The counting inner product on ${\mathbb{C}}^{\mathbb{Z}\mathrm{/}N\mathbb{Z}}$

Let $N\ge 1$ and let ${\mathbb{C}}^{\mathbb{Z}\mathrm{/}N}$ be the complex vector space of all functions $\mathbb{Z}\mathrm{/}N\mathbb{Z}\to \mathbb{C}$, with pointwise addition and scalar multiplication (The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$, Vector space over a field). For $f,g\in {\mathbb{C}}^{\mathbb{Z}\mathrm{/}N}$ define

$⟨f,g⟩:={\sum }_{x\in \mathbb{Z}\mathrm{/}N}f(x)\overline{g(x)},$

the sum being the finite sum over the finite index set $\mathbb{Z}\mathrm{/}N\mathbb{Z}$ in the additive commutative monoid of $\mathbb{C}$ (A finite sum in a commutative monoid indexed by an arbitrary finite set). Enumerating the group by its standard representatives gives the equivalent formula

$⟨f,g⟩={\sum }_{x=0}^{N-1}f([x{]}_N)\overline{g([x{]}_N)},$

where $\bar{\phantom{z}}$ is complex conjugation (Real and imaginary parts, complex conjugation, and modulus). The two displays agree. By For $n\ge 1$, every class in $\mathbb{Z}\mathrm{/}n$ has one representative $r$ with $0\le r<n$, so $\mid \mathbb{Z}\mathrm{/}n\mid =n$; while $\mathbb{Z}\mathrm{/}0$ is in bijection with $\mathbb{Z}$ the map $x\mapsto [x{]}_N$ is a bijection from the von Neumann natural $N=\{0,\dots ,N-1\}$ onto $\mathbb{Z}\mathrm{/}N\mathbb{Z}$, and the summand $f(x)\overline{g(x)}$ depends only on the class $x$; a finite commutative-monoid sum is unchanged by reindexing along a bijection (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule, part 1), so the second display is a rewrite of the first. No representative of a class is ever selected: every application of $f$ or of $g$ is an evaluation at a class.

The pairing is an inner product in the sense of Real and complex inner product spaces, with the inner product linear in the first argument and Real and complex inner-product spaces and their induced length, with the linear-first convention fixed there. Linearity in the first argument, $⟨a{f}_1+b{f}_2,g⟩=a⟨f_1,g⟩+b⟨f_2,g⟩$ for $a,b\in \mathbb{C}$, follows from the field laws of $\mathbb{C}$ ($\mathbb{C}=\mathbb{R}[x]\mathrm{/}(x^2+1)$ is a field, every element is uniquely $a+bi$, and every nonzero element has inverse $(a-bi)\mathrm{/}(a^2+b^2)$) together with the two elementary laws of a finite sum over a fixed finite index set, ${\sum }_x(u_x+v_x)={\sum }_x{u}_x+{\sum }_x{v}_x$ and ${\sum }_{x}c{u}_x=c{\sum }_x{u}_x$; each of these laws is proved from the recursion clauses of A finite sum in a commutative monoid indexed by an arbitrary finite set by induction on an enumeration of the index set. Conjugate symmetry, $⟨f,g⟩=\overline{⟨g,f⟩}$, follows from the same two laws together with the fact that complex conjugation is an involutive field automorphism, hence $\overline{g(x)\stackrel{‾}{f(x)}}=\overline{g(x)}f(x)$ and conjugation commutes with finite sums (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive).

Positive definiteness. Taking $g=f$ gives $⟨f,f⟩={\sum }_x\mathrm{\mid }f(x){\mathrm{\mid }}^2$ by $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$ (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive), a sum of nonnegative real numbers, so $⟨f,f⟩\ge 0$. For the vanishing clause, reindex by the standard representatives to identify this group sum of the real family $x\mapsto \mathrm{\mid }f(x){\mathrm{\mid }}^2$ with the sequential real sum ${\sum }_{x=0}^{N-1}\mathrm{\mid }f([x{]}_N){\mathrm{\mid }}^2$ (Finite sums and finite products, by recursion, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule); by claim 4 of Laws of finite sums and finite products a finite sum of nonnegative reals vanishes only if every term vanishes, and $\mathrm{\mid }f([x{]}_N){\mathrm{\mid }}^2=0$ happens exactly when $f([x{]}_N)=0$ (Conjugation is an involutive real-field automorphism, $z\bar{z}=\mathrm{\mid }z{\mathrm{\mid }}^2$, and modulus is definite, multiplicative, and subadditive). Since every class of $\mathbb{Z}\mathrm{/}N\mathbb{Z}$ is $[x{]}_N$ for some $0\le x<N$, this forces $f=0$.

This is the pairing induced by the counting set function on the finite set $\mathbb{Z}\mathrm{/}N\mathbb{Z}$, which weights a finite set by its cardinality and therefore gives weight $1$ to each point (Counting measure on an arbitrary set). No factor $1\mathrm{/}N$ is inserted anywhere in the definition; a normalisation constant is carried by the transform, not by the pairing.

• Why the counting normalisation and not $(1\mathrm{/}N)$-counting. Taylor's (11.5) weights the function space on ${\mathrm{\Gamma }}_n$ by $(1\mathrm{/}n)$-times counting measure and the space on ${\mathbb{Z}}_n$ by counting measure, matching the factor $1\mathrm{/}n$ carried by his forward transform (11.1); this page uses the counting pairing on both sides and puts the constant $N^{-1\mathrm{/}2}$ in the transform. Under the identification ${\omega }^j\leftrightarrow [j{]}_N$ the two conventions are related by $f^{\mathrm{\#}}=N^{-1\mathrm{/}2}{\mathcal{F}}_{N}f$, a relabelling of the same finite sums, not a change of the mathematics.