Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 18 / 1

Let $M={\mathbb{R}}^d/\Lambda$ be a flat torus, where $\Lambda$ is a full-rank Euclidean lattice, and use the nonnegative Laplace-Beltrami operator $\Delta =-{\sum }_{k=1}^d{\partial }_k^2$. The dual lattice isFor $w\in {\Lambda }^{\ast }$, the function ${\varphi }_w(x)=\mathrm{vol}(M{)}^{-1/2}{e}^{2\pi i⟨w,x⟩}$ is periodic under $\Lambda$, so it defines a function on $M$, and direct differentiation gives $\Delta {\varphi }_w=4{\pi }^2|w{|}^2{\varphi }_w$. Integrating characters on a fundamental parallelepiped proves that these functions are orthonormal. After choosing a lattice basis, the ordinary Fourier series theorem on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ proves completeness, so there are no additional eigenvalues. Thus the spectrum of a flat torus iswith multiplicity equal to the number of dual vectors of the given length; the zero eigenvalue occurs once.

It remains to show that a two-dimensional Euclidean lattice is determined up to an orthogonal transformation by its vector-length multiset. This is the two-dimensional lattice reconstruction from vector lengths. Let $v$ be a shortest nonzero vector, of length $a$. It is primitive: $v=ku$ with integer $|k|>1$ would give a shorter nonzero vector. The lattice vectors on its line are therefore precisely the integer multiples of $v$. Subtract two occurrences of each length $ka$, $k=1,2,\dots$, from the length multiset. This removes exactly one primitive lattice line, even when other vectors happen to have the same lengths. The smallest remaining length $b$ is the length of a shortest vector $w$ independent of $v$.

These two vectors form a basis of the lattice. Otherwise a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ has a lattice representative $u=sv+tw$ with $|s|,|t|\le 1/2$. It cannot have $t=0$, since primitivity of $v$ would then force $u=0$. It is therefore independent of $v$, whileThe inequality is strict: if both coefficients are nonzero, the vectors are not parallel and the triangle inequality is strict; if $s=0$, then $|u|\le b/2$. This contradicts the choice of $w$.

The covolume $A$ is also determined by the length multiset. If $N(R)$ counts lattice vectors in a radius-$R$ disk, thenTo justify the estimate, tile the plane by a bounded fundamental parallelogram. Only cells meeting a fixed-width neighbourhood of the disk boundary contribute an error; their total number is $O(R)$. The lattice basis $v,w$ has Gram matrixThus $|c|=\sqrt{a^2{b}^2-A^2}$ is determined. Replacing $w$ by $-w$ makes $c\ge 0$, so the Gram matrix is determined entirely. Equal Gram matrices give an orthogonal map between the lattice bases and hence between the lattices.

Two isospectral two-dimensional flat tori consequently have orthogonally equivalent dual lattices. Taking duals shows that their original lattices are orthogonally equivalent too, and the orthogonal map descends to an isometry of the quotient tori. Two isospectral flat tori of dimension two are isometric.