What are the finite-dimensional complex representations of \(S_3\)? How do they restrict to \(S_2\)?
Solution
Let \(H=S_2=\langle(12)\rangle\le S_3\).
1. Up to isomorphism, \(S_3\) has exactly three irreducible complex representations: the trivial representation \(\mathbf 1\), the sign representation \(\mathrm{sgn}\), and the two-dimensional standard representation \(V\).
The conjugacy classes of \(S_3\) have cycle types \[ 1^3,\qquad 2\,1,\qquad 3, \] so there are three irreducible complex representations. Their dimensions \(d_i\) satisfy \[ \sum_i d_i^2=|S_3|=6. \] The trivial and sign representations give two one-dimensional irreducibles. The remaining dimension must therefore satisfy \[ 1^2+1^2+d^2=6, \] so \(d=2\). The standard representation on \[ \{(x_1,x_2,x_3)\in\CC^3:x_1+x_2+x_3=0\} \] is irreducible and supplies this two-dimensional representation.
2. Every finite-dimensional complex representation of \(S_3\) is a direct sum of copies of these three irreducibles.
Maschke’s theorem applies because \(\operatorname{char}\CC=0\) does not divide \(|S_3|\). Hence every finite-dimensional representation is completely reducible.
3. Their restrictions to \(H\cong C_2\) are\[ \mathbf1|_H\cong\mathbf1_H, \qquad \mathrm{sgn}|_H\cong\mathrm{sgn}_H, \qquad V|_H\cong\mathbf1_H\oplus\mathrm{sgn}_H. \]
The first two are immediate. For the standard representation, the transposition \((12)\) acts on \[ V=\{x_1+x_2+x_3=0\} \] with eigenvectors \[ (1,1,-2)\quad\text{of eigenvalue }1, \qquad (1,-1,0)\quad\text{of eigenvalue }-1. \] Thus as a representation of \(H\), \(V\) splits as the direct sum of the trivial and sign characters.