Shapovalov form on Verma modules

I will formulate my question first; below is the relevant background information and notation. I have asked this question on stack, but I think the odds are better that I actually get an answere here (but do correct me if I'm wrong to post it here).

WHY is the Shapovalov form on a Verma module symmetric?

The fact that it's orthogonal with respect to its weight decomposition simplifies things of course, so that we only have to calculate expressions of the form $$\langle f^{n_1}_1\ldots f^{n_l}_lv_\lambda,f^{m_1}_1\ldots f^{m_l}_lv_\lambda\rangle$$ where $$n,m\in\mathbb{Z}^l_{\ge0}$$ are such that $$\sum^l_{i=1}n_i\beta_i=\sum^l_{i=1}m_i\beta_i$$. This seems rather hard to do because one has to commute elements of root spaces of non-simple roots. In principle, any element of the form $$f^{n_1}_1\ldots f^{n_l}_l$$ for $$n\in\mathbb{Z}^l_{\ge0}$$ can be expressed as a sum of elements of the form $$f^{n_{i_1}}_{i_1}\ldots f^{n_{i_p}}_{i_p}$$, where now $$p\in\mathbb{Z}_{\ge1}$$, and $$i_1,\ldots,i_p\in\{1,\ldots,r\}$$ (not necessarily distinct!) and $$n_{i_1},\ldots,n_{i_l}\in\mathbb{Z}_{\ge1}$$. This seems hopeless as well. In general, I don't feel like such a computation-based approach can be very helpful (let alone insightful). An elementary argument would be awesome.

Background

Let $$L$$ be a simple Lie algebra (over $$\mathbb{C}$$), let $$L=L_-\oplus H\oplus L_+$$ be a root space decomposition with $$H$$ a Cartan subalgebra of $$L$$ ($$\dim(H)=r$$ and $$\dim(L_-)=\dim(L_+)=l$$), let $$\Delta=\{\alpha_1,\ldots,\alpha_r\}$$ be a set of simple positive roots, and enumerate the positive roots as $$\beta_1,\ldots,\beta_r$$ such that the first $$r$$ are the simple onese. Denote by $$B$$ the Borel subalgebra $$H\oplus L^+$$. Let $$\{e_1,\ldots,e_r,f_1,\ldots,f_r\}$$ be a set of Chevalley generators, and let $$\{e_{r+1},\ldots,e_l,f_{r+1},\ldots,f_l\}$$ be elements of the root spaces corresponding to the positive, negative roots that are not simple.

Let $$\lambda\in H^\ast$$, denote by $$\lambda^+\in B^\ast$$ its extension to $$B$$ (by declaring it to be zero on $$L_+$$), and denote by $$\mathbb{C}_\lambda$$ the one-dimensional (over $$\mathbb{C}$$) $$\mathfrak{U}(B)$$-module defined by $$xv_\lambda=\lambda^+(x)v$$ for any $$x\in B$$ and for any $$v\in\mathbb{C}_\lambda$$, where $$v$$ is a nonzero element of $$\mathbb{C}_\lambda$$. The Verma module is then the induced module $$M_\lambda:=\mathfrak{U}(L)\otimes_{\mathfrak{U}(B)}\mathbb{C}_\lambda$$. Denote $$1\otimes 1\in M_\lambda$$ by $$v_\lambda$$.

The Verma module is a direct sum of weight spaces that are finite-dimensional, i.e. $$M_\lambda=\bigoplus_{k\in\mathbb{Z}^r_{\ge0}}M^k_\lambda$$ where $$M^k_\lambda=\{v\in M_\lambda\mid\forall\,h\in H:hv=(\lambda-\sum^r_{i=1}k_i\alpha_i)(h)v\}$$ for any $$k\in\mathbb{Z}^r_{\ge0}$$. It is easy to see that, for any $$k\in\mathbb{Z}^r_{\ge0}$$, the weight space $$M^k_\lambda$$ is the linear span of $$\{f^{n_1}_1\ldots f^{n_l}_l\mid n\in\mathbb{Z}^l_{\ge0}\,\wedge\,\sum^l_{i=1}n_i\beta_i=\sum^r_{i=1}k_i\alpha_i\}.$$ The restricted dual $$M^\vee_\lambda$$ of $$M_\lambda$$ corresponding to the Chevalley involution $$\omega\colon L\to L$$ defined by $$\forall\,i\in\{1,\ldots,r\}\colon\omega(e_i)=-f_i\,\wedge\,\omega(f_i)=-e_i,$$ is as the $$\mathfrak{U}(L)$$-module whose underlying abelian group is $$\bigoplus_{k\in\mathbb{Z}^r_{\ge0}}(M^k_\lambda)^\ast$$ and where the action of $$\mathfrak{U}(L)$$ is defined for each $$k\in\mathbb{Z}^r_{\ge0}$$ by setting $$(xf)(v)=f(-\omega(x)v)$$ for all $$x\in L$$, for all $$f\in(M^k_\lambda)^\ast$$, and for each $$v\in M^k_\lambda$$. The unique vector $$v^\ast_\lambda\in(M^0)^\ast_\lambda$$ dual to $$v_\lambda$$ is an element of the weight space $$(M^\vee_\lambda)^\lambda$$ and it's annihilated by $$L_+$$, so we have a unique morphism $$\phi\colon M_\lambda\to M^\vee_\lambda$$ of left $$\mathfrak{U}(L)$$-modules such that $$\phi(v_\lambda)=v^\ast_\lambda$$. Now we define the Shapovalov form to be the bilinear form $$\langle\cdot,\cdot\rangle:M_\lambda\times M_\lambda\to\mathbb{C},(v,w)\mapsto\phi(v)(w).$$ It has the property that $$\langle M^k_\lambda,M^{k^\prime}_\lambda\rangle=\{0\}$$ for all $$k,k^\prime\in\mathbb{Z}^r_{\ge0}$$ such that $$k\neq k^\prime$$. In addition, it satisfies $$\langle v_\lambda,v_\lambda\rangle=1$$ and $$\langle xv,w\rangle=\langle v,-\omega(x)w\rangle$$ for any $$x\in L$$ and for any $$v,w\in M_\lambda$$.

• There is an argument in Jantzen "Moduln mit einem höchsten Gewicht". Certainly our form $\langle -, -\rangle$ is the sum of a symmetric and an anti-symmetric form, both of which are contravariant. However as the highest weight space is one-dimensional, the anti-symmetric part vanishes there, hence vanishes everywhere. Aug 14, 2019 at 11:08
• @GeordieWilliamson I wasn't expecting such a beautifully simple argument. This is exactly what I was looking for - should have been able to think of it myself. Thank you! Aug 14, 2019 at 15:29

1 Answer

There is alternative way to define the Shapovalov form which makes the symmetry easy to see. By the PBW theorem you can write each element $$X$$ of $$\mathfrak{U(g)}$$ as $$X = f_{i_1} \cdots f_{i_m} \cdot h_{j_1} \cdots h_{j_n} \cdot e_{k_1} \cdots e_{k_o}$$. Now denote by $$\pi: \mathfrak{U(g)} \to \mathfrak{U(h)}$$ the projection map defined by $$X \mapsto h_{j_1} \cdots h_{j_n}.$$ Then the universal Shapovalov form $$S: \mathfrak{U(g)} \otimes \mathfrak{U(g)} \to \mathfrak{U(h)}$$ is defined by $$S(X, Y) = \pi(\tau(X)\cdot Y),$$ where $$\tau: \mathfrak{U(g)} \to \mathfrak{U(g)}$$ is anti-automorphism that is identity on $$\mathfrak{U(h)}$$ and $$\tau(f_{i_1} \cdots f_{i_m}) = e_{i_m} \cdots e_{i_1}$$.

Since $$\pi(\tau(u)) = \pi(u)$$ for any $$u \in \mathfrak{U(g)}$$, the symmetry follows.

Now for any $$\lambda \in \mathfrak{h}^*$$ we can evaluate $$\lambda$$ on the image of $$\pi$$ and composing with $$S$$ we get a $$\mathbb{C}$$-valued form, call it $$S^\lambda$$. It is not hard to show that $$S^\lambda$$ defines a symmetric, $$\tau$$-contravariant form on any highest weight module of highest weight $$\lambda$$. For details see Representations of Semisimple Lie Algebras in the BGG Category O by James Humphreys.

• I have seen this approach in Kumar, but have not dived into it and was trying to avoid it and have a direct solution. Aug 14, 2019 at 15:33