Finite Abelian Characters for Combinatorics

For $g\in G$ define ${\rho }_{\chi }(g):\mathbb{C}\to \mathbb{C}$ by ${\rho }_{\chi }(g)(z):=\chi (g)z$. This map is $\mathbb{C}$-linear and invertible with inverse $z\mapsto \chi (g{)}^{-1}z$, because $\chi (g)\in {\mathbb{C}}^{\times }$; moreover ${\rho }_{\chi }(0)$ is the identity map, since $\chi (0)=1$, and ${\rho }_{\chi }(g+h)={\rho }_{\chi }(g)\circ {\rho }_{\chi }(h)$ for all $g,h$, because $\chi (g+h)=\chi (g)\chi (h)$ and multiplication in $\mathbb{C}$ is associative. So ${\rho }_{\chi }$ is a group homomorphism $G\to \mathrm{GL}(\mathbb{C})$, hence a complex representation of $G$ on the one-dimensional space $\mathbb{C}$, and its defining formula in terms of $\chi$ determines it uniquely from $\chi$.

For clause 2 let $\rho :G\to \mathrm{GL}(V)$ be a one-dimensional complex representation, so ${\dim }_{\mathbb{C}}V=1$, and choose a basis vector $v$ of $V$. For each $g$ there is a unique scalar $\lambda (g)$ with $\rho (g)v=\lambda (g)v$, and $\rho (g)$ invertible forces $\lambda (g)\mathrm{\ne }0$; by [L9] the resulting map $\lambda :G\to {\mathbb{C}}^{\times }$ is a group homomorphism, hence an additive character of $G$. Since $\rho (g)$ is multiplication by the scalar $\lambda (g)$ on a one-dimensional space, its trace is $\lambda (g)$, so the trace character of $\rho$ is $\lambda$; and because the trace of an endomorphism is basis-independent, the additive character so obtained does not depend on the choice of basis.

Every irreducible complex representation of $G$ has degree $1$: by [L7] the fundamental theorem of algebra makes $\mathbb{C}$ an algebraically closed field, so [L8] makes every endomorphism of an irreducible complex representation $V$ of $G$ a scalar $\lambda {\mathrm{id}}_V$; since an irreducible $V$ is nonzero, the map $\lambda \mapsto \lambda {\mathrm{id}}_V$ is injective, hence ${\mathrm{End}}_G(V)=\mathbb{C}$ and $\mathbb{C}$ is a splitting field for $G$, whereupon [L8] gives ${\dim }_{\mathbb{C}}V=1$ because $G$ is abelian.

For clause 4 fix $g\in G$ and put $n:=\mathrm{ord}(g)$; since $G$ is finite, $n$ is a natural number with $n\ge 1$ and $g^n=e$. Written additively, as this page writes $G$, the power $g^n$ is the $n$-fold sum $n\cdot g$, so $n\cdot g=0$, and the additive reading of claim 3 of [L10] gives $\chi (n\cdot g)=\chi (g{)}^n$. Hence $1=\chi (0)=\chi (g{)}^n$: the value $\chi (g)$ is an $n$-th root of unity. By [L11] there is $k$ with $0\le k<n$ and $\chi (g)=\exp (2\pi ik\mathrm{/}n)$, and the modulus formula in [L11] gives $\mathrm{\mid }\chi (g)\mathrm{\mid }=\mathrm{\mid }\exp (i\cdot 2\pi k\mathrm{/}n)\mathrm{\mid }=1$.

In the standard basis $\{1\}$ of $\mathbb{C}$ the matrix of the endomorphism ${\rho }_{\chi }(g)$ is the $1\times 1$ matrix $(\chi (g))$, whose trace is $\chi (g)$; since the character is computed with the basis-independent trace, ${\chi }_{{\rho }_{\chi }}(g)=\mathrm{tr}({\rho }_{\chi }(g))=\chi (g)$ for every $g\in G$. Thus the trace character of ${\rho }_{\chi }$ is $\chi$, as clause 1 asserts.

Let $W$ be an irreducible complex representation of $G$. By step 1.3 it is one-dimensional, so by step 1.2 its trace character $\psi (g):=\mathrm{tr}({\rho }_W(g))$ is an additive character of $G$; by step 2.1 the representation ${\rho }_{\psi }$ attached to $\psi$ has trace character $\psi$. Thus $W$ and ${\rho }_{\psi }$ are degree-one representations producing the same homomorphism $\psi$, so [L9] makes them equivalent. This is clause 3.

For clause 5 let $\chi \mathrm{\ne }\psi$ be additive characters. Each ${\rho }_{\chi }$ is irreducible: it lives on the nonzero space $\mathbb{C}$ of dimension $1$, and a subrepresentation $U$ is a linear subspace with ${\dim }_{\mathbb{C}}U\le 1$, so either ${\dim }_{\mathbb{C}}U=0$ and $U=0$, or ${\dim }_{\mathbb{C}}U=1={\dim }_{\mathbb{C}}\mathbb{C}$ and $U=\mathbb{C}$ by [L5]; by [L4] this is exactly irreducibility, and by step 2.1 the character of ${\rho }_{\chi }$ is $\chi$, so ${\rho }_{\chi }$ has irreducible character $\chi$ in the sense of [L3]. If ${\rho }_{\chi }$ and ${\rho }_{\psi }$ were equivalent, then their characters would be equal by the invariance clause of [L3], so step 2.1 would give $\chi =\psi$, contrary to hypothesis; hence the two irreducible representations are inequivalent.

Clause 1 is steps 1.1 and 2.1, clause 2 is step 1.2, clause 3 is steps 1.3 and 3.1, clause 4 is step 1.4, and clause 5 is step 3.2, so all five clauses of the statement are proved. The argument uses only finite groups, finite-dimensional complex representations and finite lists of roots, and therefore invokes no choice principle.

Every additive character of $G$ is an irreducible complex character of $G$: by [L1] the character $\chi$ is the trace character of the irreducible one-dimensional representation ${\rho }_{\chi }$, and by [L2] the character of an irreducible representation is an irreducible complex character; the same holds for $\psi$.

Both $\chi$ and $\psi$ lie in $\mathrm{c}\mathrm{f}(G)$, the space on which [L4] and [L3] are stated: since $G$ is abelian, $gx{g}^{-1}=x$ for all $g,x\in G$ (in the additive writing used for $G$ on this page, $g+x-g=x$), so every function $G\to \mathbb{C}$, in particular $\chi$ and $\psi$, is constant on conjugacy classes, which is the defining property in [L5].

Let ${\chi }_1,\dots ,{\chi }_r$ be irreducible complex characters of $G$, one from each equivalence class, as in [L3]. By [L2] a character depends only on the equivalence class of its representation and by [L1] the representation ${\rho }_{\chi }$ is irreducible, so $\chi ={\chi }_i$ for some index $i$; likewise $\psi ={\chi }_j$ for some index $j$.

The normalized inner product of [L4] is linear in its first argument, as that published definition records of the form $⟨\phi ,\psi ⟩=\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{g\in G}\phi (g)\overline{\psi (g)}$ on class functions; this is the statement's final sentence.

If $\chi =\psi$, then $\chi ={\chi }_i$ for an index $i$ as in step 1.3, and [L3] applied to the pair $(i,i)$ gives $⟨\chi ,\psi ⟩=⟨{\chi }_i,{\chi }_i⟩={\delta }_{ii}=1$.

If $\chi \mathrm{\ne }\psi$, then the indices of step 1.3 satisfy $i\mathrm{\ne }j$: equality $i=j$ would give $\chi ={\chi }_i={\chi }_j=\psi$, a contradiction. Hence [L3] applied to the pair $(i,j)$ gives $⟨\chi ,\psi ⟩=⟨{\chi }_i,{\chi }_j⟩={\delta }_{ij}=0$.

By [L4] the inner product just computed is the displayed normalized sum, $⟨\chi ,\psi ⟩=\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{g\in G}\chi (g)\overline{\psi (g)}$; combining with steps 2.1 and 2.2, this quantity is $1$ when $\chi =\psi$ and $0$ otherwise, which is the orthogonality clause of the statement.

Let ${\chi }_0(g):=1$ for every $g\in G$. Then ${\chi }_0$ is an additive character, by [L6], and it is the trivial additive character. If $\chi \mathrm{\ne }{\chi }_0$, then step 3.1 with $\psi ={\chi }_0$ gives $\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{g\in G}\chi (g)\overline{{\chi }_0(g)}=0$; since $\overline{{\chi }_0(g)}=\bar{1}=1$ by [L7], this reads $\frac{1}{\mathrm{\mid }G\mathrm{\mid }}{\sum }_{g\in G}\chi (g)=0$, and multiplying by the nonzero complex number $\mathrm{\mid }G\mathrm{\mid }$ gives ${\sum }_{g\in G}\chi (g)=0$. That is the zero-sum clause.

Step 3.1 proves the orthogonality values, step 4.1 the vanishing sum for a nontrivial character, and step 1.4 the linear-first convention; these are all the claims of the statement.