Let \(\rho: G \to \operatorname{GL}(V)\) be a representation of the finite group \(G\).
Show that the map \(\hat{\rho}: g \in G \mapsto \rho(g^{-1})^t\) defines a representation, where \(t\) means the transpose of a matrix.
Let \(\chi_\rho\) and \(\chi_{\hat{\rho}}\) be the characters of \(\rho\) and \(\hat{\rho}\). Show that \(\chi_{\hat{\rho}}(g) = \overline{\chi_\rho(g)}\) (i.e. the complex conjugate) for all \(g \in G\).
Let \(V\) be a simple \(G\)-module. Show: If \(W\) is a simple \(G\)-module such that the trivial representation occurs in \(V \otimes W\), then \(W\) must be isomorphic to the representation defined in (a).
Solution
(a).
1. \(\hat\rho(gh) = \rho((gh)^{-1})^t = \rho(h^{-1}g^{-1})^t = (\rho(h^{-1})\rho(g^{-1}))^t = \rho(g^{-1})^t \rho(h^{-1})^t = \hat\rho(g)\hat\rho(h)\).
\(\rho\) is a homomorphism, and \((AB)^t = B^t A^t\).
2. \(\hat\rho(e) = \rho(e)^t = I\).
\(\rho(e) = I\).
3. Hence \(\hat\rho\) is a representation.
1 and 2.
(b).
1. \(\chi_{\hat\rho}(g) = \operatorname{tr}(\hat\rho(g)) = \operatorname{tr}(\rho(g^{-1})^t) = \operatorname{tr}(\rho(g^{-1}))\).
the trace is invariant under transpose.
2. \(\operatorname{tr}(\rho(g^{-1})) = \chi_\rho(g^{-1})\).
definition of character.
3. \(\chi_\rho(g^{-1}) = \overline{\chi_\rho(g)}\).
since \(G\) is finite, \(\rho(g)\) has finite order, so its eigenvalues are roots of unity; the eigenvalues of \(\rho(g^{-1}) = \rho(g)^{-1}\) are the inverses (conjugates) of the eigenvalues of \(\rho(g)\), so the traces are conjugate.
4. Hence \(\chi_{\hat\rho}(g) = \overline{\chi_\rho(g)}\).
<1>1–<1>3.
(c).
1. The trivial representation occurs in \(V \otimes W\) iff \(\langle \chi_V \chi_W, 1 \rangle \neq 0\).
the multiplicity of the trivial character in a representation is \(\langle \chi, 1 \rangle = \frac{1}{|G|}\sum_g \chi(g)\).
2. \(\langle \chi_V \chi_W, 1 \rangle = \frac{1}{|G|}\sum_g \chi_V(g)\chi_W(g) = \langle \chi_V, \overline{\chi_W} \rangle\).
definition of the inner product, and \(\overline{\chi_W(g)} = \chi_W(g^{-1})\).
3. Since \(V\) and \(W\) are simple, \(\langle \chi_V, \overline{\chi_W} \rangle \neq 0\) iff \(\chi_V = \overline{\chi_W}\).
orthogonality of irreducible characters.
4. \(\overline{\chi_W} = \chi_{\hat W}\) by part (b).
(b).
5. Hence \(\chi_V = \chi_{\hat W}\), so \(V \cong \hat W\), i.e. \(W \cong \hat V\).
3 and 4, and irreducible representations are determined by their characters.
6. Q.E.D.
3 (a), 4 (b), 5 (c).