Knowlpedia

Character of a Tensor Product

For complex representations, the character of a tensor product is the pointwise product of characters.

Let GG be a finite group and let V,WV,W be finite-dimensional complex with actions ρV,ρW\rho_V,\rho_W. Their V⊗WV\otimes W is a representation via

ρV⊗W(g)  =  ρV(g)⊗ρW(g).\rho_{V\otimes W}(g) \;=\; \rho_V(g)\otimes \rho_W(g).

The is χV(g)=tr(ρV(g))\chi_V(g)=\mathrm{tr}(\rho_V(g)).

Proposition

For all g∈Gg\in G,

χV⊗W(g)  =  χV(g) χW(g).\chi_{V\otimes W}(g) \;=\; \chi_V(g)\,\chi_W(g).

Equivalently, χV⊗W=χV⋅χW\chi_{V\otimes W} = \chi_V\cdot \chi_W as functions G→CG\to \mathbb{C}.

This follows from the linear algebra identity

tr(A⊗B)=tr(A) tr(B),\mathrm{tr}(A\otimes B) = \mathrm{tr}(A)\,\mathrm{tr}(B),

using the .

Examples

Example 1: Cyclic group CnC_n

For G=Cn=⟨g⟩G=C_n=\langle g\rangle, 1D characters satisfy multiplication under tensor product. Let Va,VbV_a,V_b be 1D reps with g↦ζag\mapsto \zeta^a and g↦ζbg\mapsto \zeta^b where ζ=e2πi/n\zeta=e^{2\pi i/n}. Then

χVa⊗Vb(gm)=χVa(gm)χVb(gm)=ζamζbm=ζ(a+b)m,\chi_{V_a\otimes V_b}(g^m)=\chi_{V_a}(g^m)\chi_{V_b}(g^m)=\zeta^{am}\zeta^{bm}=\zeta^{(a+b)m},

so Va⊗Vb≅Va+bV_a\otimes V_b \cong V_{a+b}.

Example 2: S3S_3: tensoring by the sign representation

Let ε\varepsilon be the 1D sign representation of S3S_3, and let σ\sigma be the 2D standard irreducible. On the three conjugacy classes (e),(transposition),(3-cycle)(e),(\text{transposition}),(\text{3-cycle}),

χε=(1,−1,1),χσ=(2,0,−1).\chi_\varepsilon=(1,-1,1),\qquad \chi_\sigma=(2,0,-1).

Then

χσ⊗ε=χσχε=(2,0,−1)=χσ,\chi_{\sigma\otimes \varepsilon}=\chi_\sigma\chi_\varepsilon=(2,0,-1)=\chi_\sigma,

so σ⊗ε≅σ\sigma\otimes \varepsilon \cong \sigma.

Example 3: S3S_3: σ⊗σ\sigma\otimes\sigma

Using χσ⊗σ=χσ2\chi_{\sigma\otimes\sigma}=\chi_\sigma^2 pointwise gives

χσ⊗σ=(2,0,−1)2=(4,0,1).\chi_{\sigma\otimes\sigma}=(2,0,-1)^2=(4,0,1).

Decomposing into irreducibles using the known irreducible characters 1=(1,1,1)\mathbf{1}=(1,1,1), ε=(1,−1,1)\varepsilon=(1,-1,1), σ=(2,0,−1)\sigma=(2,0,-1), we check:

(4,0,1)=(1,1,1)+(1,−1,1)+(2,0,−1),(4,0,1) = (1,1,1) + (1,-1,1) + (2,0,-1),

so

σ⊗σ≅1  ⊕  ε  ⊕  σ.\sigma\otimes\sigma \cong \mathbf{1}\;\oplus\;\varepsilon\;\oplus\;\sigma.