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The standard 2-dimensional representation of S3 inside the permutation representation on C3 is irreducible

Example

Let S3 act on C3 by permuting the standard basis vectors e1,e2,e3. The line L:=C(1,1,1) is invariant, and its invariant complement U:={(x,y,z)C3:x+y+z=0} is the standard 2-dimensional representation. This representation is irreducible.

Facts & Assumptions

Given: The permutation representation of S3 on C3.

[L1]

A permutation action on a finite set gives a permutation representation on the free vector space with that basis (The trivial representation, the regular representation, and permutation representations from finite G-sets).

[L2]

A representation is irreducible exactly when it has no proper nonzero subrepresentation (Subrepresentations, direct sums of representations, and irreducibility).

Verification

technique · direct
1.1

By [L1], the action of S3 on {e1,e2,e3} extends to C3 by permuting coordinates. The vector (1,1,1) is fixed by every permutation, so L is invariant. The coordinate sum x+y+z is also permutation-invariant, so U is invariant and C3=LU.

L1given
2.1

Suppose WU is a nonzero invariant line, and choose 0v=(a,b,c)W. Because W is invariant under the transposition τ=(12), one has τ(v)=λv for some scalar λ, and τ2=1 forces λ=±1. If λ=1, then a=b and the relation a+b+c=0 gives v=a(1,1,2). If λ=1, then a=b and c=0, so v=a(1,1,0).

step 1.1L2givenalgebra
3.1

The 3-cycle σ=(123) sends (1,1,2) to (1,2,1) and (1,1,0) to (0,1,1), and neither image is a scalar multiple of the original vector. Thus neither of the two possibilities from step 2.1 can span an invariant line. So U has no proper nonzero invariant line, and by [L2] it is irreducible.

step 2.1L2givenalgebra

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