Finite Abelian Characters for Combinatorics — Examples

Compute $\zeta$: iterating the addition law [L5] gives ${\zeta }^5=\exp (5\cdot 2\pi i\mathrm{/}5)=\exp (2\pi i)=1$, because $2\pi i\in 2\pi i\mathbb{Z}=\ker (\exp )$; and $\zeta \mathrm{\ne }1$, since otherwise $2\pi i\mathrm{/}5\in \ker (\exp )=2\pi i\mathbb{Z}$ would give $1\mathrm{/}5\in \mathbb{Z}$, which is false. So $\zeta$ is a fifth root of unity different from $1$, and by [L6] the fifth roots of unity are exactly the five distinct values $1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4=\exp (2\pi ik\mathrm{/}5)$, $0\le k<5$.

Well-definedness. Suppose $[s]=[t]$ in $\mathbb{Z}\mathrm{/}5\mathbb{Z}$, so $5\mid (s-t)$ by [L4]; write $s=t+5m$ with $m\in \mathbb{Z}$. Then the integer power laws together with ${\zeta }^5=1$ give ${\zeta }^{rs}={\zeta }^{rt}({\zeta }^5{)}^{rm}={\zeta }^{rt}$, so the prescription ${\chi }_r([s]):={\zeta }^{rs}$ does not depend on the chosen representative and defines a function ${\chi }_r:\mathbb{Z}\mathrm{/}5\mathbb{Z}\to \mathbb{C}$ for each $r$.

The five are distinct and exhaustive. If ${\chi }_r={\chi }_t$, evaluating at $[1]$ gives ${\zeta }^r={\zeta }^t$, that is $\exp (2\pi ir\mathrm{/}5)=\exp (2\pi it\mathrm{/}5)$, and [L5] gives $2\pi i(r-t)\mathrm{/}5\in 2\pi i\mathbb{Z}$, so $5\mid (r-t)$; as $0\le r,t\le 4$ this forces $r=t$. Conversely let $\chi$ be any additive character and put $\lambda :=\chi ([1])$; five applications of multiplicativity in [L1], together with $[1]+[1]+[1]+[1]+[1]=[5]=[0]$ in [L4], give ${\lambda }^5=\chi ([1]{)}^5=\chi ([5])=\chi ([0])=1$, so $\lambda$ is a fifth root of unity and by step 1.1 there is a unique $k\in \{0,1,2,3,4\}$ with $\lambda ={\zeta }^k$. For $0\le s\le 4$ one has $[s]=s\cdot [1]$, whence $\chi ([s])={\lambda }^s={\zeta }^{ks}={\chi }_k([s])$; therefore $\chi ={\chi }_k$, and every additive character of $\mathbb{Z}\mathrm{/}5\mathbb{Z}$ occurs among the five.

Each ${\chi }_r$ is an additive character. By [L4], $[s]+[t]=[s+t]$, so ${\chi }_r([s]+[t])={\zeta }^{r(s+t)}={\zeta }^{rs}{\zeta }^{rt}={\chi }_r([s]){\chi }_r([t])$; every value is one of the fifth roots of unity listed in step 1.1 and hence nonzero. So ${\chi }_r:\mathbb{Z}\mathrm{/}5\mathbb{Z}\to {\mathbb{C}}^{\times }$ is a group homomorphism, that is, an additive character of $\mathbb{Z}\mathrm{/}5\mathbb{Z}$, by [L1].

The table and its orthogonality. By [L2] the five additive characters are exactly the irreducible complex characters of $\mathbb{Z}\mathrm{/}5\mathbb{Z}$, so the array of values $T_{rs}={\chi }_r([s])={\zeta }^{rs}$ is the character table of the group: its rows, for $r=0,1,2,3,4$, are $(1,1,1,1,1)$, $(1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4)$, $(1,{\zeta }^2,{\zeta }^4,\zeta ,{\zeta }^3)$, $(1,{\zeta }^3,\zeta ,{\zeta }^4,{\zeta }^2)$ and $(1,{\zeta }^4,{\zeta }^3,{\zeta }^2,\zeta )$. By [L4] the group has order $5$ and its five elements are $[0],\dots ,[4]$, so row orthogonality [L3] applied to ${\chi }_r$ and ${\chi }_t$ gives exactly $\frac{1}{5}{\sum }_{s=0}^4{\chi }_r([s])\overline{{\chi }_t([s])}={\delta }_{rt}$. As a direct check of an off-diagonal entry, for $r=1$ and $t=0$ the summands are ${\zeta }^s\bar{1}={\zeta }^s$, and ${\sum }_{s=0}^4{\zeta }^s=0$ by [L7], since step 1.1 lists $1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4$ as the five fifth roots of unity; on the diagonal ${\chi }_r([s])\overline{{\chi }_r([s])}=\mathrm{\mid }{\zeta }^{rs}{\mathrm{\mid }}^2=1$ by [L6] and [L8], so each diagonal average is $\frac{1}{5}\cdot 5=1$.

Steps 2.1 and 3.1 show that each ${\chi }_r$ is a well-defined additive character, step 2.2 that they are pairwise distinct and are all of them, and step 3.2 computes the table and verifies its orthogonality; this proves every claim of the example.