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.



The first Betti number of a finite covering space of a closed 3-manifold
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 $\beta\_1(M)$.


