I'm studying Varadarajan's Lie algebra, and I think one of the proofs uses this fact without explicitly proving it:

If $\phi:g\rightarrow gl(V)$ is a finite dimensional representation of an arbitrary finite dimensional Lie algebra over a characteristic zero field $k$, if $k'$ is the algebraic closure of $k$, then $\phi'$ may be extended to a representation over $k'$. Is it true that $\phi$ is semisimple iff $\phi'$ is semisimple? (One direction is obvious. I'm confused about the other direction)

My field theory is weak, any help would be appreciated!

(I'm aware of this post Complete reducibility of a field extension of an lie algebra representation but this post it's actually something different.)

  • $\begingroup$ This is a corollary of Cartan's criterion for semisimple Lie algebras. $\endgroup$
    – ureui
    Jun 6 at 7:06
  • 1
    $\begingroup$ @ureui Can you please elaborate on this? $\endgroup$ Jun 6 at 15:19
  • $\begingroup$ Just checking: Is your $\phi'$ defined as the natural map $g_{k'} \rightarrow gl(V_{k'})$, i.e. is it actually a representation of the scalar extension $g_{\color{red}{k'}}$? $\endgroup$ Jun 7 at 19:20
  • $\begingroup$ Yes! the natural extension. $\endgroup$ Jun 8 at 7:02
  • $\begingroup$ I assume the direction you call "obvious" is the one "upwards" from $k$ to $k'$, but I do not even perceive that as obvious ... $\endgroup$ Jun 9 at 17:21

One possibility (might be simpler approaches).

Consider $A$ to be the enveloping algebra of $\mathfrak{g}$ (it is a $k$-algebra). Then we think of $V$ as an $A$-module. But in fact, if $B$ is the quotient of $A$ by the kernel of the action, $V$ is a $B$-module and $B$ is a finite-dimensional algebra over $k$. So you in fact have a question about finite-dimensional algebras over a field. Is a finite-dimensional $B$-module $V$ semisimple if and only if $V_{k^{\prime}}$ is semisimple $B_{k^{\prime}}$-module? One possible approach is to consider the Jacobson radical, since a module is semisimple if and only if it is killed by the Jacosbon radical. Then you can ask whether $J(B_{k^{\prime}})$ is equal to $J(B)_{k^{\prime}}$ - that will suffice (here $J(-)$ is the Jacobson radical). I see, for example, in Lam, "A first course in noncommutative rings", Theorem (5.17), says it is so.

Again, might be other approaches.


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