Does a spectral gap lift to covering spaces? Let $M$ be a complete Riemannian manifold. Denote $\Delta_M\ge0$ the unique self-adjoint extension of the Laplace-Beltrami operator in $L^2(M)$ and $\sigma(\Delta_M)\subset [0,\infty)$ its spectrum. Further define:
$$\lambda(M):=\inf\{\mu\in\sigma(\Delta_M)\vert~\mu\neq 0\}$$
Question: Let $N$ be a complete Riemannian manifold with $\lambda(N)>0$. If  $p\colon \hat N\rightarrow N$ is a finite sheeted Riemannian covering, do we also have $\lambda(\hat N)>0$?

I have asked the  same question on math.stackexchange without receiving an answer. Please see there for some examples and my attempts to answer the question.
 A: This is not an answer, but just somewhere useful you could start looking. Its good to think about when the quotient of fundamental groups has more than one generator.
The Rayleigh principle should imply that $\lambda_1(M,g)\geq \lambda_1(\tilde{M},\tilde{g})$, since your $\lambda_1(M,g)=\inf_{f\in H^1(M), ||f||_{2,g}=1}\int_M|df|^2$, and this integral is multiplicative under finite covers. (So this infimum upstairs can only be smaller, as there may be $H^1(M)$ functions which weren't lifts by $p$).
In this paper, 
https://projecteuclid.org/euclid.tmj/1178224610,
they prove that when the covering $p:\tilde{M}\rightarrow M$ satisfies $\pi_1(M)/p_\ast\pi_1(\tilde{M})\simeq \mathbb{Z}_k$, there exists a metric on $M$ such that $\lambda_1(M,g)=\lambda_1(\tilde{M},\tilde{g})$. But this may not hold general finite covers.
A: I think the answer is yes. You may look at any book on spectral theory.
Here is a comment for the dimension 2 case. Let $N=\Gamma\backslash\mathbb{H}$ be a hyperbolic surface, where $\mathbb{H}$ is the upper half-plane and $\Gamma$ is a finite-index subgroup of $SL(2,\mathbb{Z})$. Here $\Gamma$ acts on $\mathbb{H}$ by Mobius transformation. It is well-known from the spectral theory (see Lang, $SL(2,\mathbb{R})$, for example) the spectrum bellow 1/4 is discrete. Now take $\Gamma'$ be a finite index subgroup of $\Gamma$. Then the projection map $\pi:\Gamma'\backslash\mathbb{H}\rightarrow\Gamma\backslash\mathbb{H}$ is a finite cover. Moreover, by the above, the spectrum of $\Gamma'\backslash\mathbb{H}$ is still discrete bellow 1/4.
I think the compact case is even simpler since the spectrums are all discrete.
A: If one manifold is a finite cover of another, then their spectral gaps coincide. This can be seen, for instance, from the interpretation of the spectral gap as the exponential diagonal rate of decay of the heat kernel.
