Simplified partial trace of two operators If I have two operators A and B living in the Composite Hilbert Space $H_I \bigotimes H_{II} $ and I want to take the partial trace of $C=AB$ over the subspace $H_I$, i.e., $Tr_I[AB]$, is there any identity that can help me do this in terms of $Tr_I[A]$ and $Tr_I[B]$. Actually what I am interested in is the partial trace of the commutator $[A,B]$.
 A: Let $H$ and $K$ be Hilbert spaces with bases $|e_a\rangle$  and $|f_i \rangle$, respectively.
Let $A,B: H \otimes K \to H \otimes K$ be two operators, and let $C=A\circ B$ be their composition.  This means that they are of the form
$$ A ~=~|e_a\rangle \otimes |f_i \rangle ~ A^{ai}{}_{bj}~ \langle e^b| \otimes \langle f^j |,  $$
$$B ~=~|e_b\rangle \otimes |f_j \rangle ~ B^{bj}{}_{ck}~ \langle e^c| \otimes \langle f^k |, $$
$$ C ~=~|e_a\rangle \otimes |f_i \rangle ~ A^{ai}{}_{bj}~ B^{bj}{}_{ck}~ \langle e^c| \otimes \langle f^k |,  $$
where there are implicitly summed over repeated indices. The partial traces over $H$ are 
$$ Tr_{H}A~=~ |f_i \rangle ~ A^{ai}{}_{aj}~ \langle f^j|, $$
$$Tr_{H}B ~=~ |f_j \rangle ~ B^{bj}{}_{bk}~ \langle f^k| , $$
$$ Tr_{H}C ~=~ |f_i \rangle ~ A^{ai}{}_{bj}~ B^{bj}{}_{ak}~ \langle f^k|.  $$
$Tr_{H}C$ contains in general off-diagonal information, that are not included in $Tr_{H}A$ and $Tr_{H}B$, so $Tr_{H}C$ can in general not be written as a function of $Tr_{H}A$ and $Tr_{H}B$ only. 
Similar reasoning applies to the commutator $A\circ B-B\circ A$.
