MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathfrak{g}$ be a quadratic (finite dimensional) lie algebra and $\rho:\mathfrak{g}\rightarrow \mathfrak{gl}(W)$ be an anti-symmetric representation of $\mathfrak{g}$ on a finite dimensional inner product space $W$. Moreover we have: $$\forall \xi \in \mathfrak{g}, \forall u,v\in W\qquad <\rho(\xi).u,v>=-<\rho(\xi).v),u>.$$ For any arbitrary basis $\{\xi_i\}_{i=1}^{n}$ of $\mathfrak{g}$, define a linear map $H:W\rightarrow W$, with $H(u)=\sum_{i=1}^{n}\rho(\xi_i).(\rho(\xi_i).u))$. Then $H$ is a well-defined symmetric linear transformation on $W$.

I need to know that:

What can we say about eigenvalues of $H$ and their relation with the properties of representation $\rho$? Are there some geometric interpretations for the eigenvalues?

share|cite|improve this question
The inner-product vector space $W$ is a geometric space. And I think the eigenvalues have some special geometric meaning. – Ramandi Feb 23 '14 at 12:46
Thank you Dear Ali Taghavi,من چی را درست نوشته بودم پس؟ – Ramandi Feb 23 '14 at 13:57
If $\mathfrak g$ is reductive and $\{\xi_i\}$ is orthonormal wrt $\mathrm{ad}$-invariant inner product, $H$ is the image of the Casimir element of $\mathfrak g$ under $\rho$ ( The Casimir element is in the center of the universal enveloping algebra. It follows from Schur's lemma that in case your representation is irreducible, $H$ acts as a scalar on $W$. – Claudio Gorodski Feb 23 '14 at 14:02
Fantastic, I'm Thankful DearClaudio Gorodoski – Ramandi Feb 23 '14 at 15:04

Your Answer


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

Browse other questions tagged or ask your own question.