Let \(V\) be the inner product space in Exercise 7 and \(G\) the group of unitary matrices in \(V\). If \(B\) is in \(G\), let \(C_B\) denote the linear operator on \(V\) defined by \[ C_B(A) = BAB^{-1}. \] Show that
\(C_B\) is a unitary operator on \(V\);
\(C_{B_1B_2} = C_{B_1}C_{B_2}\);
there is no unitary transformation \(U\) on \(V\) such that \[ UL_BU^{-1} = C_B \] for all \(B\) in \(G\).
Solution
Goal. Prove the three facts about the conjugation operators \(C_B\).
1. (a) \(C_B\) is unitary.
1.1. \(C_B(A) = BAB^{-1} = BAB^*\) (since \(B\) is unitary, \(B^{-1} = B^*\)).
\(B\) is unitary.
1.2. \(\|C_B(A)\| = \|BAB^*\| = \|A\|\).
the Frobenius (Hilbert–Schmidt) norm is invariant under left and right multiplication by unitaries: \(\|BAB^*\|^2 = \operatorname{tr}((BAB^*)^*(BAB^*)) = \operatorname{tr}(BA^*B^*BAB^*) = \operatorname{tr}(BA^*AB^*) = \operatorname{tr}(A^*A) = \|A\|^2\).
1.3. Hence \(C_B\) is an isometry, so it is unitary.
a linear isometry on a finite-dimensional inner product space is unitary.
2. (b) \(C_{B_1 B_2} = C_{B_1} C_{B_2}\).
2.1. \(C_{B_1 B_2}(A) = (B_1 B_2) A (B_1 B_2)^{-1} = B_1 B_2 A B_2^{-1} B_1^{-1}\).
definition.
2.2. \(C_{B_1}(C_{B_2}(A)) = C_{B_1}(B_2 A B_2^{-1}) = B_1 B_2 A B_2^{-1} B_1^{-1}\).
apply \(C_{B_2}\) then \(C_{B_1}\).
2.3. Hence \(C_{B_1 B_2} = C_{B_1} C_{B_2}\).
2.1 and 2.2 agree.
3. (c) No unitary \(U\) satisfies \(U L_B U^{-1} = C_B\) for all \(B\).
3.1. \(L_B\) is left multiplication by \(B\): \(L_B(A) = BA\).
definition (from Exercise 7).
3.2. \(L_B\) and \(C_B\) are not conjugate by a single \(U\) for all \(B\).
the map \(B \mapsto L_B\) is a representation of \(G\) (left regular), while \(B \mapsto C_B\) is the conjugation representation; these are not equivalent as representations of \(G\) (the left regular representation and the conjugation representation have different characters — e.g. the trace of \(L_B\) is \(n \operatorname{tr}(B)\) while the trace of \(C_B\) is \(|\operatorname{tr}(B)|^2\)).
3.3. Hence no such \(U\) exists.
if \(U L_B U^{-1} = C_B\) for all \(B\), then \(L\) and \(C\) would be equivalent representations, contradicting 3.2.
4. Q.E.D.
1, 2, 3 prove (a), (b), (c).