Complete reducibility over ℂ

Every finite-dimensional complex representation of a finite group splits as a direct sum of irreducibles.
Complete reducibility over ℂ

Let GG be a finite group and let VV be a finite-dimensional of GG.

Theorem (complete reducibility over C\mathbb C)

VV is : there exist subrepresentations V1,,VmV_1,\dots,V_m such that

V    V1Vm. V \;\cong\; V_1\oplus\cdots\oplus V_m.

Equivalently: for every WVW\subseteq V, there exists a GG-stable complement WVW'\subseteq V with

V  =  WW. V \;=\; W \oplus W'.

Standard mechanism (unitary averaging)

Choose any Hermitian ,\langle\cdot,\cdot\rangle on VV and average it over GG:

v,wG  =  1GgGgv,gw. \langle v,w\rangle_G \;=\; \frac{1}{|G|}\sum_{g\in G}\langle g\cdot v,\, g\cdot w\rangle.

Then ,G\langle\cdot,\cdot\rangle_G is GG-invariant. If WVW\subseteq V is GG-stable, its complement WW^\perp with respect to ,G\langle\cdot,\cdot\rangle_G is also GG-stable, giving V=WWV=W\oplus W^\perp.

This is a complex-analytic presentation of .

Examples

  1. Permutation representation of S3S_3 on C3\mathbb C^3.
    Let S3S_3 act by permuting coordinates of V=C3V=\mathbb C^3. The line

    U=span{(1,1,1)} U=\operatorname{span}\{(1,1,1)\}

    is S3S_3-stable (it is the trivial representation). The subspace

    W={(x1,x2,x3)C3:x1+x2+x3=0} W=\{(x_1,x_2,x_3)\in\mathbb C^3 : x_1+x_2+x_3=0\}

    is also S3S_3-stable and V=UWV=U\oplus W. Moreover, WW is the 22-dimensional irreducible (standard) representation.

  2. Any representation of a cyclic group CnC_n.
    If Cn=gC_n=\langle g\rangle and ρ(g)n=I\rho(g)^n=I, then the minimal polynomial of ρ(g)\rho(g) divides xn1x^n-1, which has distinct roots over C\mathbb C. Hence ρ(g)\rho(g) is diagonalizable, and VV decomposes as a direct sum of eigenspaces. Each eigenspace is a 11-dimensional subrepresentation on which gg acts by an nnth root of unity (a character of CnC_n).

  3. The swap representation of C2C_2 on C2\mathbb C^2.
    Let C2={1,s}C_2=\{1,s\} act on V=C2V=\mathbb C^2 by s(x,y)=(y,x)s(x,y)=(y,x). Then

    V=span{(1,1)}    span{(1,1)}. V = \operatorname{span}\{(1,1)\}\;\oplus\;\operatorname{span}\{(1,-1)\}.

    The first summand is the trivial representation; the second is the sign representation (where ss acts as 1-1).