# Eigenvalues of Cartan subalgebra and Casimir Operator

• As I understand it, given a compact, semi-simple lie-algbera $\mathfrak{g}$, there exists a basis for $\mathfrak{g}$ such that the the components of the killing form $\kappa$ are $\kappa_{\alpha\beta} = \delta_{\alpha \beta}$.

• You can also express a semisimple $\mathfrak{g}$ in terms of the Cartan-Weyl basis: $$\mathcal{B}_{\text{Cartan-Weyl}} = \{H_{i}: \; i =1...d-r \}\cup \{ E_{\alpha} \; \alpha = 1...r\}$$ Where $\{H_{i}\}$ is the basis of the cartan sub-algebra $\mathfrak{h} \subset \mathfrak{g}$ which simultaneously diagonalizes $ad_{h} \; \forall h \in \mathfrak{h}$, and $E_{\alpha}$ the associated step operators.

• Are these two bases necessarily the same?

I was asked to show that given a Casimir operator $C^{(2)} = g_{\alpha\beta}T^{\alpha}T^{\beta}$ on $\mathfrak{g}$,that the expectation value, $c^{(2)}$ of the operator in a hermitian rep'n was greater than $\sum_{i=1}^{d-r} h_{i}^{2}$, where $h_{i}$ is the eigenvalue of $H_{i}$. Apparently the solution goes as:

$$\langle \psi| \rho(T^{\alpha})^{2} |\psi \rangle = \langle \psi| \sum_{i=1}^{d-r}\rho(H_{i})^{2} + \sum_{\alpha=1}^{k}\rho(E_{\alpha})^{2} |\psi\rangle$$

But this obviously assumes that the basis in which $\kappa_{\alpha\beta} = \delta_{\alpha\beta}$ is the Cartan-Weyl basis. Is this generally true?

• What is $T^{\alpha}$ and $\rho()$? – Frederic Thomas Apr 16 '18 at 15:55
• $T^{\alpha}$ is a generator of the LA, $\rho$ is a representation, $\rho: \mathfrak{g} \rightarrow End(V)$ into some vector space $V$ – thesundayscientist Apr 17 '18 at 13:11
• I guess, once chosen the basis in a way that $\kappa_{\alpha\beta}=\delta_{\alpha\beta}$, the inequality can be proven, it's fine, as Casimir operators are invariant, it's also valid in another basis. And with the 2. equation you posted you have it is already "proven", as a physicist I would be happy with it. If you want to know it well proven, actually I would post it in the Math Stack exchange. Anyway, I think that this type of question should be posted there ^_ ^. – Frederic Thomas Apr 17 '18 at 14:20