Editing Dirichlet character (section)

== Orthogonality ==

The two orthogonality relations are<ref>See [[#Relation to group characters]] above.</ref>
:<math>\sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} \chi(a)=
\begin{cases}
\phi(m)&\text{ if  }\;\chi=\chi_0\\
0&\text{ if  }\;\chi\ne\chi_0
\end{cases}
</math> &nbsp; &nbsp; and &nbsp; &nbsp;  <math>\sum_{\chi\in\widehat{(\mathbb{Z}/m\mathbb{Z})^\times}}\chi(a)=
\begin{cases}
\phi(m)&\text{ if  }\;a\equiv 1\pmod{m}\\
0&\text{ if  }\;a\not\equiv 1\pmod{m}.
\end{cases}
</math>

The relations can  be written in the symmetric form
:<math>\sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} \chi_{m,r}(a)=
\begin{cases}
\phi(m)&\text{ if  }\;r\equiv 1\\
0&\text{ if  }\;r\not\equiv 1
\end{cases}
</math> &nbsp; &nbsp; and &nbsp; &nbsp;  <math>\sum_{r\in(\mathbb{Z}/m\mathbb{Z})^\times} \chi_{m,r}(a)=
\begin{cases}
\phi(m)&\text{ if  }\;a\equiv 1\\
0&\text{ if  }\;a\not\equiv 1.
\end{cases}
</math>

The first relation is easy to prove: If <math>\chi=\chi_0</math> there are <math>\phi(m)</math> non-zero summands each equal to 1. If <math>\chi\ne\chi_0</math>there is<ref>by the definition of <math>\chi_0</math></ref>  some <math>a^*,\; (a^*,m)=1,\;\chi(a^*)\ne1.</math> &nbsp;Then
:<math>\chi(a^*)\sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} \chi(a)=\sum_{a}\chi(a^*) \chi(a)=\sum_{a} \chi(a^*a)=\sum_{a} \chi(a),
</math><ref name="permute">because multiplying every element in a group by a constant element merely permutes the elements. See [[Group (mathematics)]]</ref> &nbsp; implying
:<math>(\chi(a^*)-1)\sum_{a} \chi(a)=0.</math> &nbsp; Dividing by the first factor gives <math>\sum_{a} \chi(a)=0,</math> QED. The identity <math>\chi_{m,r}(s)=\chi_{m,s}(r)</math> for <math>(rs,m)=1</math> shows that the relations are equivalent to each other. 

The second relation can be proven directly in the same way, but requires a lemma<ref>Davenport p. 30 (paraphrase) To prove [the second relation] one has to use ideas that we have used in the construction [as in this article or Landau pp. 109-114], or appeal to the basis theorem for abelian groups [as in Ireland & Rosen pp. 253-254]</ref>
:Given <math>a \not\equiv 1\pmod{m},\;(a,m)=1,</math> there is a <math> \chi^*,\; \chi^*(a)\ne1.</math>

The second relation has an important corollary: if <math>(a,m)=1,</math> define the function
:<math>f_a(n)=\frac{1}{\phi(m)} \sum_{\chi} \bar{\chi}(a) \chi(n). </math> &nbsp; Then
:<math>f_a(n)
= \frac{1}{\phi(m)} \sum_{\chi} \chi(a^{-1}) \chi(n)
= \frac{1}{\phi(m)} \sum_{\chi} \chi(a^{-1}n)
= \begin{cases} 1, &  n \equiv a \pmod{m} \\ 0, & n\not\equiv a\pmod{m},\end{cases}</math>
That is <math>f_a=\mathbb{1}_{[a]}</math> the [[indicator function]] of the residue class <math>[a]=\{ x:\;x\equiv a \pmod{m}\}</math>. It is basic in the proof of Dirichlet's theorem.<ref>Davenport chs. 1, 4; Landau p. 114</ref><ref>Note that if <math>g:(\mathbb{Z}/m\mathbb{Z})^\times\rightarrow\mathbb{C} </math> is any function
<math>g(n)=\sum_{a\in(\mathbb{Z}/m\mathbb{Z})^\times} g(a)f_a(n)</math>; see [[Fourier transform on finite groups#Fourier transform for finite abelian groups]]</ref>