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 $v,w\in M\_\lambda$.



Shapovalov form on Verma modules
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](https://bookstore.ams.org/gsm-94/) by James Humphreys. 


