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.

  • $\begingroup$ Do you have a simple example of $N$ non compact and $\lambda(N)>0$? $\endgroup$ – RaphaelB4 Oct 29 '18 at 8:55
  • $\begingroup$ If you equip $N=\mathbb{R}^n$ with a complete Riemannian metric that has sectional curvatures $K< -1$, then $\lambda (N) > 0$ (see here). $\endgroup$ – Jan Bohr Oct 29 '18 at 16:00

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.

  • $\begingroup$ The result you've mentioned is very interesting, because my intuition for the closed case would have been that the first nonzero eigenvalue is always smaller on the covering space. I will have a closer look at the paper, but (apart from the closedness assumption) the problematic bit seems to be that they construct a special metric for which the spectrum upstairs can be controlled, while in my question we want control for an arbitrary metric. $\endgroup$ – Jan Bohr Nov 4 '18 at 13:17

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.

  • $\begingroup$ Can you suggest a reference that discusses the interpretation of the spectral gap in terms of exponential decay rate of the heat kernel? Of course preferably on non-compact manifolds. $\endgroup$ – Jan Bohr Jul 17 '18 at 10:29
  • 2
    $\begingroup$ In the linked math.stackexchange question I calculate that $\lambda(S^1_L) = (2\pi/L)^2$ (Where $S^1_L$ is a circle with length $L$). Doesn't this give rise to a counterexample to your claim that finite covers have the same spectral gap? $\endgroup$ – Jan Bohr Jul 17 '18 at 10:36
  • $\begingroup$ Do you exclude the 0 eigenvalue for compact manifolds.? $\endgroup$ – R W Jul 17 '18 at 10:42
  • $\begingroup$ No, $0$ is allowed. Note that I've defined $\lambda(M)$ as the infimum of all nonzero spectral points. I think one can ask a related question, namely: If $0\notin \sigma(\Delta_N)$, do we also have $0\notin \sigma(\Delta_{\hat N})$? However I'm also interested in the general case, where $0\in \sigma(\Delta_N)$, but it is isolated in the spectrum. $\endgroup$ – Jan Bohr Jul 17 '18 at 10:53
  • $\begingroup$ I have found the following result: $\vert \mathrm{tr}(e^{-t\Delta_M})- \dim \ker \Delta_M \vert \le C e^{-t \lambda(M)}$. But it only seems to hold if $\Delta_M$ is discrete. $\endgroup$ – Jan Bohr Jul 17 '18 at 11:28

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.

  • $\begingroup$ Well, I also think that it is true. Unfortunately most references on spectral geometry seem to deal at most with the case that $N$ is compact and $\hat N$ is non-compact. If you have in your mind a useful result from spectral theory (that I could find in any book), I'm happy to hear about it. $\endgroup$ – Jan Bohr Jul 17 '18 at 10:59
  • $\begingroup$ As for the 2D-case: I haven't looked into the proof of the claim but I suspect it to rather establish a result that incidentally also holds for the covering space rather than actually lifting the gap. However, it's a good example to have in mind, because the only test examples I could wrap my head around so far were compact (where existence of a spectral gap is clear). $\endgroup$ – Jan Bohr Jul 17 '18 at 11:02

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