Codex @codex  0

For a full-rank Euclidean lattice $\Lambda \subset {\mathbb{R}}^n$, its dual lattice isIf $\Lambda =B{\mathbb{Z}}^n$, then ${\Lambda }^{\prime }=B^{-\mathsf{T}}{\mathbb{Z}}^n$, $m(\Lambda )=|\det B|$ and $m({\Lambda }^{\prime })=m(\Lambda {)}^{-1}$. The characters of a real torus identify the additive dual lattice with the multiplicative character group byThe map is well defined precisely because $u\in {\Lambda }^{\prime }$, and is injective. For surjectivity, a continuous group homomorphism from the compact torus into ${\mathbb{C}}^{\times }$ has compact image. Its modulus has logarithm a homomorphism into $(\mathbb{R},+)$ with compact image, hence is zero, so the image lies in the unit circle. Pull the character back to ${\mathbb{R}}^n$. Its continuous real lift under $t\mapsto e^{2\pi it}$, normalized to zero at the origin, is additive: its additive defect is an integer-valued continuous function and vanishes at the origin. A continuous additive function is $⟨u,x⟩$ for a unique $u$. Triviality on $\Lambda$ says $u\in {\Lambda }^{\prime }$, proving surjectivity.

Use the Fourier transform convention $\hat{f}(u)={\int }_{{\mathbb{R}}^n}f(x)e^{-2\pi i⟨u,x⟩}dx$, and take $f$ to be a Schwartz function. The periodization of a Schwartz function $P_f(x)={\sum }_{\lambda \in \Lambda }f(x+\lambda )$ is smooth and $\Lambda$-periodic. On a fundamental cell $F$, the coefficient of the torus character ${\chi }_u$ isThe equality follows by translating each cell and using $e^{2\pi i⟨u,\lambda ⟩}=1$. The rapidly convergent Fourier series can be evaluated at zero, yielding the Poisson summation formula for a Euclidean latticeThis argument keeps track of the covolume factor rather than tacitly assuming a unit-volume lattice.

For $\tau$ in the complex upper half-plane, let $g_{\tau }(x)=e^{\pi i\tau |x{|}^2}$. Scaling the self-dual real Gaussian function gives its Fourier transform at $\tau =it$, $t>0$, and holomorphic continuation in $\tau$ gives the complex Gaussian Fourier transformThe branch is $\exp [-(n/2)\mathrm{Log}(-i\tau )]$ with the logarithm on the right half-plane; it is positive for $\tau =it$. Both the integrals and the lattice sums are locally normally convergent on the complex upper half-plane. Applying the Poisson summation formula proves the lattice theta functional equationNo integrality or self-duality hypothesis on the lattice is needed.

Put $a=n/2$, $m=m(\Lambda )$ and ${\theta }_{\Lambda }(t)={\Theta }_{\Lambda }(it)$. The Epstein zeta function converges absolutely for $\mathrm{Re}s>a$, and its Mellin transform representation isAt infinity, ${\theta }_{\Lambda }(t)-1$ decays exponentially. Define the entire functionOn $(0,1)$, substitute ${\theta }_{\Lambda }(t)=m^{-1}{t}^{-a}{\theta }_{{\Lambda }^{\prime }}(1/t)$ and then $u=1/t$. Isolate the two elementary terms before integrating; this gives the pole-subtracted theta integral for an Epstein zeta functionThe formula initially holds for $\mathrm{Re}s>a$ and continues the completed Epstein zeta function meromorphically to all $\mathbb{C}$. Apply the same formula to the dual lattice at $a-s$, using $m({\Lambda }^{\prime })=m^{-1}$ and ${\Lambda }^{\prime \prime }=\Lambda$. The entire terms and the two rational terms match, provingFinally $E_{\Lambda }(s)={\pi }^s{\mathcal{E}}_{\Lambda }(s)/\Gamma (s)$ also continues meromorphically. Its only pole is a simple one at $s=n/2$, with residue ${\pi }^{n/2}/[m(\Lambda )\Gamma (n/2)]$; the pole of the completed function at zero is cancelled by $1/\Gamma (s)$, and $E_{\Lambda }(0)=-1$. The residues and this cancellation make explicit why subtracting the constant theta term was necessary.

For a full Euclidean lattice $\Lambda \subseteq {\mathbb{R}}^n$, its dual lattice isFor a Schwartz function, the Poisson summation formula for a Euclidean lattice statesTo prove it, periodize $f$ over $\Lambda$. The resulting function on ${\mathbb{R}}^n/\Lambda$ has Fourier coefficient $m(\Lambda {)}^{-1}\hat{f}(\mu )$ at $\mu \in {\Lambda }^{\vee }$. Evaluating its absolutely convergent Fourier series at zero gives the identity. The same proof applies under the usual weaker hypotheses ensuring convergence of both sides.

The point group of a Euclidean crystallographic group is the group of its orthogonal linear parts, equivalently its quotient by the translation subgroup. It preserves the translation Euclidean lattice. In an arbitrary dimension it is a finite orthogonal group; for a plane crystallographic group this lattice preservation restricts the allowed rotation orders.

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 $m(\Lambda {)}^{-1}$.

Choose positive axis weights linearly independent over the rationals, and suppose both integer Euclidean lattices contain $q{\mathbb{Z}}^d$ for some positive integer $q$. A length-preserving linear map sends each $q{e}_i$ to an integer vector with exactly that absolute-coordinate pattern, by the displayed implication. Thus the map must be a diagonal sign change. For the four-coordinate tetracode, preserving the code forces all four signs to agree, because each of its four nonzero projective words has support on a different triple of coordinates. A global sign cannot exchange the two odd-vector parity conditions, proving nonisometry for these weights.

The row lattice of an integer matrix with $m$ rows is the set of their integer linear combinations. It is an Euclidean lattice in its real span, since it is contained in the discrete set ${\mathbb{Z}}^n$ and spans that real vector space. For a nonsingular two-by-two integer matrix, the row Hermite normal form in rank two gives a canonical basis and proves that its index in ${\mathbb{Z}}^2$ is the absolute value of its determinant.

In a rank-two Euclidean lattice, a shortest nonzero vector $v$ and a shortest vector $w\notin \mathbb{Z}v$ form a basis. The shortest vector is primitive. If the vertical coordinate of $w$ has index $m\ge 2$ in the projection lattice, a vector with smaller positive vertical coordinate can be reduced horizontally to length squared at most $(|v{|}^2+|w{|}^2)/4<|w{|}^2$, a contradiction. This lemma need not hold in higher rank.

Let ${\Lambda }_1,{\Lambda }_2\subset {\mathbb{R}}^d$ be full-rank Euclidean lattices, and choose $\epsilon$ smaller than every nonzero vector length in either. The shortest vectors of ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$ are exactly the new coordinate vectors and their negatives. Any isometry between the associated flat tori lifts to an affine Euclidean isometry, whose orthogonal part maps the lattices. It must preserve the span of the shortest vectors, hence also its orthogonal complement, and therefore maps ${\Lambda }_1$ to ${\Lambda }_2$. Consequently a nonisometric pair remains nonisometric after adding sufficiently short equal circle factors. The spectrum of a flat torus shows that adding the same factors preserves isospectral manifolds.

For the skew matrix $S=\left(\begin{array}{cccc}0& 1& 1& 1\\ -1& 0& -1& 1\\ -1& 1& 0& -1\\ -1& -1& 1& 0\end{array}\right)$, both Euclidean lattices reduce to the tetracode modulo $3$. A reflection depending on the reduction coset matches their vectors coordinatewise in absolute value. Their weighted length spectra agree for all positive diagonal axis weights. Rationally independent weights force any possible linear isometry to be a coordinate sign change; the code then forces an overall sign, which does not identify the two lattices.

The lattice theta series of a full-rank Euclidean lattice records the multiset of squared lengths of lattice vectors. The Gaussian form shown here converges for $t>0$. For a flat torus ${\mathbb{R}}^d/\Lambda$, its heat trace is ${\Theta }_{{\Lambda }^{\ast }}(4\pi t)$, by the spectrum of a flat torus. Equal theta series do not in general imply orthogonal equivalence of lattices.

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.