What does Lie(SU(n) ⊗ SU(m)) looks like?

[mpham26uchicago]

$\text{Lie}(SU(n) \otimes SU(m))$ is not $\mathfrak{su}(n) \otimes \mathfrak{su}(m)$?

I think the tensor product of the Lie algbra makes the resulting exponential too large. So the quest continues...

KRONCKER SUM is the answer!!! The Kronecker sum is the matrix sum defined by

$$A \oplus B = A \otimes I_b + I_a \otimes B$$

where $A, B$ are square matrices of order $a, b$ respectively. The Kronecker sum satisfies the nice property

$$\exp(A) \otimes \exp(B) = \exp(A\oplus B)$$

So in our case, $SU(n) \otimes SU(m) = \exp(\mathfrak{su}(n)) \otimes \exp(\mathfrak{su}(n)) = \exp(\mathfrak{su}(n) \oplus \mathfrak{su}(m))$

Thus we have

$$\text{Lie}(SU(n) \otimes SU(m)) = \mathfrak{su}(n) \otimes I_m + I_n \otimes \mathfrak{su}(m)$$

[mpham26uchicago]

Solved!