Section 7.3 Orthogonality relations
Proof.
Fix an isomorphism \(G \cong \bbZ/m_1\bbZ \times \dotsb \times \bbZ/m_k\bbZ\text{.}\) Let \(g\) correspond to the tuple \((a_1,\dotsc,a_k)\text{.}\) There is some \(r\) such that \(a_r \not\cong 0 \pmod{m_r}\text{.}\) Define the character \(\psi\) as
\begin{equation*}
\psi = (\chi_0,\chi_0,\dotsc,\chi_0,\chi_1,\chi_0,\dotsc,\chi_0)\text{,}
\end{equation*}
where, for all \(i \neq r\text{,}\) \(\chi_0\) is the principal character on the \(i\)th factor \(\bbZ/m_i \bbZ\text{,}\) and for \(i=r\text{,}\) \(\chi_1\) the character defined previously on the factor \(\bbZ/m_r \bbZ\text{.}\)
Then we have
\begin{equation*}
\begin{split}
\psi(b_1,\dotsc,b_k) \amp = \chi_0(b_1)\chi_0(b_2) \dotsm \chi_0(b_{r-1}) \chi_1(b_r) \chi_0(b_{r+1}) \dotsm \chi_0(b_k) \\
\amp = \chi_1(b_r) \\
\amp = e(b_r / m_r)
\end{split}\text{.}
\end{equation*}
Specifically, \(\psi(g) = e(a_r / m_r) \neq 1\) because \(a_r \not\cong 0 \pmod{m_r}\text{.}\)
Remark 7.3.2.
This lemma, for more general (infinite) groups, is known as the Peter-Weyl theorem (more precisely, it is part of the Peter-Weyl theorem).Now we can consider sums of character values over an arbitrary (not cyclic) finite abelian group.
Theorem 7.3.3.
Let \(G\) be a finite abelian group. Then:- If a character \(\chi\) of \(G\) is given, then\begin{equation} \sum_{g \in G} \chi(g) = \begin{cases} |G|, \amp \text{if } \chi=\chi_0, \\ 0, \amp \text{otherwise} \end{cases}\text{.}\tag{7.3.1} \end{equation}
- If an element \(g\) of \(G\) is given, then\begin{equation} \sum_{\chi \in \hat{G}} \chi(g) = \begin{cases} |G|, \amp \text{if } g = 1, \\ 0, \amp \text{otherwise} \end{cases}\text{.}\tag{7.3.2} \end{equation}
Proof.
If \(\chi = \chi_0\) then \(\chi(g) = 1\) for all \(g \in G\text{.}\) Likewise if \(g = 1\) then \(\chi(g) = 1\) for all \(\chi \in \hat{G}\text{.}\) In those cases the sum is \(|G|\) (we use \(|\hat{G}| = |G|\)).
Now, given \(\chi\text{,}\) let \(S_\chi = \sum_{g \in G} \chi(g)\text{.}\) We have, for any \(h \in G\text{,}\)
\begin{equation}
\chi(h) S = \sum_{g \in G} \chi(h)\chi(g) = \sum_{g \in G} \chi(hg)
= \sum_{x \in G} \chi(x) = S_\chi\text{,}\tag{7.3.3}
\end{equation}
that is, \((1-\chi(h))S_\chi = 0\text{.}\) If \(\chi\) is non-principal then (by definition) there is some \(h \in G\) such that \(\chi(h) \neq 1\text{.}\) With this choice of \(h\) it follows that \(S_\chi = 0\text{.}\)
Similarly, given \(g \in G\text{,}\) let \(S_g = \sum_{\chi \in \hat{G}} \chi(g)\text{.}\) For any \(\psi \in \hat{G}\text{,}\) consider
\begin{equation}
\psi(g) S_g = \psi(g) \sum_{\chi \in \hat{G}} \chi(g)
= \sum_{\chi \in \hat{G}} (\psi\chi)(g)
= \sum_{\eta \in \hat{G}} \eta(g) = S_g\text{.}\tag{7.3.4}
\end{equation}
Here we are using that \(\hat{G}\) is a group, so \(\{\psi\chi \colon \chi \in \hat{G}\} = \hat{G}\text{.}\) We get that \((1-\psi(g))S_g = 0\text{.}\) By the previous lemma, if \(g \neq 1\text{,}\) there is some \(\psi \in \hat{G}\) so that \(\psi(g) \neq 1\text{.}\) Then \(S_g = 0\text{,}\) as claimed.
Theorem 7.3.4.
Let \(G\) be a finite abelian group. Then:- For any \(\chi,\psi \in \hat{G}\) we have:\begin{equation} \sum_{g \in G} \chi(g) \overline{\psi(g)} = \begin{cases} |G|, \amp \text{if } \chi=\psi, \\ 0, \amp \text{otherwise} \end{cases}\text{.}\tag{7.3.5} \end{equation}
- For any \(g, h \in G\) we have:\begin{equation} \sum_{\chi \in \hat{G}} \chi(g) \overline{\chi(h)} = \begin{cases} |G|, \amp \text{if } g=h, \\ 0, \amp \text{otherwise} \end{cases}\text{.}\tag{7.3.6} \end{equation}
Proof.
Recall that
\begin{equation}
\overline{\psi(g)} = (\overline{\psi})(g) = \psi^{-1}(g) = (\psi(g))^{-1}\text{.}\tag{7.3.7}
\end{equation}
So \(\chi(g)\overline{\psi(g)} = (\chi\psi^{-1})(g)\text{.}\) By the previous result, the sum \(\sum_{g \in G} (\chi\psi^{-1})(g)\) is either equal to \(|G|\) if \(\chi\psi^{-1} = \chi_0\text{,}\) the principal character, or else is equal to \(0\text{.}\) Since \(\chi_0\) is the identity element in \(\hat{G}\text{,}\) then \(\chi\psi^{-1} = \chi_0\) if and only if \(\chi=\psi\text{.}\)
Similarly, \(\chi(g)\overline{\chi(h)} = \chi(gh^{-1})\text{.}\) Therefore the sum \(\sum_{\chi \in \hat{G}} \chi(gh^{-1})\) is either equal to \(|G|\) if \(gh^{-1} = 1\text{,}\) or else is equal to \(1\text{.}\)
Remark 7.3.5.
We won’t pursue this idea much further, but it is worth mentioning: We can define “inner products” on \(G\) and on \(\hat{G}\) by
\begin{equation}
\langle g \mid h \rangle = \frac{1}{|G|} \sum_{\chi \in \hat{G}} \chi(g)\overline{\chi(h)},
\qquad
\langle \chi \mid \psi \rangle = \frac{1}{|G|} \sum_{g \in G} \chi(g) \overline{\psi(g)}\text{.}\tag{7.3.8}
\end{equation}
These are not exactly inner products because \(G\) and \(\hat{G}\) are not vector spaces. So we can consider the vector spaces with \(G\) and \(\hat{G}\) as bases; or, even better, the “group algebras” of \(G\) and \(\hat{G}\text{.}\) Without getting into the definitions of those, we’ll just remark that the previous theorem means that the elements of \(G\) form an orthonormal basis for their group algebra, and likewise, the elements of \(\hat{G}\) form an orthonormal basis for their group algebra.
