Sigma$, and for $M$ the quotient by the fixed point free involution $(z,x)\mapsto (-z,\sigma x)$. Then $\beta _1(M)= 3$, $[\pi _1(\tilde{M} ):\pi _1(M)]=2$ while $\beta _1(\tilde{M} )= g+1$ is arbitrarily large. Of course you can do the same with any odd value for $\beta_1(M)$.
• Thanks for your answer. However, I couldn't see that $\beta_1(\tilde{M})=3$. Is it a torus bundle over the circle? – user156937 Jul 17 '15 at 9:37
• Also, I think that $1\times$ the fixed points of $\sigma$ is the set of fixed points of your involution on $S^1\times \Sigma$. – user156937 Jul 17 '15 at 9:50
• For the first question, use that $H^1(M,\mathbb{Q})$ is isomorphic to the invariant part of $H^1(\tilde{M},\mathbb{Q} )$. For the second, observe that $-1\neq 1$. – abx Jul 17 '15 at 13:35