Row orthogonality for additive characters of a finite abelian group

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.