Commutators of tensor product of Pauli matrices Given tensor product of rank-2 Pauli matrices $\sigma^a$. Each $\sigma^a$ is related to the generator of SU(2) Lie algebra.
We know they satisfy
$$[\sigma^a, \sigma^b ] = 2 i \epsilon^{abc} \sigma^c$$
Do you know any equality/identity to simplify:
$$
[\sigma^a \otimes \sigma^c, \sigma^b \otimes \sigma^d] = ?
$$
also
$$
[\sigma^a \otimes \sigma^c  \otimes \sigma^e, \sigma^b \otimes \sigma^d  \otimes \sigma^f] = ?
$$
$$
[\sigma^a \otimes \sigma^c  \otimes \sigma^e \otimes \sigma^g, \sigma^b \otimes \sigma^d  \otimes \sigma^f \otimes \sigma^h] = ?
$$
so that the final answers have no commutators?
Commutator is defined by default as
$$
[A,B]:=AB-BA
$$
 A: If you check out Kronecker Product you will see that it has the mixed-product property:
$$
(\mathbf {A} \otimes \mathbf {B} )(\mathbf {C} \otimes \mathbf {D} )=(\mathbf {AC} )\otimes (\mathbf {BD} ).
$$
Using this property and the fact that
$$
\sigma^a\sigma^b = \delta_{ab}I+i\epsilon_{abc}\sigma^c
$$
you can expand the product $(\sigma^a \otimes \sigma^c)(\sigma^b \otimes \sigma^d)$ as
\begin{align}
(\sigma^a \otimes \sigma^c)(\sigma^b \otimes \sigma^d) 
&= (\sigma^a\sigma^b)\otimes(\sigma^c\sigma^d) \\ &= (\delta_{ab}I+i\epsilon_{abe}\sigma^e)\otimes(\delta_{cd}I+i\epsilon_{cdf}\sigma^f) \\
&=\delta_{ab}\delta_{cd}I+i\epsilon_{abe}\delta_{cd}(\sigma^e\otimes I)+i\epsilon_{cdf}\delta_{ab}(I \otimes \sigma^f)-\epsilon_{abe}\epsilon_{cdf}(\sigma^e\otimes\sigma^f).
\end{align}
Since the first and last terms in this expression are symmetric when the indices $ab$ and $cd$ are permuted, the first commutator you ask for simplifies to
$$
[\sigma^a \otimes \sigma^c, \sigma^b \otimes \sigma^d] = 2i\epsilon_{abe}\delta_{cd}(\sigma^e\otimes I)+2i\epsilon_{cdf}\delta_{ab}(I \otimes \sigma^f).
$$
Note that the two terms are mutually exclusive since if $\delta_{cd}=1$, then $\epsilon_{cdf}=0$, and likewise for the pair of indices $ab$.
