property of adjoint

Question

Let $\mathfrak{g}$ be real Lie algebra of matrices over $\mathbb{C}$ that is closed under conjugate transpose $(\cdot)^*$. Why is it the case that for each $X \in \mathfrak{g}$, $\mathrm{ad}(X^*) = \mathrm{ad}(X)^*$? We are using the inner product $<X,Y> := \mathrm{Re} \mathrm{tr}(XY^*)$ on $\mathfrak{g}$.

Answer 1

Just write out the definitions and simplify. The definition of adjoints with respect to an inner product says that the equation you want is equivalent to saying that, for all $Y$ and $Z$, $$ \langle\text{ad}(X^*)(Y),Z\rangle=\langle Y,\text{ad}(X)(Z)\rangle. $$ Now use the definition of ad to rewrite this as $$ \langle[X^*,Y],Z\rangle=\langle Y,[X,Z]\rangle. $$ Next insert the definition of your inner product to rewrite this as $$ \text{Retr}([X^*,Y]\cdot Z^*)=\text{Retr}(Y,[X,Z]^*). $$ Next use the fact that, in a Lie algebra of matrices, the Lie bracket is the ordinary commutator, and remember that the transpose of a product is the product of the transposes in the reverse order, to put the desired equation into the equivalent form $$ \text{Retr}(X^*YZ^*-YX^*Z^*)=\text{Retr}(YZ^*X^*-YX^*Z^*). $$ Next, note that Retr must mean real part of the trace (not real part of the transpose, which can't be an inner product because it isn't a scalar). Also remember that the trace is linear, so that your desired equation amounts to $$ \text{Retr}(X^*YZ^*)-\text{Retr}(YX^*Z^*)= \text{Retr}(YZ^*X^*)-\text{Retr}(YX^*Z^*). $$ Finally, remember that the trace of a matrix product $AB$ is the same as the trace of $BA$. Apply that with $A=X^*$ and $B=YZ^*$ to see that the first terms on the two sides of your desired equation match. The second terms are identical, so the equation holds.