if S and T are Hilbert-Schmidt between two Hilbert spaces U and V, is $\operatorname{trace}_U(S^{\ast}T) = \operatorname{trace}_V(T^{\ast}S)$?

Question

Let $U$ and $V$ be two real Hilbert spaces and $S$, $T$ two Hilbert-Schmidt operators between $U$ and $V$. Denote their adjoints by $S^{\ast}$ and $T^{\ast}$, both viewed after identification of $U$ and $V$ with their respective duals as Hilbert-Schmidt operators between $V$ and $U$. Is it true that $$\operatorname{trace}_U(S^{\ast}T) = \operatorname{trace}_V(T^{\ast}S)$$ holds (at least when the Hilbert spaces are separable)? Formulated differently, is the set of Hilbert-Schmidt operators between different Hilbert spaces still a Hilbert space when equipped with the inner product $\langle S,T \rangle=\operatorname{trace}_U(S^{\ast}T)$?

Answer 1

Assuming both $U$ and $V$ are separable, infinite dimensional Hilbert spaces, it is well known that there exists an isometry $\varphi :V\to U$.

Considering the map $$ \Phi: T\in \mathscr B(U, V) \mapsto \varphi T \in \mathscr B(U, U), $$ notice that $\Phi$ restricts to a bijective linear map from $L_2(U, V)$, the space of Hilbert-Schmidt operators, onto $L_2(U, U)$.

Moreover, for every $S, T\in L_2(U, V)$, one has that $$ \text{trace}_U(S^*T) = \text{trace}_U(S^*\varphi ^*\varphi T) = \text{trace}_U\big (\Phi(S)^*\Phi(T)\big ), \tag 1 $$ so $\Phi$ is compatible with the usual inner-product on $L_2(U, U)$ and the proposed inner-product on $L_2(U, V)$.

This answers the last question posed by the OP, namely that the set of Hilbert-Schmidt operators between different Hilbert spaces is still a Hilbert space with the indicated inner-product.

However, the relation $$ \text{trace}_U(S^{\ast}T) = \text{trace}_V(T^{\ast}S), \tag 2 $$ is not true, as it compares a map that is conjugate-linear on $S$, with one that is linear on $S$. Nevertheless, we may use (1) and the fundamental property of the trace to write $$ \text{trace}_U(S^*T) = \text{trace}_U\big (\Phi(S)^*\Phi(T)\big ) = \text{trace}_U\big (\Phi(T)\Phi(S)^*\big ) = $$$$ = \text{trace}_U(\phi TS^*\varphi ^*) = \text{trace}_V(TS^*), $$ where the last step is a consequence of the unitary invariance of the trace.