I think that the following statement is true, but I do not know how to prove it.


Let $\mathfrak{g}\_1$ and $\mathfrak{g}\_2$ be two real simple Lie algebras. If $M$ is a (infinite dimensional) complex simple $\mathfrak{g}\_1\oplus\mathfrak{g}\_2$-module, then $M\cong M\_1\otimes M\_2$ where $M\_1$ and $M\_2$ are simple modules of $\mathfrak{g}\_1$ and $\mathfrak{g\_2}$ respectively.


This ought to be a well known result if it is true. As is known to all, it is true for finite groups and compact groups.



Simple modules for direct sum of simple Lie algebras
This is true more generally for any two finite-dimensional Lie algebras over a field. Recall that a $\frak{g}$-module is the same as a module over $U (\frak{g})$, the universal enveloping algebra of $\frak{g}$, which is an *associative* algebra. Moreover, $U ({\frak{g}\_1\oplus\frak{g}\_2})\simeq U({\frak{g}\_1})\otimes U ({\frak{g}\_2})$, so this question reduces to showing that a simple module $M$ over the tensor product $A\_1\otimes A\_2$ decomposes as $M\_1\otimes M\_2$, where $M\_i$ is a simple module over a universal enveloping algebra $A\_i$. If my memory serves, there is proof along these lines in Dixmier, "Universal enveloping algebras". Another standard reference is McConnell and Robson, "Noncommutative Noetherian rings".


