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Let $G_1$ and $G_2$ be finite groups and let $G = G_1 ×G_2$. Suppose $ρ: G_1 → GL_m(\Bbb C)$ and $ϕ: G_2 → GL_n(\Bbb C)$ are representations. Let $V =M_{mn}(\Bbb C)$ be the vector space of $m×n$-matrices over $\Bbb C$. Define $τ : G → GL(V )$ by $τ(g_1,g_2)(A) = ρ_{g_1}Aϕ_{g_2}^T$ where $B^T$ is the transpose of a matrix $B$.

  1. Show that $τ$ is a representation of $G$.
  2. Prove that $χ_τ (g_1, g_2) = χ_ρ(g_1)χ_ϕ(g_2)$.
  3. Show that if $ρ$ and $ϕ$ are irreducible, then $τ$ is irreducible.
  4. Prove that every irreducible representation of $G_1×G_2$ can be obtained in this way.

The solution of 1 is pretty clear.

I am facing problem from 2 onwards. For a representation $\phi : G \to GL(V)$ where $V \cong \Bbb C^n$

$\chi _{\phi}(g)=Tr(\phi_g)=\sum_{i=1}^n<\phi_g(e_i),e_i>$ but what to do here I was writting in terms of $E_{ij}$ bit not getting satisfactory answer. Please help from 2 onwards.

1 Answers 1

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You have the right idea: for 2, we note that the inner product over $M_{mn}$ is given by $\langle A,B \rangle = tr(AB^*)$, with $*$ denoting the conjugate transpose. In other words, our inner product is simply the "dot product" of matrices. Note also that $E_{ij} = e_ie_j^T$. We then have $$ \chi_\tau(g_1,g_2) = \sum_{i= 1}^m \sum_{j=1}^n \langle \tau(g_1,g_2)E_{ij}, E_{ij} \rangle = \sum_{i= 1}^m \sum_{j=1}^n \langle \rho_{g_1}e_ie_j^T\phi_{g_2}^T, e_ie_j^T \rangle = \\ \sum_{i= 1}^m \sum_{j=1}^n \langle \rho_{g_1}e_i e_j^T\phi_{g_2}^T, e_ie_j^T \rangle = \\ \sum_{i= 1}^m \sum_{j=1}^n Tr( \rho_{g_1}e_ie_j^T\phi_{g_2}^Te_je_i^T) = \\ \sum_{i= 1}^m \sum_{j=1}^n Tr( e_i^T\rho_{g_1}e_ie_j^T\phi_{g_2}^Te_j) = \\ \sum_{i= 1}^m \sum_{j=1}^n [e_i^T\rho_{g_1}e_i][e_j^T\phi_{g_2}^Te_j] = \\ \sum_{i= 1}^m \sum_{j=1}^n \langle\rho_{g_1}e_i,e_i\rangle \langle \phi_{g_2} e_j,e_j\rangle = \\ \chi_\rho(g_1) \chi_{\phi}(g_2) $$ For 3: I would try to prove by contrapositive. That is: show that if $\tau$ is reducible but $\phi$ is irreducible, then $\rho$ is reducible. Not sure about 4.

  • 0
    Would you explain the 4th to 5th line? How the trace breaks into two parts as trace is not in general multiplicative.2017-02-09
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    Matrix multiplication is associate, and the matrix whose trace we take in the 4th line is really $1\times 1$, which is to say a scalar.2017-02-09
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    Oh, I miscounted, that was 5 to 6. From 4 to 5: $Tr(AB)= Tr(BA)$, so the trace of a product is invariant under "cyclic permutations".2017-02-09
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    Alternatively, we can go straight from lines 3 to 5 if we note that both summands extract the $i,j$ entry of line 3's left-side of the inner product.2017-02-09
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    I am saying about $$Tr( e_i^T\rho_{g_1}e_ie_j^T\phi_{g_2}^Te_j) = \\ \sum_{i= 1}^m \sum_{j=1}^n [e_i^T\rho_{g_1}e_i][e_j^T\phi_{g_2}^Te_j] $$ this line.2017-02-09
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    Well, then the first comment applies. The trace of a scalar is that scalar, and we're free to rebracket any product.2017-02-09
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    What is the scalar here? In $$\sum_{i= 1}^m \sum_{j=1}^nTr( e_i^T\rho_{g_1}e_ie_j^T\phi_{g_2}^Te_j) = \\ \sum_{i= 1}^m \sum_{j=1}^n [e_i^T\rho_{g_1}e_i][e_j^T\phi_{g_2}^Te_j] $$. Both $i,j$ are varying. I think you are trying to say $Tr(cA)=cTr(A)$ then what is $c$ scalar?2017-02-09
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    I'm saying that $e_i^T\rho_{g_1}e_ie_j^T\phi_{g_2}^Te_j$ is a $1 \times 1$ matrix (and thus equal to its trace).2017-02-09
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    yes yes. True. Thank you.2017-02-09