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Every continuous group homomorphism ${\mathbb{R}}^n/\Lambda \to {\mathbb{C}}^{\times }$ is $x+\Lambda \mapsto e^{2\pi i⟨u,x⟩}$ for a unique $u$ in the dual lattice. Compactness forces unit modulus, and lifting the character to the simply connected cover makes its phase a real linear functional. These characters are the frequencies of torus Fourier series.

The completion ${\mathcal{E}}_{\Lambda }(s)={\pi }^{-s}\Gamma (s)E_{\Lambda }(s)$ satisfies ${\mathcal{E}}_{\Lambda }(s)=m(\Lambda {)}^{-1}{\mathcal{E}}_{{\Lambda }^{\vee }}(n/2-s)$. Its simple poles at zero and $n/2$ have residues $-1$ and $m(\Lambda {)}^{-1}$. The pole-subtracted theta integral for an Epstein zeta function gives both the continuation and this dual lattice symmetry.

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.

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.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with its quotient Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its volume is the lattice covolume. The dual lattice indexes its complete Fourier series of Laplacian eigenfunctions.

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.

The central problem of spectral geometry is how much of a Riemannian metric is determined by the eigenvalues, with multiplicities, of its Laplace-Beltrami operator. For a membrane with fixed boundary, the squared vibration frequencies are proportional to the Dirichlet eigenvalues. Kac's 1966 paper brought this inverse problem into focus: the full sequence of frequencies might encode a domain's shape, but recovering a few geometric quantities is much weaker than recovering an isometry class.

The first major line of development extracts geometric information from spectral sums. On a smooth closed $d$-dimensional manifold, the heat trace has the short-time expansionwhere ${\lambda }_j$ are the nonnegative eigenvalues of $P=-\Delta$ and $R$ is scalar curvature. The coefficients are heat invariants, formed by integrating local curvature expressions. They recover dimension, Riemannian volume and total scalar curvature. For smooth planar domains with the Dirichlet boundary condition,Thus area and boundary length are audible. The constant term also contains topological information through Gauss-Bonnet theorem, with corner corrections when the boundary is polygonal. The Weyl law is the corresponding leading eigenvalue-counting result. Neither the leading term nor the whole list of local heat coefficients automatically reconstructs global geometry.

In the 1970s, the wave trace introduced a complementary link with global closed geodesics. The distributionis spectral, while its singularities are governed by periodic geodesic motion. The 1975 Duistermaat-Guillemin work made this relationship precise. Its trace formula places singularities at geodesic lengths and computes their coefficients under suitable clean or nondegenerate hypotheses. When a coefficient is nonzero, the corresponding length is audible. Possible cancellation must be addressed before identifying the entire length spectrum with the singular support. This approach led to spectral rigidity questions, rather than just the recovery of integral curvature quantities.

The second main line constructs counterexamples to metric determination. A precedent was the 1964 flat-torus example: different sixteen-dimensional Euclidean lattices could give the same Laplacian eigenvalues. Subsequent lattice constructions lowered the dimension; Conway and Sloane's 1992 work exhibited four-dimensional Euclidean lattices with matching lattice theta series. For a flat torus ${\mathbb{R}}^d/\Lambda$, the spectrum is the multiset $4{\pi }^2|\xi {|}^2$ for $\xi \in {\Lambda }^{\ast }$, so equality of lattice theta series of the dual lattices gives isospectrality even when the Euclidean lattices are not related by an orthogonal transformation.

A systematic geometric construction arrived with the Sunada theorem in 1985. For a common Riemannian cover with finite isometry group $T$, almost conjugate subgroups have the same number of invariant vectors in each Laplacian eigenspace. Their freely acting quotients are therefore isospectral manifolds. This reduces an analytic equality to a finite-group representation identity. It produces many examples on surfaces and in higher dimensions, but nonconjugacy of the subgroups must be supplemented by an argument excluding extra quotient isometries. Transplantation theorem provides an explicit version: piecewise eigenfunctions are recombined by a fixed invertible linear transformation that respects the gluing and boundary conditions.

The planar problem itself was answered negatively by Gordon, Webb and Wolpert in 1992: noncongruent simply connected polygonal plane domains have equal Dirichlet spectra. This is stronger than counterexamples among abstract manifolds or flat tori, since these are genuine two-dimensional planar membranes. It does not settle every restricted version, such as asking for two domains within a specified smooth or convex class.

By the 1990s, isospectrality was also known to coexist with continuous metric variation and changes in local geometry. For example, a 1997 construction gives continuous isospectral families on products of spheres and tori whose members need not be locally isometric; in some examples the maximum scalar curvature changes. Conversely, rigidity and compactness results show that an isospectral class is not unrestricted. The 1988 surface compactness theorem makes fixed-spectrum sets of closed orientable surface metrics compact modulo diffeomorphisms. Compactness does not mean there is only one metric, or even finitely many metrics. By 2005, the subject therefore combined local heat invariants, global geodesic information, group-theoretic constructions, explicit transplantation and geometric rigidity. The spectrum determines substantial geometry, while generally failing to determine the full metric.

Use the following skew-symmetric matrix, whose columns reduce modulo three to four of the specified nonzero tetracode words:Set $B_{+}=3I+S$, $B_{-}=-3I+S$, and define the full Euclidean lattices $L_{\pm }^0=B_{\pm }{\mathbb{Z}}^4$. Their determinants have absolute value $144$, sinceFor arbitrary positive real numbers $a_1,\dots ,a_4$, put $D=\mathrm{diag}(\sqrt{a_1},\dots ,\sqrt{a_4})$ and $L_{\pm }=D{L}_{\pm }^0$. This is the tetracode length-isospectral lattice construction. We prove equality of the length multisets by an explicit bijection, not just equality of the shortest lengths.

Modulo three, both lattices have image $\mathrm{im}S=T$. The matrix $S$ has rank two over ${\mathbb{F}}_3$; its four column lines are precisely the four lines in $T$. Also $S^2=0$ modulo three, so $\ker S=\mathrm{im}S=T$. Each nonzero tetracode word has exactly one zero coordinate. Every $v\in L_{\pm }^0$ therefore has some coordinate divisible by three. Moreover, all four coordinates of $v$ have the same parity, because every entry of either $B_{\pm }$ is odd.

For $v\in L_{+}^0$, choose the least index $i$ with $3\mid v_i$ and set $w=v-2v_i{e}_i$. The coordinate reflection bijection for tetracode lattices will show that $w\in L_{-}^0$. First, $w\equiv v(\mathrm{mod}3)$, so $Sw\equiv 0(\mathrm{mod}3)$. Second, positive-lattice membership implies $Sv\equiv 3v\equiv -v(\mathrm{mod}4)$. HenceThe expression in parentheses has $i$th coordinate zero and every other coordinate $v_j+v_i{S}_{ji}$ even, by the common parity and $S_{ji}=\pm 1$. Thus $(I-S)w\equiv 0(\mathrm{mod}4)$. Together the congruences modulo three and four say $(-3I-S)w\equiv 0(\mathrm{mod}12)$, which is precisely $B_{-}^{-1}w\in {\mathbb{Z}}^4$.

Conversely, if $v\in L_{-}^0$, the same reflection has unchanged reduction modulo three, and now $Sv\equiv v(\mathrm{mod}4)$. ThusIts $i$th coordinate is $4v_i$ and its other coordinates are even multiples of two, so it vanishes modulo four; it vanishes modulo three as well. The reflected vector is in $L_{+}^0$. The chosen index is unchanged by reflection, including for vectors reducing to zero, so this map is an involution between the two lattices. It preserves every $|v_j|$ and therefore preserves $\Vert Dv{\Vert }^2={\sum }_j{a}_j{v}_j^2$. The two lattices have the same length spectrum, including every multiplicity, for every positive choice of weights.

To prove a genuine failure of isometry, chooseThese four numbers are linearly independent over $\mathbb{Q}$: changing the signs of the independent square roots in the multiquadratic field $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ isolates each coefficient of a rational relation. If an orthogonal transformation $O$ maps $L_{+}$ to $L_{-}$, thenand $P^T\mathrm{diag}(a_j)P=\mathrm{diag}(a_j)$. Comparing rational coefficients of the independent $a_j$ in the diagonal entries gives $P_{ji}^2={\delta }_{ji}$. Consequently $P$ is diagonal with entries $\pm 1$.

Such a coordinate sign change must preserve $T$ modulo three. For each zero coordinate $i$, there are exactly two opposite nonzero tetracode words. Preserving their line requires the other three signs to be equal. Taking two different zero coordinates forces all four signs to be equal. Thus $P=I$ or $P=-I$, either of which would imply $L_{+}^0=L_{-}^0$. ButThis contradiction proves $L_{+}$ and $L_{-}$ are not isometric. Allowing affine isometries changes nothing: translating the image of zero back to zero reduces them to the orthogonal case.

For the flat tori, take period lattices equal to the dual lattices:The standard spectrum of a flat torus lemma states that its characters of a real torus are indexed by the dual of its period lattice. Here they are $e^{2\pi i⟨v,x⟩}$ for $v\in L_{\pm }$, with Laplacian eigenvalue $4{\pi }^2\Vert v{\Vert }^2$. These characters give a complete orthogonal basis, so our length-preserving bijection proves that $X_{+}$ and $X_{-}$ are isospectral manifolds. The second standard lemma is that an isometry of flat tori lifts to an affine isometry of their Euclidean universal covers. Its orthogonal part maps the period lattices to one another, and therefore maps their dual lattices to one another. This would give the impossible isometry $L_{+}\cong L_{-}$. Hence $X_{+}$ and $X_{-}$ are isospectral non-isometric flat four-dimensional tori.

The musical isomorphisms are the pointwise maps induced by the Riemannian metric:In local coordinates they are $(X^{♭}{)}_i=g_{ij}{X}^j$ and $({\alpha }^{♯}{)}^i=g^{ij}{\alpha }_j$. The inverse metric makes these inverse bundle maps, hence inverse maps between smooth vector fields and smooth differential one-forms.

For a covariant two-tensor $\beta$, its metric trace is ${\mathrm{tr}}_g\beta =g^{ij}{\beta }_{ij}={\sum }_i\beta (e_i,e_i)$ in a Riemannian orthonormal frame. The contraction is independent of the chosen frame. There is a terminology distinction here: if $\beta$ is literally an alternating differential two-form, this trace is zero. The object needed below is the covariant two-tensor $D(X^{♭})$, generally not alternating. Interpreting that tensor as an alternating form would make the stated divergence construction impossible.

Let $D$ be the Levi-Civita connection. The Riemannian gradient, divergence and positive Laplace-Beltrami operator areThus $g({\mathrm{grad}}_{g}f,Y)=Yf$, and in coordinatesThis choice makes the Laplacian eigenvalues nonnegative on a closed Riemannian manifold. The alternative convention $+\mathrm{div}\mathrm{grad}$ reverses all eigenvalue signs, but none of the comparisons below.

For the Riemannian submersion, let $F=f\circ \pi$ and choose orthonormal horizontal lifts $E_i$ of a base frame and an orthonormal vertical frame $V_{\alpha }$. The horizontal part of $D_{E_i}{E}_i$ projects to the base Levi-Civita connection; this follows directly from its Koszul formula applied to basic horizontal fields, because their inner products and projected brackets are pulled back from the base. Therefore the horizontal trace of the Riemannian Hessian of $F$ is the pulled-back base Hessian trace. Since $V_{\alpha }F=0$, its vertical terms areWriting $H={\sum }_{\alpha }(D_{V_{\alpha }}{V}_{\alpha }{)}^{\mathrm{hor}}$ for the unnormalised fibre mean curvature vector givesThe totally geodesic submanifold condition on the fibres makes $H=0$. Consequently eigenfunction pullback by a Riemannian submersion givesHere the Riemannian submersion is onto, so the map is injective. Its image is exactly the members of the upstairs eigenspace constant on each whole fibre: such a function descends smoothly by local sections of the submersion, and the displayed identity forces the descended function to have the same eigenvalue. In particular, whenever these eigenspaces are finite dimensional, the downstairs multiplicity is no larger than the upstairs multiplicity. The upstairs eigenspace may also contain functions varying along fibres.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with the metric induced by the Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its dual lattice is ${\Lambda }^{\ast }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. The characters $e_w(x)=e^{2\pi i⟨w,x⟩}$ descend to the quotient and form an orthogonal Fourier basis of $L^2$ on it. Direct differentiation gives the spectrum of a flat torus:with multiplicity equal to the number of dual lattice vectors of each length. In particular zero has multiplicity one.

Let the given four-dimensional isospectral manifolds be ${\mathbb{R}}^4/{\Lambda }_1$ and ${\mathbb{R}}^4/{\Lambda }_2$. For $d=4+k$, take their products with the same $k$ circles of length $\epsilon$, equivalently the lattices ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$. The dual lattices are ${\Lambda }_i^{\ast }\oplus {\epsilon }^{-1}{\mathbb{Z}}^k$, so their Laplacian eigenvalues areThe given equality of base spectra with multiplicities makes these product spectra identical.

Choose $\epsilon$ smaller than the shortest nonzero vector length in either original Euclidean lattice. By short-vector cancellation for flat tori, the shortest nonzero vectors in the product lattice are exactly $\pm \epsilon e_j$ in its new factor. To spell out the nonisometry argument, any isometry of the product flat tori lifts to an isometry of their universal Euclidean covers. This is affine, and its orthogonal linear part maps one product lattice onto the other. It must map the span of their shortest vectors, namely the new $k$-dimensional factor, onto itself; orthogonality then preserves the original four-dimensional factor and maps ${\Lambda }_1$ onto ${\Lambda }_2$. That would give an isometry of the original flat tori, contrary to the assumption. Isospectral nonisometric flat tori therefore exist in every dimension $d\ge 4$.

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.

A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice, with the Riemannian metric induced from the Euclidean metric. Translation by a lattice vector is a Riemannian isometry, so this metric is well defined and has zero curvature. Its volume is the covolume of $\Lambda$.

The dual lattice is ${\Lambda }^{\vee }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. For $w\in {\Lambda }^{\vee }$, the function $e_w(x)=\exp (2\pi i⟨w,x⟩)$ descends to the flat torus, and direct differentiation givesConversely, write $\Lambda =B{\mathbb{Z}}^d$ and $x=By$. Ordinary Fourier series on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ have frequencies $n\in {\mathbb{Z}}^d$, corresponding to $w=B^{-\mathsf{T}}n\in {\Lambda }^{\vee }$. After normalization they form a complete orthonormal basis in $L^2$, and the Fourier coefficients of a smooth function decay rapidly. Applying the Laplace-Beltrami operator term by term shows that a smooth eigenfunction with eigenvalue $\lambda$ has nonzero coefficients only where $4{\pi }^2|w{|}^2=\lambda$. Thus the spectrum of a flat torus isThe formula counts multiplicity over either the real or complex numbers: the two frequencies $w,-w$ give the real sine and cosine functions.

For the rigidity assertion in dimension two, it suffices to reconstruct a rank-two Euclidean lattice $L={\Lambda }^{\vee }$ from its vector-length multiset. The multiset first determines its covolume $A$. Indeed, a bounded fundamental parallelogram, and comparison of the cells meeting a disk with slightly larger and smaller disks, give the lattice-point asymptotics by fundamental cellsThe spectrum determines $N_L(R)$, including multiplicities, so it determines $A$.

Let $a$ be the smallest nonzero vector length, and choose $v\in L$ with $|v|=a$. This vector is primitive: $v=ku$ with $|k|>1$ would contradict minimality. From the length multiset subtract the two vectors $kv,-kv$ at every positive length $ka$, for $k=1,2,\dots$. The shortest remaining length $b$ is exactly the length of a shortest vector $w\notin \mathbb{Z}v$. This subtraction is legitimate even if several directions have length $a$: it subtracts just two copies, and then $b=a$. It also proves that $b$ is independent of our choice of shortest direction. Replace $w$ by $w-kv$ to arrangeThe replacement stays outside $\mathbb{Z}v$ and cannot be shorter than $b$, so a reduced choice still has length $b$.

The shortest-independent-vector basis lemma says that $v,w$ are a basis of $L$. To prove it, choose coordinates with $v=(a,0)$. Because $v$ is primitive, $L/\mathbb{Z}v\cong \mathbb{Z}$; let $h>0$ be the smallest positive vertical coordinate in $L$. Write $w=(x,mh)$, with $m\ge 1$ after changing its sign. If $m\ge 2$, a vector $u$ of vertical coordinate $h$, reduced horizontally modulo $v$, hasThis contradicts the definition of $b$, since $u\notin \mathbb{Z}v$. Hence $m=1$ and $v,w$ generate $L$.

The Gram matrix of this basis is $\left(\begin{array}{cc}{a}^2& c\\ c& b^2\end{array}\right)$, with $c=v\cdot w$, and its determinant is $A^2$. ThereforeChanging the sign of one basis vector allows $c\ge 0$. The numbers $a,b,A$, all audible from the spectrum, consequently determine the Gram matrix, and hence $L$, up to an orthogonal transformation. Taking dual lattices gives the same conclusion for $\Lambda$. Isospectral flat two-dimensional tori are isometric. The argument uses the special basis property of rank-two Euclidean lattices, so it does not claim higher-dimensional rigidity.