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Let $(M,g)$ be a compact Riemannian manifold, and let $\Delta$ be the Laplace-Beltrami operator. Let $\lambda_1 >0$ be the first positive eigenvalue. That is, there exists a non-trivial function $\phi$ satisfying $\Delta \phi + \lambda_1\phi = 0$, and that $\lambda_1$ is the smallest such real number. Then one can always normalize the eigenfunction such that $\max_M\phi = 1$ and $\min_M = -k\geq -1$. I have two questions:

  1. Can one further choose an eigenfunction such that the maximum is attained at a unique (or at worse an isolated) point? For example, for a sphere $\mathbb{S}^2$ with the round metric, the eigenfunctions are given by $\cos d$, where $d$ is the distance to a fixed point (say the north pole N). The north pole in this case is the unique maximum. On the other hand if we take the product $\mathbb{S}^2\times\mathbb{S}^2$ with the product of the round metrics, and if we let $d_1$ and $d_2$ be the distance functions (from the respective north poles) on each factor, then each $u_i = \cos d_i$ is an eigenfunction, but the point where they attain maximums are clearly is unique (for instance $u_1$ takes maximum value on $\{N\}\times\mathbb{S}^2$). But in this case one can take $u = (\cos d_1 + \cos d_2)/2$ to be a noramlized eigenfunction with a unique maxima. So my question is whether such a normalized eigenfunction exist on all compact Riemannian manifolds $(M,g)$.
  2. If further the Ricci curvature is bounded below, say $\mathrm{Ric}_g\geq n-1$, where $n$ is the dimension, it is well known that $\lambda_1\geq n$. then does there exist an eigenfunction $u$ such that $$\mathrm{Hess}(u) \geq -\frac{\lambda_1}{n}ug?$$ If the answer to this is affirmative, then such an eigenfunction will clearly have an isolated maximum (since the Hessian will be negative definite at the maximum), hence answering the first question also in affirmative. It is of course well known (Obata) that $\lambda_1 = n$ if and only if $(M,g)$ is isometric to the round sphere. Moreover, in this case we necessarily have that for any eigenfunction $u$, both sides of the above inequality must in fact be equal. This follows from a simple integration by parts argument. On the other hand, suppose $\lambda_1>n$, then once again taking the example of $\mathbb{S}^2\times\mathbb{S}^2$ it is clear that the above inequality of the Hessian will not hold for all eigenfunctions. My guess is that the Hessian inequality is too strong, and is likely wrong. But a counter example would be nice.
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  • $\begingroup$ By Uhlenbeck's theorem, generically the first nonzero eigenfunction is unique up to scaling. So there's not really any choice in your first question. (Uhlenbeck also shows generically eigenfunctions are Morse, which I think answers the question about isolated maxima.) $\endgroup$ – Neal May 26 '20 at 13:04
  • $\begingroup$ @Neal - Thanks for the reference on Uhlenbeck's theorem. In my case of course the Riemannian metric is given, and I dont seem to have any room to perturb it to make use of Uhlenbeck's theorem. But in any case, as a side note, if the Riemannian metric is Kahler, then would Uhlenbeck's theorem provide a nearby Kahler metric that has a one dimensional eigenspace? $\endgroup$ – Poincare-Lelong Jun 19 '20 at 9:37
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The answer to question 1 is no. On a two-dimensional torus $\mathbb{S}^1 \times \mathbb{S}^1$, the first nontrivial eigenspace consists of $\mathrm{Span}\{\cos(x),\sin(x),\cos(y),\sin(y)\}$. Each eigenfunction in this eigenspace is separable, i.e. it is the product of a lower-dimensional function and a constant function.

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