For a linear algebraic group over an algebraically closed field of characteristic zero $G$, with unipotent radical $U$, we have that $G/U$ is reductive.

When $G$ is solvable, then Lie's theorem says that irreducible representations are 1-dimensional, so the unipotent radical acts trivially, and so the irreducible representations of $G$ are in bijection with the irreducible representations of $G/U$ under the pullback functor.

I figure there must be a generalization of this to the case of general $G$. Chriss/Ginzburg proves something like this in their book using geometric methods but I feel like there should also be a more group-theoretic proof which specializes to the argument using Lie's theorem when $G$ is solvable.

It would suffice to show that on any irreducible representation of $G$, the unipotent radical acts trivially. I can't find a proof of this statement, however. Is there a quick and easy way to bootstrap Lie's theorem, or can someone point me to a reference?

Apologies if this is too easy for MO.


1 Answer 1


Let $V$ be an irreducible representation of $G$. Consider it as a representation of $U$. Since $U$ is unipotent, $V^U\neq 0$. On the other hand, since $U$ is normal in $G$, $V^U$ is a subrepresentation. But the irreducibility of $V$ implies then that $V^U=V$, which is what you wanted.

  • $\begingroup$ I decide to write the question in MO mathoverflow.net/questions/256763, I tried your approach and I get some form like $\beta=(−1)^{(f(q1))_{Z2 \times Z2} (n2)_{Z2}}$. I numerically check that it should be $(-1)^{\#}$, and the inflation trivialize the cocycle, but analytically it is hard to obtain the exponent # based on your previous answer -maybe I misunderstood. $\endgroup$
    – miss-tery
    Dec 8, 2016 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.