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On a compact Riemannian manifold $M$ (we assume Dirichlet boundary condition if $\partial M \neq \emptyset$), the Laplace-Beltrami operator $-\Delta$ has a discrete spectrum $0 < \lambda_1 \leq \lambda_2 < .....$ going to $\infty$. Consider the eigenspace corresponding to the first eigenvalue $\lambda_1$. My question is, what is the dimension of this eigenspace?

It seems that the answer should be one, and the otherwise one can locate two nodal domains corresponding to one of the first eigenfunctions, and then $\lambda_1$ is the first eigenvalue of each of these nodal domains, which should violate domain monotonicity.

On the other hand, suppose we are given one (even non-sign-changing) eigenfunction $\varphi$ corresponding to $\lambda$. Suppose the manifold $M$ is such that it has a non-trivial isometry $T$. Then $T\varphi (x) := \varphi (Tx)$ is also an eigenfunction, but in general one does not expect that $T\varphi = \varphi$.

What am I missing here?

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    $\begingroup$ The multiplicity can be larger than 1, as shown by the Bolza surface. $\endgroup$ – Bullet51 Aug 24 at 16:40
  • $\begingroup$ @Bullet51 But the result seems correct for domains in $\mathbb{R}^n$, right? For example, see Evans' book on PDE, page 336. What goes wrong if one tries to repeat this proof for manifolds? $\endgroup$ – user144878 Aug 24 at 16:50
  • $\begingroup$ @Bullet51 By domain monotonicity, I meant that since the eigenvalues are defined variationally by Raleigh quotients, if you have two domains $\Omega_1 \subsetneq \Omega_2$, then $\lambda_1(\Omega_1) \geq \lambda_1(\Omega_2)$ simple because $H^1_0(\Omega_1) \subsetneq H^1_0(\Omega_2)$. $\endgroup$ – user144878 Aug 24 at 17:08
  • $\begingroup$ Is there a kind of domain monotonicity which assures strict inequality? $\endgroup$ – Bullet51 Aug 24 at 17:21
  • $\begingroup$ Suppose strict inequality does not hold. Then you can find $\Omega_1 \subsetneq \Omega_2$ such that they both have the same $\lambda_1$. Now, take the first Dirichlet eigenfunction on $\Omega_1$, and extend it by $0$ to $\Omega_2$. This is now a first eigenfunction for $\Omega_2$ which vanishes on an open set. Does this violate unique continuation? $\endgroup$ – user144878 Aug 24 at 17:24
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Clearly one needs to assume that $M$ is connected (otherwise the first eigenvalue is not necessarily simple). Then indeed the first eigenvalue is simple, the corresponding eigenfunction has a constant sign and if $M$ admits an isometry, so does this eigenfunction.

The proof is standard variational: the first eigenfunction delivers a minimum to $\int |\nabla u|^2 \,dx $ under Dirichlet b.c. under assumption $\int u^2\,dx=1$. Then $v=|u|$ delivers the same minimum and thus is also an eigenfunction. But this cannot happen unless $v>0$ in $M\setminus \partial M$. Then $u$ must have a constant sign (we use that $M$ is connected), say $u>0$.

Then, if $\lambda_1$ is not simple, there is another eigenfunction $w$ which is orthogonal to $u$, but this is impossible since $w$ also has a constant sign.

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    $\begingroup$ But, in the simple case of a circle $S^1$, let's say parametrized by $t \in [0, 2\pi]$ with the endpoints identified, both $\sin t$ and $\cos t$ seem to be first eigenfunctions. Is this incorrect? $\endgroup$ – user144878 Aug 24 at 16:46
  • $\begingroup$ in the case of $\mathbb{S}^1$ the first eigenfunction is a constant, and the first eigenvalue is $0$ $\endgroup$ – Victor Ivrii Aug 24 at 16:51
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    $\begingroup$ But it seems that $λ_1$ is used to denote the first eigenvalue greater that $0$. $\endgroup$ – Bullet51 Aug 24 at 17:02
  • $\begingroup$ $0$ is an eigenvalue only if $M$ is closed. Then the next eigenvalue could be of any chosen multiplicity, but nobody would call it "the first eigenvalue". $\endgroup$ – Victor Ivrii Aug 24 at 17:05
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    $\begingroup$ If $\partial M=\emptyset$ then $0=\lambda_1 < \lambda_2\le \lambda_3\le \ldots$. The same is true for Neumann b.c. even if $\partial M\ne \emptyset$. There are plenty of self-adjoint operators in math and applications, many of them have only point spectrum (or at least near the bottom), but the lowest eigenvalue is not necessarily even non–negative. Your confusion is the consequence of bad notations. $\endgroup$ – Victor Ivrii Aug 24 at 17:58

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