Eigenfunctions of the Laplace-Beltrami operator of a torus

Question

The eigenfunctions of the Laplace-Beltrami operator of the flat torus $\mathbb{T}^2=\mathbb{R}^2/\mathbb{Z}^2$ and their multiplicity are well-known.

What happens if we change the sides of the torus and we consider two sides that are not commensurable, for example $\mathbb{T}_{1,\sqrt{2}}^2=\mathbb{R}^2/(\mathbb{Z}\times\sqrt{2}\mathbb{Z}$)?

What are the eigenfunctions of the Laplace-Beltrami operator of $\mathbb{T}_{1,\sqrt{2}}^2$? and their multiplicity?

Thank you in advance

Answer 1

Berger, Gauduchon, and Mazet studied this case in "Spectre d'une variété Riemannienne" and showed that isospectral flat tori of dimension 2 (and 1) are isometric. In order to do it, they provide a full description of the spectrum of every flat torus of dimension 2. However this result is not true for every dimension as it is shown in the same book. The more general related question is "Can one hear the shape of a drum?". For the case of tori this is: For any pair of flat tori of dimension $m$, does isospectrality imply the existence of an isometry between the tori? The answer is Yes for dimensions 1, 2 and 3. Not necessarily for higher dimensions. Note that the case for dimension 3 was proved in 1997 by Schiemann.

Berger et al. proved the following:

Let $\Gamma$ be a lattice of maximum rank in $\mathbb R^2$, i.e. there is $\bf{e}_1,\bf{e}_2$, a basis of $\mathbb R^2$, such that $ \Gamma :=\left\{\left. p_1 \mathbf{e}_1 + p_2 \mathbf{e}_2 \right| p_1,p_2 \in \mathbb Z\right\} $. Then $T_\Gamma = \mathbb R^2 /_\Gamma$ is the flat torus induced by $\Gamma$.

Now consider the existing and uniquely determined lattice $\Gamma^*:=\{\bf{y}\in\mathbb R^n |\langle\bf{x},\bf{y}\rangle \in\mathbb Z,\forall\bf{x}\in\Gamma\}$. The lattice $\Gamma^*$ is called the dual lattice of $\Gamma$.

The following Theorem holds:  Let $T_\Gamma$ be a flat torus, then

• $\lambda \in \text{spec}(T_\Gamma)$ if and only if there is an $\bf{x} \in \Gamma^*$ such that $\lambda = \|\bf{x}\|^2 4 \pi^2$. -The geometric multiplicity $m_\lambda = \dim(V_\lambda(T_\Gamma))$ is even if $\lambda \neq 0$ and \begin{equation*} m_\lambda = \left|\left\{\bf{x}\in\Gamma^* \left|\| \bf{x}\|^2 = \frac{\lambda}{4 \pi^2}\right.\right\} \right|. \end{equation*}

• For every $\lambda \in \text{spec}(T_\Gamma)$, the set $$ B_\lambda:=\left\{ f_{\bf{x}} \left| \bf{x} \in \Gamma^*, \|\bf{x} \|^2= \frac{\lambda}{4 \pi^2} \right. \right\}$$ is a basis of the eigenspace $V_\lambda (T_\Gamma)$.

Answer 2

This is a nice exercise. More generally consider a flat torus $\mathbb{R}^n/\Gamma$ where $\Gamma$ is a lattice. I claim that the eigenfunctions of the Laplacian in this case are given by

$$f_w : \mathbb{R}^n \ni v \mapsto e^{2 \pi i \langle v, w \rangle} \in \mathbb{C}$$

where $w$ runs over all elements of the dual lattice $\Gamma^{\vee}$ consisting of all vectors such that $\langle v, w \rangle \in \mathbb{Z}$ for all $v \in \Gamma$. This is a necessary and sufficient condition for a function defined as above to be $\Gamma$-invariant. The eigenvalue of $f_w$ is $-4 \pi^2 \langle w, w \rangle$ by direct computation; thus the multiplicities are controlled by the multiplicities of the lengths of vectors in $\Gamma^{\vee}$.

In this particular case, if $\Gamma$ is spanned by $(1, 0)$ and $(0, \sqrt{2})$ then $\Gamma^{\vee}$ is spanned by $(1, 0)$ and $\left( 0, \frac{1}{\sqrt{2}} \right)$. A vector $\left( n, \frac{m}{\sqrt{2}} \right) \in \Gamma^{\vee}$ (so that $n, m \in \mathbb{Z}$) has length $n^2 + \frac{m^2}{2}$, so the multiplicity of an eigenvalue $-4 \pi^2 \langle w, w \rangle$ is determined by how many ways you can write $\langle w, w \rangle$ in that form.

To show that this construction gives all eigenfunctions you can show that the span of these eigenfunctions is dense in continuous functions on $\mathbb{R}^n/\Gamma$ by Stone-Weierstrass. Any remaining eigenfunctions can be chosen to be orthogonal to these so there can't be any left.

Incidentally, this is one relatively straightforward way to construct examples of Riemannian manifolds which are isospectral but not isometric: it suffices to find two lattices with the same theta function which are not isometric (using the fact that, by Poisson summation, the theta function of a lattice determines the theta function of its dual lattice). This observation is due to Milnor; see page 30 of these notes by Elkies.

Answer 3

I wanted to follow up on Qiaochu's answer by discussing the multiplicity of the eigenvalues. As Qiaochu mentions, for any $w = (m_1, m_2/\sqrt{2})$, $m_1,m_2 \in \mathbb{Z}$, in the dual lattice, the function $\mathbb{R}^2 \rightarrow \mathbb{C}, \, v=(v_1,v_2) \mapsto f_w(v)$ given by $$f_w(v) = \exp\left(2 \pi i\left(v_1m_1+\frac{v_2m_2}{\sqrt{2}}\right)\right)$$ is $\Gamma$-periodic and descends to a map on the torus. Also, it is an eigenfunction of $\Delta$, $$\Delta f_w = -4 \pi \left(m_1^2 + \frac{m_2^2}{2}\right)f_w,$$ with eigenvalue $-4\pi(m_1^2+m_2^2/2).$

We see that the eigenvalues are of the form $-4\pi(n/2) = -2\pi n$, with multiplicity $$\#\left\{(m_1,m_2) \in \mathbb{Z}^2 : 2m_1^2 + m_2^2 = n\right\}.$$ Note that $2m_1^2+m^2$ is a positive definite quadratic form with discriminant $D=-8$. Note that the class number $h(D) = h(-8) = 1$ of this discriminant is $1$, which you can also see by examining the narrow class number of the corresponding number field $\mathbb{Q}(\sqrt{-2})$. Dirichlet worked out the number of representations of quadratic forms with class number $h(D)=1$. In particular, for $D=-8$, we have that the multiplicity of the eigenvalue $-2\pi n$ is $$\#\left\{(m_1,m_2) \in \mathbb{Z}^2 : 2m_1^2 + m_2^2 = n\right\} = 2 \sum_{d | n} \left(\frac{-8}{d}\right),$$ where $(\frac{\cdot}{\cdot})$ is the Kronecker symbol.

A good reference for further information about the number of representations of quadratic forms is Chapter 11 of Iwaniec's Topics in Classical Automorphic Forms.