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Consider a one-parameter real analytic family of metrics $g_t$ on a compact manifold $M$ converging to a metric $g$ in $C^k$-norm, for some $k$. It is known that the Laplace spectrum of $g_t$ will converge to the Laplace spectrum of $g$. However, is it true that $\lambda_n(t)$ will converge to $\lambda_n$, where $0 \leq \lambda_1(t) \leq \lambda_2(t) \leq \cdots$ denotes the Laplace spectrum of $g_t$, and $0 \leq \lambda_1 \leq \lambda_2 \leq \cdots$ denotes the Laplace spectrum of $g$? Thanks!

Addendum: I just found the following paper. Regarding the theorem on page 1, I wish to ask more generally: the theorem says that if the operators $A(t)$ are $C^M$, where $M$ could be $\omega, \infty$ or Holder class, then the eigenvalues of $A(t)$ can be parametrized (under certain conditions) to be $C^M$. But does that mean the following: if the operators $A(t)$ are positive, self-adjoint and have discrete spectrum, and for each $t$, we have $0 \leq \lambda_1(t) \leq \lambda_2(t) \leq....$ as the spectrum of $A(t)$, then are $\lambda_i(t)$ parametrized to be $C^M$?

Edit: I am new to MO and MSE and unfortunately was not aware of the practices. I have edited the MSE question requesting people to post their answers here instead.

Further edit: As commented below, the existence of the limit metric is not that important. All I want to know is, if we rearrange the eigenvalues at each step according to increasing order, will the rearranged functions still be $C^M$?

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    $\begingroup$ Crossposted on MSE. $\endgroup$ Oct 21 '15 at 20:29
  • $\begingroup$ I don't think so. If you take a family of shrinking rectangles with Neumann b.c., then you have very reasonable Laplacian convergence towards the Laplacian on an interval (e.g., in resolvent norm, hence much more than mere spectral convergence), but half the eigenvalues on the rectangles (those corresponding to the transversal direction) will simply blow up as the rectangles get thinner and thinner. This setting can likely be turned into a more precise counterexample to your conjecture, e.g. by replacing rectangles by tori. $\endgroup$ Oct 21 '15 at 22:31
  • $\begingroup$ @DelioMugnolo I might be wrong, but I think the OP is trying to say that the limit metric $g$ is also on the same manifold $M$, ruling out degeneracy. For a $C^k$ metric convergence, spectral convergence is clear. Besides, I think the existence of a limit metric might not be that pertinent to this question. To me at least, the main question seems to be whether the eigenvalues can be parametrized to be $C^M$ while respecting their order of magnitude. I think they can still be parametrized continuously, but I am not sure about the other cases. $\endgroup$
    – HSM
    Oct 21 '15 at 22:42
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    $\begingroup$ @student: so it's considered rude to crosspost on MO and MSE (people can waste time re-inventing the wheel). If you're not sure, post on MSE first. Then if you don't get anything in a couple of days, move to MO. In this case, I think MO is the correct venue. So what I would suggest is editing the MSE posting to say that it has been posted here; and to please post any answers on this site. $\endgroup$ Oct 21 '15 at 22:44
  • $\begingroup$ IMO, the main problem is when two eigenvalues cross each other. In the ordering given by the OP, two otherwise smooth trajectories would develop kinks. $\endgroup$
    – Fan Zheng
    Oct 22 '15 at 2:07
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The answer is no, for the reason Fan mentioned above. When two parameterized curves of eigenvalues cross, the resulting ordered eigenvalues will only be Lipschitz.

For a simple example, consider the square $[0,1] \times [0,b]$ with Dirichlet boundary conditions. The eigenvalues are $(m\pi)^2 + (n\pi/b)^2$ for $m,n \in \mathbb{N}$, so the first eigenvalue is $\pi^2(1 + 1/b^2)$, which depends smoothly on $b$. However, the second eigenvalue is $\pi^2(4+1/b^2)$ for $b\leq 1$ and $\pi^2(1+4/b^2)$ for $b>1$, which is not differentiable at $b=1$.

The result you cited will apply to any eigenvalue that is simple, in particular the first one. (This is trivial for a closed manifold, since the first eigenvalue is always 0, but for the Dirichlet problem on a compact manifold with boundary the conclusion is more interesting.)

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