# Action of the Casimir on highest weight modules for Kac-Moody algebra

Let $$g$$ be a Kac-Moody algebra with a symmetrizable Cartan matrix, and let $$\{u_j\}$$ and $$\{u^j\}$$ be bases of $$g$$ dual with respect to a nondegenerate invariant bilinear form $$(\cdot|\cdot)$$ on $$g$$, and consistent with the triangular decomposition of $$g$$. Let $$L(\Lambda)$$ be an integrable representation of $$g$$ with highest weight $$\Lambda$$, and let $$v_\Lambda$$ be its highest weight vector. Denote the Casimir $$\Omega=\sum_j u_j\otimes u^j$$.

Question. Why $$\Omega(v_\Lambda\otimes v_\Lambda)=(\Lambda|\Lambda)v_\Lambda\otimes v_\Lambda$$? Could someone give some explanation or some references?

• Can you perhaps tell us which article/book you got the equation from? And why is the explanation there not satisfactory? Feb 19 '21 at 13:44
• Thank you for your response. The relation is correct. But I want to know the reason, since i am just a beginer in Lie algebra. The relation is just Lemma 3 in D. H. Peterson and V. G. Kac, Infinite flag varieties and conjugacy theorems, Proc, Natl. Acad. Sci, USA, Vol. 80,1983, pp. 1778-1782. Feb 19 '21 at 13:58
• Calling the form invariant means $g$-invariant, in the sense that $\operatorname{ad}$ is skew-symmetric with respect to it? Feb 19 '21 at 14:35
• The paper @tudong referenced: Kac and Peterson - Infinite flag varieties and conjugacy theorems. Feb 19 '21 at 14:36
• Yes, the form invariant means $(ad x \cdot y|z)+(y,adx\cdot z)=0$. Feb 20 '21 at 0:38

You should be a bit careful, as this isn't precisely the action of the Casimir on $$v \otimes v$$, but instead follows from it.
For each positive root $$\alpha$$, let $$e_\alpha^{(1)}, \dots, e_\alpha^{(n_\alpha)}$$ be a basis of the root space $$\mathfrak{g}_\alpha$$, and let $$\{f_\alpha^{(i)}\}$$ be the corresponding dual space for $$\mathfrak{g}_{-\alpha}$$. Define $$\Omega_\alpha:= \sum_{i=1}^{n_\alpha} f_\alpha^i e_\alpha^i$$. Also, let $$\{h_i, h^i\}$$ be a dual basis for $$\mathfrak{h}$$, and set $$\Omega_0:= \sum_i h_ih^i$$. Then the Casimir element is given by $$\Omega= 2\nu^{-1}(\rho)+\Omega_0 +2\sum_{\alpha \in \Delta^+} \Omega_\alpha$$ where $$\rho$$ is a Weyl vector defined by $$\rho(\alpha_i^\vee)=1$$ for all simple coroots $$\alpha_i^\vee$$, and $$\nu: \mathfrak{h} \to \mathfrak{h}^\ast$$ is the isomorphism determined by the nondegenerate bilinear form.
A key fact about the Casimir operator is that, for any $$v \in L(\mu)$$ for any dominant weight $$\mu$$, we have that $$\Omega(v)=(\mu | \mu+2\rho)v$$. Now, by just expanding out over the tensor product using the above definition of $$\Omega$$, we get for any $$v \in L(\Lambda)$$ $$\Omega(v \otimes v) = (\Omega(v)) \otimes v + v \otimes (\Omega(v)) +2 \sum_{\alpha \in \Delta \sqcup \{0\}} \sum_{i=1}^{n_\alpha} e_\alpha^i(v) \otimes f_\alpha^i(v).$$
Now, if $$v \in G(v_\Lambda)$$ is a vector in the $$G$$-orbit of $$v_\Lambda$$ (one such example is precisely $$v_\Lambda$$), we have $$v \otimes v \in L(2\Lambda)$$. Then the left-hand side is precisely $$(2\Lambda | 2\Lambda +2\rho) (v\otimes v)$$. The right-hand side is similarly $$2(\Lambda | \Lambda+2\rho)(v \otimes v) + 2 (\text{the term you are interested in}).$$
Solving for the term you want gives precisely $$(\Lambda | \Lambda)(v\otimes v)$$.
• Thanks for your explanation. Could you say something more about the relation $\Omega(v\otimes v)=(\Omega(v))\otimes v+v\otimes (\Omega(v))+2\sum_{\alpha\in\Delta^+\cup\{0\}}\sum_{i=1}^{n_\alpha}e_\alpha^i(v)\otimes f_\alpha^i(v)$. I am confused in this relation. And why it is $\Delta^+$? It seems to me it will be $\Delta$. In this case, $e_\alpha^i$ and $f_\alpha^i$ are the dual basis of $g$, just as $\{u_j\}$ and $\{u^j\}$. Feb 24 '21 at 0:05
• @tudong You are correct, the summation in the final term of $\Omega(v \otimes v)$ should have been over $\Delta$ and not $\Delta^+$; this has been edited. And yes, these are the dual basis, with the convention that $e_{-\alpha}=f_{\alpha}$. To see why the Casimir distributes the way that it does, you can work out for any $e,f$ in $\mathfrak{g}$ that $ef(v \otimes v) = e(fv \otimes v + v \otimes fv)=efv \otimes v +fv \otimes ev +ev \otimes fv +v \otimes efv$; doing this for each term and appropriately handling the scalar action coming from the basis of $\mathfrak{h}$ will give it to you. Feb 24 '21 at 0:46