Full-rank lattices, covolume, and the dual lattice

Let $n\ge 1$. A full-rank lattice in ${\mathbb{R}}^n$ is a subgroup $\mathrm{\Lambda }\subseteq {\mathbb{R}}^n$ of the form

$\mathrm{\Lambda }=A{\mathbb{Z}}^n=\{{\sum }_{i=1}^n{k}_i{a}_i:k=(k_1,\dots ,k_n)\in {\mathbb{Z}}^n\},$

where $A$ is an invertible real $n\times n$ matrix with columns $a_1,\dots ,a_n$; here $Ak$ is the matrix product (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes) and the column matrix $Ak$ is identified with the vector it represents. A full-rank lattice is exactly a full Euclidean lattice in the sense of Full Euclidean lattice and covolume: invertibility of $A$ makes its columns linearly independent over $\mathbb{R}$ (A finite square real matrix is invertible if and only if its determinant is nonzero), so they form a real basis of ${\mathbb{R}}^n$ and $\mathrm{\Lambda }$ is their $\mathbb{Z}$-span, and conversely the basis matrix of every full Euclidean lattice is invertible. The covolume of $\mathrm{\Lambda }$ is

$\mathrm{covol}(\mathrm{\Lambda }):=\mathrm{\mid }\det A\mathrm{\mid }>0,$

with the determinant and the real absolute value of For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix, and the dual lattice is

${\mathrm{\Lambda }}^{\ast }:=\{\xi \in {\mathbb{R}}^n:\xi \cdot \lambda \in \mathbb{Z}\text{ for every }\lambda \in \mathrm{\Lambda }\}=A^{-T}{\mathbb{Z}}^n,$

where $\xi \cdot \lambda$ is the Euclidean inner product (The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$), $A^{-T}:=(A^{-1}{)}^T=(A^T{)}^{-1}$ is formed with the transpose of The transpose $A^{\mathsf{T}}$ of a matrix and the inverse of Invertible matrices and the general linear group ${\mathrm{GL}}_n(F)$ (Transpose is linear and involutive, and $(AB{)}^{\mathsf{T}}=B^{\mathsf{T}}{A}^{\mathsf{T}}$ for the identity $(A^{-1}{)}^T=(A^T{)}^{-1}$, and A finite square real matrix is invertible if and only if its determinant is nonzero for the existence of $A^{-1}$).

Well-definedness: the presentation does not matter. Suppose $B{\mathbb{Z}}^n=A{\mathbb{Z}}^n$ for a second invertible real matrix $B$. Then $U:=A^{-1}B$ has integer entries: each column $b_j$ of $B$ lies in $A{\mathbb{Z}}^n$, say $b_j=A{u}_j$ with $u_j\in {\mathbb{Z}}^n$, so $U=(u_1\cdots u_n)\in M_n(\mathbb{Z})$ (Finite rectangular matrices over a commutative ring, their entries, rows and columns, Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose). Exchanging the roles of $A$ and $B$ shows likewise that $U^{-1}=B^{-1}A$ has integer entries. Hence $\det U\in \mathbb{Z}$ and $\det U^{-1}\in \mathbb{Z}$, and $\det U\cdot \det U^{-1}=\det (U{U}^{-1})=\det I_n=1$ (For same-sized finite square matrices over a commutative ring, $\det (AB)=\det (A)\det (B)$, Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes); by $(\mathbb{Z},\cdot ,1)$ is a commutative monoid whose group of units is $\{1,-1\}$; equivalently $u\mid 1$ holds exactly for $u=1$ and $u=-1$ the only way two integers can multiply to $1$ is $\det U=\det U^{-1}=\pm 1$ (with the same sign). Consequently $\mathrm{\mid }\det B\mathrm{\mid }=\mathrm{\mid }\det A\mathrm{\mid }\mathrm{\mid }\det U\mathrm{\mid }=\mathrm{\mid }\det A\mathrm{\mid }$, so the covolume is independent of the presentation. For the dual, $B^{-T}{\mathbb{Z}}^n=A^{-T}{U}^{-T}{\mathbb{Z}}^n$ by Transpose is linear and involutive, and $(AB{)}^{\mathsf{T}}=B^{\mathsf{T}}{A}^{\mathsf{T}}$, and $U^{-T}=(U^{-1}{)}^T$ as well as $U^T$ have integer entries (The transpose $A^{\mathsf{T}}$ of a matrix); matrices with integer entries send ${\mathbb{Z}}^n$ into ${\mathbb{Z}}^n$ under matrix multiplication (Entrywise ring-matrix operations, rectangular matrix products, identity matrices and transpose), so $U^{-T}{\mathbb{Z}}^n\subseteq {\mathbb{Z}}^n$ and, applying the same statement to the integer matrix $U^T=(U^{-T}{)}^{-1}$, also ${\mathbb{Z}}^n\subseteq U^{-T}{\mathbb{Z}}^n$. Hence $U^{-T}{\mathbb{Z}}^n={\mathbb{Z}}^n$ and $B^{-T}{\mathbb{Z}}^n=A^{-T}{\mathbb{Z}}^n$.

The dual in closed form. For $\xi =A^{-T}m$ with $m\in {\mathbb{Z}}^n$ and $\lambda =Ak$ with $k\in {\mathbb{Z}}^n$ one has $\xi \cdot \lambda =(A^{-T}m)\cdot (Ak)=m\cdot k\in \mathbb{Z}$, because $(A^{-T}m)\cdot (Ak)=m\cdot (A^{-1}Ak)$; so $A^{-T}{\mathbb{Z}}^n$ is contained in the defining set. Conversely, if $\xi \cdot \lambda \in \mathbb{Z}$ for every $\lambda \in \mathrm{\Lambda }$, then testing $\lambda =a_j=A{e}_j$ gives $\xi \cdot a_j=(A^T\xi {)}_j\in \mathbb{Z}$ for every $j$, so $A^T\xi \in {\mathbb{Z}}^n$ and $\xi =A^{-T}(A^T\xi )\in A^{-T}{\mathbb{Z}}^n$. This also shows that the second description of ${\mathrm{\Lambda }}^{\ast }$ depends only on $\mathrm{\Lambda }$. The dual is itself a full-rank lattice: $A^{-T}$ is invertible, its columns being a real basis, and $\mathrm{covol}({\mathrm{\Lambda }}^{\ast })=\mathrm{\mid }\det A^{-T}\mathrm{\mid }=\mathrm{\mid }\det A^{-1}\mathrm{\mid }=\mathrm{\mid }\det A{\mathrm{\mid }}^{-1}=\mathrm{covol}(\mathrm{\Lambda }{)}^{-1}$, using $\det A^{-T}=\det A^{-1}$ (For every square matrix over a commutative ring, $\det (A^{\mathsf{T}})=\det (A)$) and $\det A^{-1}=(\det A{)}^{-1}$ (If $A$ is invertible over a commutative ring, then $\det (A^{-1})=\det (A{)}^{-1}$). For $\mathrm{\Lambda }={\mathbb{Z}}^n$, that is $A=I_n$, one gets ${\mathrm{\Lambda }}^{\ast }={\mathbb{Z}}^n$ because $I_n^{-T}=I_n$ (Rectangular matrix multiplication and the identity matrix $I_n$, including zero-sized shapes).

Characters. For ${\lambda }^{\ast }\in {\mathbb{R}}^n$ put ${\chi }_{{\lambda }^{\ast }}(x):=e^{2\pi i{\lambda }^{\ast }\cdot x}$, with the complex exponential of The complex exponential by its power series. Then ${\chi }_{{\lambda }^{\ast }}(x+\lambda )={\chi }_{{\lambda }^{\ast }}(x)e^{2\pi i{\lambda }^{\ast }\cdot \lambda }$ for every $\lambda \in \mathrm{\Lambda }$, so ${\chi }_{{\lambda }^{\ast }}$ is $\mathrm{\Lambda }$-periodic exactly when $e^{2\pi i{\lambda }^{\ast }\cdot \lambda }=1$ for every $\lambda \in \mathrm{\Lambda }$, that is ($\ker (\exp )=2\pi i\mathbb{Z}$, and $\exp z=\exp w$ exactly when $z-w\in 2\pi i\mathbb{Z}$) exactly when ${\lambda }^{\ast }\cdot \lambda \in \mathbb{Z}$ for every $\lambda \in \mathrm{\Lambda }$ --- precisely the condition ${\lambda }^{\ast }\in {\mathrm{\Lambda }}^{\ast }$. Thus ${\lambda }^{\ast }\mapsto {\chi }_{{\lambda }^{\ast }}$ is a bijection from ${\mathrm{\Lambda }}^{\ast }$ onto the set of exponential characters ${\chi }_{{\lambda }^{\ast }}$ of the torus ${\mathbb{R}}^n\mathrm{/}\mathrm{\Lambda }$, and it is a group isomorphism for pointwise multiplication because ${\chi }_{{\lambda }^{\ast }+{\eta }^{\ast }}={\chi }_{{\lambda }^{\ast }}{\chi }_{{\eta }^{\ast }}$; this is the lattice case of the characters of The Pontryagin dual with the compact-open topology. The assertion is about the characters of the displayed exponential form; no claim is made here that every continuous character of ${\mathbb{R}}^n\mathrm{/}\mathrm{\Lambda }$ is of that form.

No choice principle is used anywhere in this item.