The cyclic group of order $r$

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 23 May 2010

The cyclic group of order $r$

The cyclic group $G\left(r,1,1\right)$ of order $r$ is the group generated by a single element $g$ with the relation $g^r=1.$ The group $G\left(r,1,1\right)=\left\{1,g,g^2,\dots ,g^{r-1}\right\}$ can be relaised as the group $\mathbb{Z}/r\mathbb{Z}=\left\{0,1,\dots ,r-1\right\}$ or as the group of $r$th roots of unity in $ℂ,$$$G\left(r,1,1\right)=\left\{e^{2\pi ik/r}|k=0,1,\dots ,r\right\}.$$ Since $G\left(r,1,1\right)$ is abelian every element is in a conjugacy class by itself. Since $ℂG\left(r,1,1\right)$ is commutative every irreducible representation is one dimensional. A representation $\rho :ℂG\left(r,1,1\right)\to ℂ$ i completely determined by the value $\rho \left(g\right)$ and the relation $g^r=1$ forces $\rho {\left(g\right)}^r=1.$ So $\rho$ is an irreducible representation of $\mathbb{Z}/r\mathbb{Z}$ iff $\rho \left(g\right)$ is an $r$th root of unity. This proves the following theorem.

1. The irreducible representations $S^{\lambda }$ of the cyclic group $G\left(r,1,1\right)$ are indexed by $0\le \lambda \le r-1.$
2. $\dim S^{\lambda }=1,$ for all $\lambda =0,1,\dots ,r-1.$
3. The irreducible representations (and the irreducible characters) of $G\left(r,1,1\right)$ are given by the maps $$\begin{array}{rcll}{\rho }^k:& ℂG\left(r,1,1\right)& \to & ℂ\\ & g& \mapsto & e^{2\pi i/r}\end{array}0\le k<r-1.$$
4. The irreducible $G\left(r,1,1\right)$-module $S^{\lambda }$ is given by the one dimensional vector space $S^{\lambda }=ℂv_{\lambda }$ with $G\left(r,1,1\right)$-action given by $$g^k{v}_{\lambda }={\xi }^{k\lambda }{v}_{\lambda },\text{where }\xi =e^{2\pi i/r}.$$

Fix $u_0,\dots ,u_{r-1}\in ℂ.$ The cyclic algebra $H_{r,1,1}=H_{r,1,1}\left(u_0,\dots ,u_{r-1}\right)$ is the algebra given by a single generator $X$ which satisfies the relation $$\left(X-u_0\right)\left(X-u_1\right)\dots \left(X-u_{r-1}\right)=0.$$ ie $H_{r,1,1}$ is the quotient of the polynomial ring $ℂ\left[X\right]$ by the ideal generated by the polynomial in ???. The algebra $H_{r,1,1}$ has basis $\left\{X^k|0\le k\le r-1\right\}$ and, if $u_k={\xi }^k,$ where $\xi =e^{2\pi i/r},$ then $H_{r,1,1}=ℂG\left(r,1,1\right)$ is the group algebra of the cyclic group.

The proof of the following theorem is exactly the same as the proof of Theorem ???. The hypotheses in the statement are exactly what is needed to guarantee that the maps $\rho k$ defined in c) are distinct (and nonzero???).

Assume that $u_0,\dots ,u_{r-1}\in ℂ$ are all distinct (and nonzero???).

1. The irreducible representations $H^{\lambda }$ of the cyclic algebra $H_{r,1,1}$ are indexed by $0\le \lambda \le r-1.$
2. $\dim H^{\lambda },$ for all $\lambda =0,1,dots,r-1.$
3. The irreducible representations (and the irreducible characters) of $H_{r,1,1}\left(u_0,\dots ,u_{r-1}\right)$ are given by the maps $$\begin{array}{rcll}{\rho }^k:& H_{r,1,1}\left(u_0,\dots ,u_{r-1}\right)& \to & ℂ\\ & X& \mapsto & u_i\end{array},0\le k\le r-1.$$
4. The irreducible $H_{r,1,1}$-module $H^{\lambda }$ is given by the one dimensional vector space $H^{\lambda }=ℂv_{\lambda }$ with $H_{r,1,1}$-action given by $$X{v}_{\lambda }=u_{\lambda }{v}_{\lambda }.$$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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