46 votes
Accepted

Explicit eigenvalues of the Laplacian

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}^...
22 votes
Accepted

Can the Laplace operator on $n-$ manifolds be represented as a sum of $n$ second order derivational operators

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 ...
20 votes

Explicit eigenvalues of the Laplacian

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^{...
17 votes
Accepted

Are these three different notions of a graph Laplacian?

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 ...
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16 votes

Why decompose a function with eigenvectors of Laplace operator?

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 ...
  • 6,044
15 votes
Accepted

is it possible to have two non-isomorphic non-regular graphs with the same adjacent spectrum and the same laplacian spectrum?

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....
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13 votes

Explicit eigenvalues of the Laplacian

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 ...
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12 votes
Accepted

Gagliardo Nirenberg inequality for the laplacian

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 (...
12 votes
Accepted

Create a graph with a specified number of spanning trees

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 $m>2$; ...
10 votes

Can the Laplace operator on $n-$ manifolds be represented as a sum of $n$ second order derivational operators

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 ...
  • 3,023
10 votes
Accepted

How to understand the combinatorial Laplacian $\Delta$ which is defined on the graph?

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 $\mathbb{R}$ (...
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9 votes
Accepted

Lower bound on the first eigenvalue of the Laplace-Beltrami on a closed Riemannian manifold

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 (...
  • 1,089
9 votes

Line graphs called "graph derivatives": any intuition?

If one considers a graph $G=(V,E)$ and a function $f:V\to\mathbb{R}$, it makes sense to look at the finite differences $f(v_i)-f(v_j)$ for neighboring vertices $v_i,v_j\in V$ as a sort of discrete ...
  • 2,499
8 votes

First eigenvalue of the Laplacian on a regular polygon

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 ...
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8 votes
Accepted

Laplace-Beltrami and averaging

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 = \...
  • 3,023
8 votes
Accepted

Difference between the Laplacian and the sub-Laplacian of a Lie group

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 ...
  • 3,023
7 votes
Accepted

heat kernel on n-sphere

Yes, it is true that $\theta$ on $S_n$ is dominated by $\theta$ on $\mathbb{R}^n$. Let $(B_t)$ be a Brownian motion on the sphere. The radial process $\theta_t=d(x,B_t)$ is a Jacobi process, that is ...
7 votes

Connectivity of weighted graph and zero Laplacian eigenvalues

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 ...
  • 11.8k
7 votes

Explicit eigenvalues of the Laplacian

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 ...
  • 2,158
7 votes

Spectra of the Laplacian operator on the spherical space-form

In addition of what you mentioned, the multiplicity of $k(k+2)$ in the Laplace spectrum of $S^3$ is given by $\dim H_k$, where $H_k$ is the space of harmonic homogeneous complex-valued polynomials of ...
  • 2,158
7 votes
Accepted

Eigenvalues and eigenfunctions of the Laplace operator on entire plane

The point spectrum coincides with the spectrum minus 0 if $p>2n/(n-1)$ and it is empty in the remaining cases ($n$ is the dimension). This is proved in G. Talenti: "Spectrum of the Laplace ...
6 votes
Accepted

Fundamental solution of Discrete Laplace in the plane

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 ...
  • 6,044
6 votes

Spectral properties of the Laplace operator and topological properties

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, ...
  • 718
6 votes

Create a graph with a specified number of spanning trees

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 ...
6 votes

General questions on the eigenfunctions of Laplacian and Dirac operators

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 ...
6 votes
Accepted

Laplace-Beltrami and the isometry group

For the maximally symmetric case with the second order operator, the proof is very simple. A second order differential operator acting on scalars can be written in the form $$ b^{ij} \nabla^2_{ij} +...
  • 31.9k
6 votes

Graph Laplacian Operator

Your operator is a rank one perturbation of the multiplication operator $(Mf)(x) = (x/2)f(x)$, which has (purely) absolutely continuous spectrum equal to $[0,1/2]$. Since the ac spectrum is invariant ...
6 votes

On equation $\Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemannian manifold

This is an expansion on Igor's comment and fixes my previous mistake (see also Willie's answer). A direct computation gives $$ D(f) = 2f^{ab}\nabla_{(a}X_{b)} + X_a(\nabla^b\nabla_b\nabla^a-\nabla^a\...
6 votes
Accepted

On equation $\Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$ on a Riemannian manifold

The following formula is known among the experts but hard to find in the literature, so I figure I will document it here. Throughout $(M,g)$ denote an arbitrary pseudo-Riemannian manifold, and $\nabla$...
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