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For convex domains $\Omega \subset \mathbb{R}^n$ with Dirichlet boundary conditions, it's known that any first Laplacian eigenfunction is log-concave. In particular, it has a unique maximum.

These are functions $f$ satisfying $-\Delta f = \lambda_1 f$ for the smallest possible $\lambda_1 > 0$ and $f|_{\partial \Omega} = 0$.

Q: Is there such a concavity result for homogeneous spaces, ie. Riemannian manifolds with a transitive action of the isometry group? Does $f$ at least have a unique maximum?

I imagine that this might even be apparent from an explicit formula for these eigenfunctions, but I haven't been able to find one in general.

I can at least check that it holds for spherical harmonics. Actually for spheres I think it can be derived from the domain result by separation of variables (take $\Omega = \mathbb{R}^n$). Perhaps such a method can work for any homogeneous space with a nice enough embedding, such as as the boundary of a convex domain?

I'd appreciate any thoughts or relevant references!

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  • $\begingroup$ For the sphere $\|x\|=1$, $\lambda_1=0$ and $f$ is constant if we consider the Laplace-Beltrami. $\endgroup$ May 9, 2022 at 17:52
  • $\begingroup$ @GiorgioMetafune I'm interested in the first non-trivial eigenvalue, $\lambda_1 > \lambda_0 = 0$. $\endgroup$
    – user404153
    May 9, 2022 at 17:56
  • $\begingroup$ But in that case the eigenfunctions change sign. $\endgroup$ May 9, 2022 at 17:59
  • $\begingroup$ @GiorgioMetafune so? You can shift them to be positive if you want. $\endgroup$
    – user404153
    May 9, 2022 at 18:03

1 Answer 1

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The flat torus $\mathbb{T} = \mathbb{R}^2/\Lambda$ gives a counterexample: The first nontrivial eigenvalue is of the form $\lambda_1 = \xi_1^2+\xi_2^2$, where $\xi = (\xi_1,\xi_2)$ is a nonzero element of the dual lattice $\Lambda^*$ of smallest norm, and the correspnding eigenfunctions are of the form $f(x_1,x_2) = a\cos(\xi_1 x_1 + \xi_2 x_2 + b)$ for some constants $(a,b)$. This function has a whole circle of maxima.

In general, on a compact manifold, an eigenfunction with nonzero eigenvalue must change sign because its average value on the manifold must be zero. (Integration by parts.)

Oh, another example occurred to me that you might find more interesting: Let $M = \mathrm{SO}(3)$ with its standard biïnvariant Riemannian metric. The first nontrivial eigenvalue has multiplicity 9, and the corresponding eigenfunctions are the 9 entries $a_{ij}$ of the standard matrix embedding of $\mathrm{SO}(3)$ into the space of 3-by-3 matrices. Each $a_{ij}$ has its maximum and minimum values equal to $\pm 1$, but it attains each on a circle embedded in $\mathrm{SO}(3)$. Meanwhile $f = -a_{11}-a_{22}-a_{33}$ has a maximum value of $1$, attained on a copy of $\mathbb{RP}^2$ embedded in $\mathrm{SO}(3)$, and a minimum value of $-3$, attained only at $I_3\in\mathrm{SO}(3)$.

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  • $\begingroup$ Thanks! That's a good point. I suppose this counter-example is due to the product structure (your function is a product with a zero eigenvalue on one factor), and it question in the title still has some hope of a positive answer for non-product spaces. $\endgroup$
    – user404153
    May 9, 2022 at 18:16
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    $\begingroup$ @user404153: Actually, if the lattice $\Lambda$ is not a rectangular lattice, the torus is not a product (except locally). For example, suppose that $\Lambda$ is the hexagonal lattice, generated by three unit vectors that sum to zero. That torus is not a product of two circles. $\endgroup$ May 9, 2022 at 18:27
  • $\begingroup$ Hm, fair enough. Thanks again! $\endgroup$
    – user404153
    May 9, 2022 at 18:36
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    $\begingroup$ @user404153: Just to be completely clear. That last sentence in my comment should have been "That torus is not metrically the product of two circles." Obviously, it is topologically (and smoothly) the product of two circles. $\endgroup$ May 9, 2022 at 19:08

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