Orthogonality of Group Characters

The following relations are often called the orthogonality relations of characters.

Theorem

Let \(G\) be a finite abelian group with \(|G| = n\). For each \(f \in \hat{G}\)

\[ \sum_{g \in G} f(g) = \begin{cases} n & f = f_0 = \mathrm{id} \\ 0 & \text{otherwise} \\ \end{cases}\]

where \(f_0\) is the principal character.

Proof

If \(f\) is the principal character then clearly

\[ \sum_{g \in G} f(g) = \sum_{g \in G} 1 = |G| = n.\]

If \(f\) is not the principal character, then there is some \(a \in G\) for which \(f(a) \neq 1\). Fix such an \(a\) and consider that

\[\begin{align*} f(a) \sum_{g \in G} f(g) &= \sum_{g \in G} f(a)f(g) \\ &= \sum_{g \in G} f(ag). \\ \end{align*}\]

Now, because \(a\) is an element of the group, and is thus invertible, the function \(x \mapsto ax\) is a bijection. Thus

\[ \sum_{g \in G} f(ag) = \sum_{g \in G} f(g)\]

given they count the same terms.

This allows us to deduce that

\[ f(a) \sum_{g \in G} f(g) = \sum_{g \in G} f(g) \implies (f(a) - 1) \sum_{g \in G} f(g) = 0.\]

This expression is in \(\mathbb{C}\) and hence it is equal to zero if either \(f(a) = 1\) or \(\sum_{g \in G} f(g) = 0\), however \(f(a) \neq 1\) by choice of \(a\) and therefore

\[ \sum_{g \in G} f(g) = 0.\]
Lemma

For any finite abelian group \(G\), and element \(g \neq \mathrm{id}\), there exists a character \(f_i\) such that \(f_i(g) \neq 1\).

Proof

If \(f_i(g) = 1\) then \(f_i(g^k) = f_i(g)^k = 1\) and the character \(f_i\) evaluates to \(1\) on the cyclic subgroup generated by \(g\). If this group is non-trivial, then it has a non-principal character, which can be extended to a character of \(G\), however this would contradict the fact that all characters evaluate to \(1\) on this cyclic subgroup.

%% TODO change this to use splitting into product of cyclic groups and fix all the characters on each group to be the principal character except 1

Theorem

Let \(G\) be a finite abelian group with \(|G| = n\). For each \(g \in G\)

\[ \sum_{f \in \hat{G}} f(g) = \begin{cases} n & g = \mathrm{id} \\ 0 & \text{otherwise}. \\ \end{cases}\]
Proof

Suppose that \(g = \mathrm{id}\). Then from the properties of a group homomorphism

\[ \sum_{f \in \hat{G}} f(g) = \sum_{f \in \hat{G}} f(\mathrm{id}) = \sum_{f \in \hat{G}} 1 = |\hat{G}| = n\]

noting that \(|\hat{G}| \cong |G|\).

Now suppose that \(g \neq \mathrm{id}\). Therefore, by the previous lemma, there exists a character \(f_i\) such that \(f_i(g) \neq 1\). Then we have that

\[ f_i(g)\sum_{f \in \hat{G}} f(g) = \sum_{f \in \hat{G}} f_i(g)f(g).\]

Because \(f_i\) is an element of the group and has an inverse \(f_i \times f\) also varies over \(\hat{G}\) and hence

\[\begin{align*} \sum_{f \in \hat{G}} f_i(g)f(g) &= \sum_{f \in \hat{G}} (f_i \times f)(g) \\ &= \sum_{f \in \hat{G}} f(g). \\ \end{align*}\]

This lets us deduce that

\[ f_i(g)\sum_{f \in \hat{G}} f(g) = \sum_{f \in \hat{G}} f(g) \implies (f_i(g) - 1)\sum_{f \in \hat{G}} f(g) = 0.\]

Since \(f_i(g) \neq 1\) by choice of \(f_i\), we can conclude that

\[ \sum_{f \in \hat{G}} f(g) = 0.\]
Corollary

Let \(G\) be a finite abelian group with \(|G| = n\). For each \(g, h \in G\)

\[ \sum_{f \in \hat{G}} \overline{f(h)}f(g) = \begin{cases} n & g = h \\ 0 & \text{otherwise}. \\ \end{cases}\]
Proof

This follows from the previous result by replacing \(g\) with \(h^{-1}g\). That is we have

\[ \sum_{f \in \hat{G}} f(h^{-1}g) = \begin{cases} n & h^{-1}g = \mathrm{id} \\ 0 & \text{otherwise}. \\ \end{cases}\]

and

\[\begin{align*} \sum_{f \in \hat{G}} f(h^{-1}g) &= \sum_{f \in \hat{G}} f(h^{-1})f(g) \\ &= \sum_{f \in \hat{G}} f(h)^{-1}f(g) \\ &= \sum_{f \in \hat{G}} \overline{f(h)}f(g) \\ \end{align*}\]

where \(f(h)^{-1} = \overline{f(h)}\) because they are roots of unity.

Then in the condition in the cases clearly \(h^{-1}g = \mathrm{id} \iff g = h\), as required.