MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a Lie Group and $\mathfrak{g}$ be its lie algebra. Let $\mathfrak{g}$ is semisimple or reductive lie algebra, then prove that $\mathfrak{g}^*$ (dual of $\mathfrak{g}$)is invariant under $Ad(G)$?

share|cite|improve this question
What is $g$ and what is $g^*$? This looks like an exercise, no – Mariano Suárez-Alvarez Sep 8 '13 at 10:54
Dear Mariano Suárez-Alvarez@ I found this fact without proof in the book of orbit method of kirillov. If you think it is exercise, I can remove it – baba ab dad Sep 8 '13 at 11:04
Not sure what the question is. The group acts on its Lie algebra by the adjoint action. That induces a natural dual action on the dual of the Lie algebra. – Peter Dalakov Sep 8 '13 at 11:22
Dear Peter Dalakov@ please see page 3 , remark 1,… – baba ab dad Sep 8 '13 at 11:32
It is impossible to understand what you are asking from what you wrote, and the remark you point in that book has very little to do with what you wrote! (Kirillov is talking about a $V$ which you do not mention at all...) – Mariano Suárez-Alvarez Sep 8 '13 at 19:05
up vote 3 down vote accepted

The question you want to ask, in order to understand remark 1, page 3 of Kirillov's book, is why, if $\mathfrak{g}$ is a reductive or semisimple Lie group, for every representation of $\mathfrak{g}$, every $\mathfrak{g}$-invariant subspace has a $\mathfrak{g}$-invariant complement. In particular, Kirillov is assuming that some reductive or semisimple Lie algebra $\mathfrak{g}$ is faithfully represented in vector space $W=\mathbb{R}^{n \times n}$ of matrices. Then $\mathfrak{g} \to W$ is a $\mathfrak{g}$-equivariant linear injection. The dual map $W^* \to \mathfrak{g}^*$ is therefore a surjection, and has some kernel $\mathfrak{g}^{\perp}$. If $\mathfrak{g}^{\perp}$ has an invariant complement in $W$, we can identify that complement with $\mathfrak{g}^*$ by the projection $W \to \mathfrak{g}^*$. There are various definitions of reductive Lie algebras, but one of them is that a reductive Lie algebra is one all of whose finite dimensional representations split into a sum of irreducible representations. It is a theorem that every semisimple Lie algebra has this property. Therefore if $\mathfrak{g}$ is semisimple, or (more generally) reductive, then there is such a complement, which identifies $\mathfrak{g}^*$ with a subspace of $W^*$. The Killing form identifies $W$ with $W^*$, so thereby identifies $\mathfrak{g}^*$ with a linear subspace of matrices.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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