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?

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    $\begingroup$ Can you perhaps tell us which article/book you got the equation from? And why is the explanation there not satisfactory? $\endgroup$ Feb 19, 2021 at 13:44
  • $\begingroup$ 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. $\endgroup$
    – tudong
    Feb 19, 2021 at 13:58
  • $\begingroup$ Calling the form invariant means $g$-invariant, in the sense that $\operatorname{ad}$ is skew-symmetric with respect to it? $\endgroup$
    – LSpice
    Feb 19, 2021 at 14:35
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    $\begingroup$ The paper @tudong referenced: Kac and Peterson - Infinite flag varieties and conjugacy theorems. $\endgroup$
    – LSpice
    Feb 19, 2021 at 14:36
  • $\begingroup$ Yes, the form invariant means $(ad x \cdot y|z)+(y,adx\cdot z)=0$. $\endgroup$
    – tudong
    Feb 20, 2021 at 0:38

1 Answer 1


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 $(\Lambda | \Lambda)(v\otimes v)$.

  • $\begingroup$ 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 $\{u^j\}$. $\endgroup$
    – tudong
    Feb 24, 2021 at 0:05
  • $\begingroup$ @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. $\endgroup$
    – SamJeralds
    Feb 24, 2021 at 0:46

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