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