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The covolume of a full Euclidean lattice is the Lebesgue measure of a fundamental parallelepiped in its ambient inner product space. In orthonormal coordinates it is $|\det B|$ for a lattice basis matrix $B$. Scaling the lattice by $t>0$ multiplies the covolume by $t^n$; the dual lattice has reciprocal covolume.

Let $j(\mathfrak{b})$ be the Minkowski embedding of a number field applied to a nonzero fractional ideal. Under the metric $|x{|}^2={\sum }_{v\text{ real}}{x}_v^2+2{\sum }_{v\text{ complex}}|x_v{|}^2$, its covolume is $\sqrt{|D_K|}N(\mathfrak{b})$. For ordinary coordinate Lebesgue measure $\prod d{x}_v\prod d\mathrm{Re}{z}_{v}d\mathrm{Im}{z}_v$, its covolume is instead $2^{-r_2}\sqrt{|D_K|}N(\mathfrak{b})$. Here $D_K$ is the field discriminant, and the absolute norm of a fractional ideal is positive and multiplicative. The formula follows by taking the embedding determinant of an integral basis, then using the index of an integral ideal and scaling to handle a fractional ideal.

Let $C=\sqrt{|D_K|}N(\mathfrak{b})$ and $n=[K:\mathbb{Q}]$. Normalize the Minkowski embedding of a number field to the covolume-one Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C^{-1/n}j(\mathfrak{b})$, using the metric with complex coordinates weighted by $2$. Its theta function isWriting $\Vert y\Vert ={\prod }_v{y}_v^{[K_v:\mathbb{R}]}$ and using the trace dual of a fractional ideal gives $\Theta (y,\mathfrak{b})=\Vert y{\Vert }^{-1/2}\Theta (y^{-1},{\mathfrak{b}}^{\vee })$. Its small-parameter asymptotic of a lattice theta sum has leading coefficient one. For the unscaled ideal lattice, the leading coefficient is instead $C^{-1}$; one must specify the normalization when quoting this limit.

The dual lattice of a full-rank Euclidean lattice $\Lambda \subset {\mathbb{R}}^n$ isHere $⟨x,y⟩$ is the Euclidean inner product. If the columns of an invertible matrix $B$ form a basis of $\Lambda =B{\mathbb{Z}}^n$, thenIndeed, $y$ belongs to the dual lattice exactly when $B^{\mathsf{T}}y\in {\mathbb{Z}}^n$; $B^{-\mathsf{T}}$ denotes the transpose of the matrix inverse. The reciprocal covolume follows by taking the determinant. With the Fourier transform kernel $e^{-2\pi i⟨x,y⟩}$, these are precisely the frequencies of the characters of a real torus: translating $x$ by any $\lambda \in \Lambda$ leaves $e^{2\pi i⟨x,y⟩}$ unchanged. In one dimension, $(h\mathbb{Z}{)}^{\vee }=h^{-1}\mathbb{Z}$ for $h>0$: increasing the spacing in physical space decreases the spacing in frequency space.

For a square integer matrix $A$ with nonzero determinant, the image $A{\mathbb{Z}}^n$ is a subgroup of the integer lattice. Its index of a subgroup isIn particular, $A=\mathrm{diag}(2,1)$ gives all points with even first coordinate. The quotient has two cosets, represented by $(0,0)$ and $(1,0)$, and the fundamental cell has area two. A determinant of two therefore counts how much integer information is missing from the image.

The integrality hypothesis is essential. The rational matrix $R=\mathrm{diag}(2,1/2)$ has determinant one but sends $(0,1)$ to $(0,1/2)$. Its generated Euclidean lattice has the same covolume as ${\mathbb{Z}}^2$ without being the same integer lattice. The unimodular matrix criterion combines an arithmetic condition with a geometric one.

A short proof of the index formula uses Smith normal form. Write $PAQ=\mathrm{diag}(d_1,\dots ,d_n)$, with $P,Q$ unimodular matrices and positive integers $d_i$. They preserve the relevant integer structure, so the quotient has the same size as ${\prod }_i\mathbb{Z}/d_i\mathbb{Z}$, namely ${\prod }_i{d}_i=|\det A|$. No assumption that the matrix is orthogonal or its basis vectors are short is needed.

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.

Use the positive Laplace-Beltrami operator, $\Delta =-\mathrm{div}\nabla$. A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice and the metric descends from the Euclidean metric. Its dual lattice isIf $V$ is the covolume of $\Lambda$, the functions ${\varphi }_{\xi }(x)=V^{-1/2}{e}^{2\pi i⟨\xi ,x⟩}$ descend to the torus. Direct differentiation gives the spectrum of a flat torus:The multiset includes one entry for each dual vector. In particular zero has multiplicity one. Writing $\Lambda =B{\mathbb{Z}}^d$ identifies the dual vectors with $B^{-T}m$, $m\in {\mathbb{Z}}^d$. Standard Fourier series on the unit cube then prove that these functions are an orthonormal basis of $L^2$ on the torus. If $\Delta f=\lambda f$, its Fourier coefficients satisfy $(4{\pi }^2|\xi {|}^2-\lambda )\hat{f}(\xi )=0$. Thus every eigenfunction is a linear combination of the indicated shell of frequencies, with no additional eigenvalues. Compactness gives the discrete self-adjoint realization, whose eigenfunctions are smooth. For real functions the two characters at $\xi ,-\xi$ become cosine and sine, giving the same real multiplicities. The opposite Laplacian sign would negate all displayed eigenvalues without changing the rigidity conclusion.

To prove two-dimensional spectral rigidity, let $L={\Lambda }^{\ast }$ and recover its Gram matrix from its vector-length multiset. Let $a$ be the shortest nonzero length and choose $v\in L$ with $|v|=a$. This vector is primitive: a proper integer multiple would have a shorter lattice vector. Hence $L\cap \mathbb{R}v=\mathbb{Z}v$. From the spectral multiset subtract exactly two occurrences of each length $ka$, $k=1,2,\dots$. The remaining multiset consists precisely of the vectors not on this line, including correct multiplicities at coincident lengths. Its least length $b$ is the length of a shortest vector $w$ independent of $v$.

The pair $(v,w)$ is a basis of $L$. Otherwise take a lattice point in a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ and reduce its two coefficients into $[-1/2,1/2]$. The resulting nonzero lattice vector $z=\alpha v+\beta w$ is not on $\mathbb{R}v$, by primitivity of $v$. Butwith strict inequality: when both coefficients are nonzero the triangle inequality is strict for independent vectors, and when one vanishes its length is at most $b/2$. This contradicts the definition of $b$. Subtract an integer multiple of $v$ from $w$ and, if necessary, change its sign to arrange $0\le c=⟨v,w⟩\le a^2/2$; minimality of $b$ makes this reduction possible without decreasing its length.

The covolume $A$ of $L$ is also spectral. The count of lattice vectors with length at most $R$ obeysFor example, translate a bounded fundamental parallelogram of diameter bound $D$: the union of tiles with centres in $B_R$ contains $B_{R-D}$ and lies inside $B_{R+D}$. Comparing their areas proves the leading coefficient. The spectrum supplies this count. Since $(v,w)$ is a lattice basis, its parallelogram area satisfies $A^2=a^2{b}^2-c^2$. ThereforeAll three quantities are determined by the spectrum. Equal Gram matrices give an orthogonal map between the dual lattices; taking duals gives an orthogonal map between the original lattices, descending to a Riemannian isometry of the tori. Isospectral flat two-tori are isometric. This is the two-dimensional lattice reconstruction from vector lengths; it uses lengths with multiplicities, not merely the set of distinct lengths.

A normalization is important here. I use the covolume-one theta function of a fractional ideal, for which the requested limit is $1$, and also give the formula for the unnormalized Gaussian theta sum.

Let $n=[K:\mathbb{Q}]$, let $d_v=[K_v:\mathbb{R}]\in \{1,2\}$, and choose one field embedding ${\sigma }_v$ for each Archimedean place. The Minkowski embedding of a number field identifieswith a real inner product space havingIts Euclidean Lebesgue measure is $d\mu ={\prod }_{v\text{ real}}d{x}_v{\prod }_{v\text{ complex}}2d\mathrm{Re}{x}_{v}d\mathrm{Im}{x}_v$. With this convention, the covolume of a fractional ideal lattice $j(\mathfrak{b})$ iswhere $D_K$ is the field discriminant and $N(\mathfrak{b})$ is the positive absolute norm of a fractional ideal. Thus the Euclidean lattice ${\Lambda }_{\mathfrak{b}}=C_{\mathfrak{b}}^{-1/n}j(\mathfrak{b})$ has covolume $1$. DefineThis Gaussian theta sum converges absolutely whenever every $y_v>0$.

The trace dual of a fractional ideal isHere ${\mathfrak{D}}_{K/\mathbb{Q}}^{-1}$ is the inverse different. Since $N({\mathfrak{D}}_{K/\mathbb{Q}})=|D_K|$, we have $C_{{\mathfrak{b}}^{\vee }}=C_{\mathfrak{b}}^{-1}$. There is a subtle distinction between the trace pairing and the positive inner product: the dual lattice of $j(\mathfrak{b})$ is $\overline{j({\mathfrak{b}}^{\vee })}$, where the bar conjugates the complex coordinates and fixes the real ones. Indeed,Consequently ${\Lambda }_{\mathfrak{b}}^{\ast }=\overline{{\Lambda }_{{\mathfrak{b}}^{\vee }}}$. Coordinatewise complex conjugation preserves the weighted squared lengths in the Gaussian theta sum.

Here are the precise analytic formulas used in the proof. For a Schwartz function on $K_{\infty }$, take the Fourier transform to beFor a full Euclidean lattice $\Lambda$ of covolume $C$, the Poisson summation formula for a Euclidean lattice isFor $f_y(x)=\exp (-\pi {\sum }_v{d}_v{y}_v|x_v{|}^2)$, the Gaussian Fourier transform, applied in orthonormal real coordinates, givesMore generally, $\widehat{\exp (-\pi x^{T}Ax)}(z)=(\det A{)}^{-1/2}\exp (-\pi z^T{A}^{-1}z)$ for a real positive-definite matrix $A$ that is symmetric. Each complex coordinate contributes two real coordinates, which explains its exponent $d_v=2$.

Apply the Poisson summation formula for a Euclidean lattice to ${\Lambda }_{\mathfrak{b}}$, whose covolume is $1$, and use the preceding description of its dual lattice. The anisotropic theta functional equation isFor comparison, the unnormalized Gaussian theta sumhas the functional equationand its corresponding limit is $C_{\mathfrak{b}}^{-1}$ rather than $1$. Explicitly, our normalization is $\Theta (y,\mathfrak{b})=\theta (C_{\mathfrak{b}}^{-2/n}y,\mathfrak{b})$.

For the small-parameter asymptotic of a lattice theta sum, $y\to 0$ means that every coordinate tends to zero. In $\Theta (y^{-1},{\mathfrak{b}}^{\vee })$, the zero lattice vector contributes $1$. Every nonzero vector contributes a term tending to zero. Once every $y_v\le 1$, all these terms are bounded by the summable Gaussian theta sum ${\sum }_{\lambda \in {\Lambda }_{{\mathfrak{b}}^{\vee }}}{e}^{-\pi |\lambda {|}^2}$. The dominated convergence theorem therefore gives $\Theta (y^{-1},{\mathfrak{b}}^{\vee })\to 1$. Using the anisotropic theta functional equation,The condition that every coordinate tends to zero matters; $\Vert y\Vert \to 0$ alone would not justify this argument.

Let $\Lambda =B{\mathbb{Z}}^d$ be a full-rank Euclidean lattice, with covolume $m(\Lambda )=|\det B|$ and dual lattice ${\Lambda }^{\vee }=B^{-\mathsf{T}}{\mathbb{Z}}^d$. For a Schwartz function $f\in \mathcal{S}({\mathbb{R}}^d)$ use the Fourier transform conventionThen the Poisson summation formula becomesBoth series have absolute convergence. To see the normalization, apply integer-lattice Poisson summation to $g(u)=f(Bu)$. The change of variables formula gives $\hat{g}(k)=|\det B{|}^{-1}\hat{f}(B^{-\mathsf{T}}k)$, yielding both the reciprocal covolume factor and the dual lattice argument. The factor is the physical lattice-point density. In particular, for spacing $h>0$,For the Gaussian function $f(x)=e^{-\pi a\Vert x{\Vert }^2}$ with $a>0$, the product of the one-dimensional Gaussian Fourier transforms givesHere $\Vert \cdot \Vert$ is the Euclidean norm. As a normalization check, take $\Lambda =\mathrm{diag}(2,3){\mathbb{Z}}^2$. Its covolume is $6$, and its dual lattice is $\mathrm{diag}(1/2,1/3){\mathbb{Z}}^2$, soThis also follows by multiplying two Gaussian Poisson summation identities with parameters $4a$ and $9a$. If one instead uses an angular Fourier transform kernel $e^{-i⟨x,\omega ⟩}$, the reciprocal frequencies are $2\pi {\Lambda }^{\vee }$, while the prefactor remains $m(\Lambda {)}^{-1}$.

The vector-length multiset of a rank-two Euclidean lattice determines it up to an orthogonal map. Let $a$ be the shortest nonzero length. Removing two copies of each positive integer multiple of $a$ removes one primitive lattice line; the shortest remaining length $b$ is the shortest independent-vector length. Those two vectors form a lattice basis: otherwise a nonzero point in the centred basis parallelogram would be an independent lattice vector of length less than $b$. Lattice-point density determines the covolume $A$. The basis Gram matrix then has diagonal entries $a^2,b^2$ and absolute off-diagonal entry $\sqrt{a^2{b}^2-A^2}$. This proves two-dimensional spectral rigidity for a flat torus through its dual lattice.

Let $B_1,B_2$ be real square invertible matrices whose columns are lattice bases. They generate the same Euclidean lattice precisely when^[ref]

If $B_2=B_{1}U$, multiplication by $U$ merely permutes the integer coordinate vectors, so the two generated sets agree. Conversely, equality of the generated sets means every column of either basis is an integer linear combination of the other. Hence $B_2=B_{1}U$ and $B_1=B_{2}V$ for integer matrices $U,V$. Substitution followed by multiplication by $B_1^{-1}$ gives $UV=I$, so the integer inverse criterion for a unimodular matrix applies.

Equal real linear spans are a weaker condition: any two real square invertible matrices span the same ambient vector space. Even equal covolumes do not force the same Euclidean lattice. The requirement that the change matrix and its inverse matrix have integer entries provides the missing arithmetic information.