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Let $M$ a complete Riemannian manifold. It is well known that the Laplace-Beltrami operator on $M$ is essentially self-adjoint and thus has a unique self-adjoint extension $\Delta_M$ in $L^2(M)$. The spectrum $\sigma(\Delta_M)$ is contained in $[0,\infty)$ and we denote $$ \lambda(M):=\inf\{\mu\in\sigma(\Delta_M)\vert \mu\neq 0\}. $$ If $M$ is closed, then $\lambda(M)$ is positive and coincides with the smallest nonzero eigenvalue. However $\lambda(M)$ may very well be $0$, as it is the case for $M=\mathbb{R}^n$. (Non-compact examples with $\lambda(M)>0$ are provided by simply connected spaces with negative sectional curvatures bounded away from $0$ - this was shown by McKean.)

Question: Suppose $N$ is a complete Riemannian manifold with $\lambda(N)>0$ and $p\colon \hat N \rightarrow N$ is a finite sheeted Riemannian covering. Do we have $\lambda(\hat N)>0$?


1) EQUIVALENT PROBLEM

This is related to this question of mine. (Note that $\lambda(M)>0$ if and only if the (unique) closed extension of the exterior derivative $d$ in $L^2$ has closed range.)


2) TRANSFER HOMOMORPHISM AND CALCULATION ON CIRCLES

Here are some ideas and an example calculation: (Let $n$ the number of sheets of $p$).

First, define the transfer homomorphism $T\colon L^2\hat N \rightarrow L^2N$ as follows: If $v\in L^2\hat N$ and $U\subset N$ is an evenly covered open set, then $Tv:=1/n \cdot \sum_{j=1}^n s_j^*v$, where $(s_j\colon U\rightarrow \hat N)_j$ are the $n$ distinct local sections of $p$. Then

  • $Tp^*= I \in \mathcal{B}(L^2N)$

  • $p^*T=:Q\in \mathcal{B}(L^2\hat N)$ is the orthogonal projection onto the subspace of functions which are invariant under the action of deck transformations.

  • $T\hat \Delta v=\Delta T v$ for $v\in C^\infty\hat N\cap L^2\hat N$.

Naive Strategy: Let's just look at eigenvalues and let's try to prove that $\sigma_p(\hat \Delta) \subset \sigma_p(\Delta)$. Problem: If $\hat\Delta v = \lambda v$, then $\Delta Tv = \lambda Tv$, but $Tv$ might be $0$.

Example: Let $S^1_L:=\mathbb{R}/{L\mathbb{Z}}$ the circle of length $L$. Then $u_k(t) = \exp(2\pi itk/L )$ defines a smooth function on $S_L^1$ and satisfies $-u_k''=(2\pi k/L)^2 u_k$. All eigenfunctions of $\Delta_{S^1_L}$ are of this form and thus $\sigma(\Delta_{S^1_L})=\{(2\pi k/L)^2\vert ~k\in \mathbb{Z}\}$ and $\lambda(S^1_L)=(2\pi/L)^2$.

Now consider $p\colon S^1_{nL} \rightarrow S^1_{L},t + nL\mathbb{Z} \mapsto t + L\mathbb{Z} $. We first see that $\lambda(S^1_{nL})=n^{-2}\lambda(S^1_L)$, hence $\sigma_p(\hat \Delta) \subset \sigma_p(\Delta)$ cannot be true. What happens?

The transfer homomorphism is of the form $ Tv(t) =1/n\cdot \sum_{j=1}^{n} v(t+jL). $ Let $v_k(t) = \exp(2\pi itk/(nL) )$ the eigenfunctions on $S^1_{nL}$. Then $$ Tv_k =v_k \cdot 1/n \cdot \sum_{j=1}^n \exp(2\pi ijk/n )=\begin{cases} 0 & k \nmid n \\ v_k & k \mid n \end{cases}, $$ hence the first few eigenfunctions indeed lie in the kernel of $T$.


3) ABSTRACT APPROACH

Here is another thought: Maybe we can forget about the covering space setting and just focus on the action of $G=\mathrm{Aut}(p)$ on $H=L^2\hat N$. Take this as new setting:

Let $G$ a finite group that acts on a Hilbert space $H$ by isometries and suppose that $A$ is a self-adjoint operator in $H$ such that

  • If $x\in D(A)$, then for all $g\in G$ $gx\in D(A)$ and $Agx=gAx$.

Let $H^G$ the subspace of $G$-invariant elements and $A^G$ with domain $D(A^G)=D(A)\cap H^G$ the restriction of $A$ to $H^G$. This is then a self-adjoint operator in $H^G$ and the question is:

Can we determine $\sigma(A)$ from $\sigma(A^G)$?

Unfortunately the answer to this question is no:

Example: Let $H = \mathbb{C}^2$, $G=\{I,\begin{bmatrix}0&1 \\ 1&0 \end{bmatrix} \}$. Then any self-adjoint $G$-equivariant matrix has the form $A=\begin{bmatrix}a &b \\ b & a\end{bmatrix}$ for $a,b\in \mathbb{R}$. It's easy to compute $\sigma(A) = \{a+b,a-b\}$ and $\sigma(A^G)=\{a+b\}$.

So in the general setting described above we cannot hope to relate the spectra. Now the Laplace operator is local and in particular its action is somehow determined by the action on $H^G$. But to prove this I have to use cutoff functions and I am not sure, if this can be transalted into the general Hilbert space setting. But maybe that is the key.

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  • $\begingroup$ Do you expect some concrete behavior, something like $\lambda(\hat{N}) = [\hat{N} : N]^{-1} \, \lambda(N)$? $\endgroup$ – Adrián González-Pérez Jul 6 '18 at 9:17
  • $\begingroup$ I willl try to use the fact that $p$ carries over to the $L^2$-spaces in a way that preserves the norms, after scaling by $[\hat{N}:N]$, and intertwines the Laplacians of $N$ and $\hat{N}$. $\endgroup$ – Adrián González-Pérez Jul 6 '18 at 9:20
  • $\begingroup$ I've corrected my (erroneous) example calculation and added it to the question. $\endgroup$ – Jan Bohr Jul 9 '18 at 9:16
  • $\begingroup$ I have posted the question on mathoverflow. $\endgroup$ – Jan Bohr Jul 16 '18 at 8:36

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