43

Besse (1978, p.202) has the spectra of compact rank 1 symmetric spaces (CROSSes). In addition to $\mathrm S^n$ due apparently to Heine (1863, §19; 1878, §128), this gives $\mathbf{RP}^n$, $\mathbf{CP}^n$, $\mathbf{HP}^n$ and $\mathbf{OP}^2$.
Edit: Also, for $M$ a compact semisimple Lie group it is well known (due apparently to Freudenthal (1954))1 that the ...

answered Sep 24 '15 at 0:53

Francois Ziegler

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22

As Raziel wrote, the local question is whether one can find a local basis of orthonormal vector fields that are divergence-free.
It's true that, in dimension $2$, this can only be done if the metric is locally flat, which is the local obstruction. This is because this is an overdetermined problem; one has two equations for a single unknown.
However, in ...

20

You can compute the eigenvalues explicitly for flat tori $\mathbb{R}^n/\Gamma$, where $\mathbb{R}^n$ has the standard Euclidean metric and $\Gamma$ is a lattice. The eigenvectors all have the form $e^{2 \pi i \langle v, w \rangle}$ where $v \in \mathbb{R}^n$ and $w \in \Gamma^{\vee}$ lies in the dual lattice. The corresponding eigenvalue is $-4 \pi^2 \| w \|^...

answered Sep 24 '15 at 0:44

Qiaochu Yuan

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17

These are usually known as the Laplacian, the normalized Laplacian and the unsigned Laplaian. All three are positive semidefinite. If the graph is regular, they all provide the same information.
If the graph is not regular they are, in general, independent. The normalized Laplacian is the right tool for the analysis of random walks. The spectral information ...

16

The exponentials used in Fourier series are eigenvalues of shifts, and thus of any operator commuting with shifts, not just Laplacian. Similarly, spherical harmonics carry irreducible representations of $SO(3)$, and so they are eigenfunctions of any rotationally invariant operator.
If the underlying space has symmetries, it's no wonder that a basis ...

15

I was informed by Sugata Mondal at the MPI that Scott Wolpert proved the following result in his 1994 Annals paper Disappearance of cusp forms in special families:
Theorem 5.14. The eigenvalues of the Laplacian above $\tfrac14$ on a closed hyperbolic surface vary nontrivially under analytic deformations.
That is, if $g_t$ is a real-analytic path of ...

15

Yes, Brendan McKay showed that almost all trees have mates that are simultaneously cospectral in both adjacency and Laplacian spectra. And more.
http://users.cecs.anu.edu.au/~bdm/papers/SpectralTrees.pdf
Edit: I wondered briefly what the smallest pair of such trees would be, and a few minutes of Sage told me that there are two on 11 vertices.
And here ...

13

Yes there is. Here is how you do it. First find an orthogonal change in variables
$$ x_j=\sum_{jk} s_{jk}y_k $$
$(s_{jk})$ orthogonal matrix, so that in the new coordinates we have
$$ \sum_{i,j}a_{ij} x_ix_j = \sum_j \mu_j^2 y_j^2, $$
where $\mu_j^2$ are the eigenvalues of the symmetric matrix $(a_{ij})$.
Note that $\newcommand{\ii}{\boldsymbol{i}}...

answered Oct 1 '12 at 9:25

Liviu Nicolaescu

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13

Jeffrey Weeks has computed the spectra of homogeneous elliptic manifolds.
For arithmetic hyperbolic manifolds, the spectrum is in principle computable in
the sense that one may define a Selberg zeta function arithmetically, whose roots give the spectrum.
Certain other homogeneous Heisenberg manifolds have their spectra computed.

12

The answer is 'no', as you can see by taking the case of $M$ being the unit $2$-sphere in $\mathbb{R}^3$ and the geodesic $\gamma$ being a great circle, say, the horizontal great circle given by $z=0$. If you consider the harmonic polynomials of degree $2$ in $x,y,z$ restricted to the $2$-sphere, these are eigenfunctions of the Laplacian on the $2$-sphere, ...

12

A $k$-cycle works if $k>2$. For $k=1$ any tree works. I don't think $k=2$ is possible unless you allow double edges: if the graph is not a tree then it has an $m$-cycle $C$ for some $m>2$; remove edges off $C$ until the graph is connected but has no cycle other than $C$, and then removing an arbitrary edge of $C$ yields at least $m$ spanning trees.

answered Jun 18 '16 at 23:18

Noam D. Elkies

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11

By the classical regularity theory for the Poisson equation, you have
$$
\Vert \nabla \psi \Vert_{L^\infty (B_1)} \le C \bigl(\Vert \Delta \psi \Vert_{L^\infty (B_2)} + \Vert \psi \Vert_{L^\infty (B_2)} \bigr).
$$
See for example Gilbarg and Trudinger, Elliptic partial differential equations of second order, 1983, theorem 3.9, where it is proved by the ...

answered Aug 17 '15 at 12:31

Jean Van Schaftingen

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10

You can use the general local formula for the Laplace-Beltrami operator in terms of any local orthonormal frame:
$$\Delta = \sum_{i=1}^n W_i^2 +\mathrm{div}(W_i)W_i$$
where the $W_i$'s are seen as derivations on functions.
You can always find a local frame of vector fields $W_1,\ldots,W_n$ that are divergence-free at a given point $q$. In terms of this ...

10

Just to add an (in my opinion important) piece of information. Say $F$ is a function on the vertices of a graph, so $F:V \to \mathbb{R}$. Then $\nabla F$ is a function from the edges to $\mathbb{R}$ (here I see an edge as a pair of vertices $(x,y)$, so edges are oriented):
$$\nabla F (x,y) := F(y) - F(x)$$
Now this definition is very natural in many ways. ...

9

The answer is no.
To see this consider two (2D) sphere's of constant curvature 1 and a (flat) cylinder of unit length and radius $\epsilon$. Cut out a disk from each sphere and glue in the cylinder (and smooth everything out in a small neighborhood of the gluing). By Gauss Bonnet this must introduce negative curvature.
Now take a function which is ...

8

This does not appear to be known. Clearly the remaining eigenfunctions are the same as those for an equilateral triangle with Dirichlet conditions on two sides and Neumann conditions on one side. It is known that these eigenfunctions are not trigonometric, see
the discussion given here: http://www.m-hikari.com/ams/ams-password-2008/ams-password57-60-2008/...

8

The answer is actually as nice as could be. The number of spanning trees of $G_n$ is
$$
\frac{1}{3^n}2^{n 2^{n-1}}\prod_{k=0}^{n-1} \big(1-(-2)^{k-n}\big)^{\binom{n}{k}}
$$
This follows directly from the theorems in the comment above. The divisibility (what a beautiful word!) properties then follow from the formula.

8

For $N \geq 5$ it is still not known if the $N$-gon which minimizes the first eigenvalue under area constraint (which exists), is the regular one. I have done some numerical computations which suggest that the regular polygons are indeed optimal. You can see the numerical ideas here (recent version) or here (old version).

8

This holds in complete generality. I sketch the case in which, instead of the ball $B(x,h)$, you consider the metric sphere $S(x,h)$. In particular (here $\dim M = n$ and the Laplacian is $\Delta = \mathrm{div}\circ \mathrm{grad}$):
$$ (\Delta u)(x) = \lim_{h\to 0} \frac{2n}{h^2}\frac{1}{|S(x,h)|}\int_{S(x,h)} [u(y)-u(x)] dy .$$
The case of the ball is ...

7

Terry Lyons in "Instability of the Liouville property for quasi-isometric Riemannian manifolds and reversible Markov chains", Journal of Diffe Geometry, 1987, constructs examples of graphs (and Riemannian manifolds) $G, G'$ which are quasi-isometric, but one admits no (nonconstant) positive harmonic functions while the other admits bounded (nonconstant) ...

7

Start with the unweighted case. We have $L=BB^T$ where $B$ is (what I call) the incidence matrix of an orientation of $G$. So $B$ has one 1 and one $-1$ in each column with all other entries zero. The left kernel of $B$ consists of the functions on $V(G)$ that are constant on the components of $G$. Since $B$ and $BB^T$ have the same rank, we get the ...

7

As Sebastian Goette explained in his comment, the sub-Laplacian $\Delta_{sub}$ depends in general from an additional structure. And so does the Laplace-Beltrami $\Delta$ that you use to compute the difference. Let me elaborate.
SUB-LAPLACIANS
On a given smooth manifold $M$, we consider a sub-Riemannian structure $(\mathcal{D},g)$, where $\mathcal{D} \...

6

This is a classical computation' going back to the 19th century. The formula for the integral kernel of the resolvent in any dimension is a function of $r = \mbox{dist}(x,y)$.
In three dimensions it is particularly simple: $r^{-1} e^{i\lambda r}$ (up to some constant factors) where $\lambda^2 = z$. The formula in any odd dimension is only a bit harder and ...

6

You can just prove it yourself directly in local holomorphic coordinates. Indeed, the $\overline{\partial}$ Laplacian on functions is equal to $\Delta_{\overline{\partial}}f=g^{i\overline{j}}\partial_i \partial_{\overline{j}}f$. Apply this to $|\partial f|^2=g^{k\overline{\ell}}\partial_k f \partial_{\overline{\ell}}f$ (length squared of $\partial f=(df)^{(1,...

6

Generalizing the case of flat tori, one can compute explicitely the spectrum of many compact flat manifolds. See for instance
spectrum on $p$-forms Miatello and Rossetti or the survey on isospectral compact flat manifolds in Contemp. Math. 491 AMS, 83--113.
There are also more progress for lens spaces than the one by Sakai mentioned by Ziegler.
See for ...

6

For nearest neighbor Laplacian, you can find a short self-contained proof of the formula $$u(x)=\frac1{2\pi}\log |x|+c+O\left(\frac1{|x|^2}\right)$$ at these lecture notes, pages 6-7. It is a simple exercise in Laplace's method to work out further coefficients of the asymptotic expansion, if necessary. The proof there is not the shortest possible; the ...

6

To add a more general, optimistic answer to the question in the title, rather than the question in the question --- while the isomorphism class of $\pi_1(M)$ is not determined by the Laplace spectrum, it is indeed possible to extract topological information.
Let $M$ be a $d$-dimensional Riemannian manifold with Laplace spectrum $\lambda_k$. McKean and ...

6

If $G_1$ and $G_2$ are graphs let $G_1 \vee G_2$ denote their wedge sum. That is, $G_1 \vee G_2$ is obtained by taking the one-point union of $G_1$ and $G_2$. It will not matter what vertices we decide to identify. If $G_1$ and $G_2$ have $k_1$ and $k_2$ spanning trees respectively, then $G_1 \vee G_2$ has $k_1k_2$ spanning trees.
As Noam points out a $k$-...

6

No, we cannot (completely) hear the shape of a drum, even if it is
spinorial. Two metric fields with the same collection of eigenvalues
are called isospectral. There exist Dirac isospectral deformations;
continuous 1-parameter families of mutually non- isometric metrics
with the same Dirac spectrum have been constructed. They are of the
form $M_s =...

answered May 13 '17 at 17:30

Carlo Beenakker

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