Consider a connected (we define connected components by defining the set of vertices where every vertex has one neighbour) sublattice $V$ of the square lattice $V \subset\mathbb{Z}^2.$

On this we define the discrete Laplacian as $T:\ell^2(V) \rightarrow \ell^2(V)$ by

$$ (Tf)(x)=\sum_{y \text{ neighbour of } x}(f(y)-f(x)).$$

Now, my question is: Is there an infinite(!) connected sublattice $V \subset\mathbb{Z}^2$ on which this objects has eigenfunctions?

Why do I ask: It is easy to see that on $\mathbb{Z}^2$ by using the Fourier transform for example, the spectrum is purely absolutely continuous.

I would like to understand whether this is because $\mathbb{Z}^2$ is infinite (and so every eigenfunction would dissolve to infinity) or whether this is because $\mathbb{Z}^2$ is translational invariant.

Why the assumptions: If the sublattice is finite, we obtain eigenfunctions because this operator is just a matrix.

If the sublattice was not connected, it could have a finite connected component.

So to exclude this, I stated the assumptions.


Yes, I'm sure this is possible, and I think there will be many ways to do it, but I don't have a fully worked out rigorous argument right now. But let me throw out some ideas. First of all, a 1D Laplacian with a random potential (aka Anderson model) $$ (Hf)(n) = f(n+1)+f(n-1)+q(n)f(n) $$ on $\ell^2(\mathbb Z)$ will have pure point spectrum with prob $1$. For example, this is true if the $q(n)$ are iid and take two values. Now you could simulate this by taking your $V$ as $V=\mathbb Z\cup \{(2n,1): n\in A\}$, with $A\subseteq \mathbb Z$ a random set produced in the same way. It's not totally clear to me if there is an easy way to reduce this to the original model or if one would have to go through the whole analysis again to prove the result for $\Delta$ on $\ell^2(V)$, but I have no doubt about the result itself (in the classical case, too, the result is fairly robust and doesn't depend on how exactly you introduce the randomness).

What is easier to prove is absence of absolutely continuous spectrum. For example, if you take $A$ above (not random, but) as a sparse set, say $A=\{ \pm 2^n \}$, then $\sigma_{ac}=\emptyset$. This follows from my result on reflectionless limit points. Typically, the spectrum will be purely singular continuous; in particular, this will be true if $A$ is something like $A=\{ \pm 2^{n^2}\}$.

  • $\begingroup$ Since $A$ is `discrete' (that is, it contains no two consecutive numbers), then finding an eigenfunction of the corresponding graph Laplacian with eigenvalue $\lambda$ is equivalent to solving $$f(n+1) + f(n-1) - 2 f(n) = \lambda (1 + \tfrac{1}{\lambda + 1} \mathbb{1}_A(n)) f(n)$$ for integer $n$ (unless $\lambda = -1$; in this case necessarily $f(n) = 0$ for $n \in A$. This is not exactly the Anderson model, but apparently it is indeed not much different. $\endgroup$ Oct 12 '17 at 17:19
  • $\begingroup$ @MateuszKwaśnicki: Yes, thanks for making it clearer, this is what I had in mind. It is not obviously reducible to a classical Anderson model, but it ought to have the same properties anyway. $\endgroup$ Oct 12 '17 at 20:45

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