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.


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 $v,w\in M_\lambda$.

  • 9
    $\begingroup$ 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. $\endgroup$ Commented Aug 14, 2019 at 11:08
  • $\begingroup$ @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! $\endgroup$ Commented Aug 14, 2019 at 15:29

1 Answer 1


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.

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

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