If you have a \(\ZZ/5\ZZ\) action on a complex vector space, what does this action look like? What about an \(S_3\) action? A dihedral group of any order?
Solution
All representations below are finite-dimensional over \(\CC\). Since the groups are finite, Maschke’s theorem gives complete reducibility.
1. A representation of \(C_5=\langle g\rangle\) is a direct sum of one-dimensional characters.
The operator \(\rho(g)\) satisfies \[ \rho(g)^5=I, \] so its minimal polynomial divides \(x^5-1\), which has distinct roots over \(\CC\). Hence \(\rho(g)\) is diagonalizable, with eigenvalues among the fifth roots of unity. Each eigenspace is \(C_5\)-stable. Thus \[ V=\bigoplus_{k=0}^4 V_k, \] where \(g\) acts on \(V_k\) by multiplication by \(\zeta_5^k\).
2. Every representation of \(S_3\) is a direct sum of copies of the three irreducibles: the trivial representation, the sign representation, and the standard two-dimensional representation.
The group \(S_3\) has three conjugacy classes, hence three irreducible complex representations. Their dimensions must satisfy \[ \sum d_i^2=|S_3|=6. \] The two one-dimensional characters are trivial and sign, leaving one irreducible of dimension \(2\), the standard representation on \[ \{(x_1,x_2,x_3)\in\CC^3:x_1+x_2+x_3=0\}. \] Maschke’s theorem decomposes every representation into these.
3. For the dihedral group\[ D_{2n}=\langle r,s\mid r^n=s^2=1,\ srs=r^{-1}\rangle, \]all irreducible complex representations have dimension \(1\) or \(2\).
Since \(r^n=1\), the operator \(\rho(r)\) is diagonalizable with eigenvalues among the \(n\)th roots of unity. If \(v\) is a \(\zeta^k\)-eigenvector for \(r\), then \[ r(sv)=s(r^{-1}v)=\zeta^{-k}sv, \] so \(s\) pairs the \(\zeta^k\)- and \(\zeta^{-k}\)-eigenspaces.
When \(k\not\equiv -k\pmod n\), the span of \(v\) and \(sv\) gives a two-dimensional irreducible model \[ r\mapsto \begin{pmatrix} \zeta^k&0\\ 0&\zeta^{-k} \end{pmatrix}, \qquad s\mapsto \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}. \] The self-inverse eigenvalues \(1\) (and also \(-1\) when \(n\) is even) yield one-dimensional characters. Maschke’s theorem then decomposes every representation into these irreducibles.