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.

  • $\begingroup$ I assume you want $M$ to be a module for $\mathfrak{g}_1 \oplus \mathfrak{g}_2$? $\endgroup$ – user44191 Mar 15 '19 at 6:59
  • $\begingroup$ @user44191 Yes, it is what I mean. $\endgroup$ – Hebe Mar 15 '19 at 8:50

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".

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  • $\begingroup$ That is great! Thank you so much! $\endgroup$ – Hebe Mar 16 '19 at 3:31

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