Weird Invariant of Finite Reflection Groups I am looking for a proof of the following:

Let $V$ be a real vector space of dimension $n$ over $\mathbb R$ endowed with an inner product $\langle \,, \rangle$. Fix a basis $e_1, \cdots, e_n$ for $V$ and the dual basis $x_1, \cdots, x_n$ for $V^*$.
Let $G$ be a finite reflection group acting on $V$, and thus acting on the ring $\mathbb R[x_1,\cdots, x_n]$ as $g\cdot P(v)= P(g^{-1} \cdot v)$. By definition $\langle \,, \rangle$ is $G$ invariant.
Let $P(v),Q(v) \in R[x_1,\cdots, x_n]$ homogeneous and $G$-invariant polynomials, i.e. $g \cdot P(v) = P(v)$ for every $g \in G$, for every $v \in V$.
Then $$\sum_{i,j=1}^n \frac{\partial P}{\partial x_i}\frac{\partial Q}{\partial x_j}\langle e_i,e_j\rangle$$ is $G$-invariant.

Example: Let's consider the root system $\Phi=B_2$ and its associated group $G$; pick an hortonormal basis for $V=\mathbb R^2$. A basis for the ring of invariants $\mathbb R[x,y]^G$ is given by $$P(x,y)=x^2 + y^2$$ $$Q(x,y)=x^2y^2.$$
Then we compute $$\sum_{i,j=1}^n \frac{\partial P}{\partial x_i}\frac{\partial P}{\partial x_j}\langle e_i,e_j\rangle = 2x(2x) + 2y(2y) = 2(x^2 + y^2) = 2P(x,y).$$ $$\sum_{i,j=1}^n \frac{\partial P}{\partial x_i}\frac{\partial Q}{\partial x_j}\langle e_i,e_j\rangle = 2x(2xy^2) + 2y(2x^2y) = 8x^2y^2 = 8Q(x,y)$$ $$\sum_{i,j=1}^n \frac{\partial Q}{\partial x_i}\frac{\partial Q}{\partial x_j}\langle e_i,e_j\rangle = (2xy^2)^2 + (2x^2y)^2 = 4x^2y^2(x^2+y^2) = 4Q(x,y)P(x,y).$$ These are all polynomials in $P$ and $Q$, and thus invariants.
Thoughts: when the basis is orthonormal, this is just the inner product between gradients, so a geometric interpretation of this can tell us that in any fixed point $v$ the angle between the two gradients is not going to change, so it is invariant. I don't know how to produce an algebraic proof in the general setting.
 A: Let $R= \mathbb{R}[x_1,\ldots,x_n]$. Since $V$ carries a $G$-invariant inner product $\langle-,-\rangle$, the map $e_i \mapsto x_i=\langle e_i,-\rangle$ is an $G$-isomorphism $V \to V^*$, and $\langle x_i,x_j\rangle := \langle e_i,e_j\rangle$ is a $G$-invariant inner product on $V^*$.
Then the map
$$ \phi: R \to R \otimes V^*$$
given by $f \mapsto \sum \frac{\partial f}{\partial x_i}\otimes x_i$ is $G$-invariant.  Let $\phi'$ be the same map but with the order of the tensor factors swapped, so it has image in $V\otimes R$.  Now consider the following composition of $G$-homomorphisms:
$$ R\otimes R \stackrel{\phi\otimes\phi'}\to R\otimes V^* \otimes V^* \otimes R \stackrel{\operatorname{id}\otimes \langle-,-\rangle\otimes \operatorname{id}}\to R\otimes R \stackrel{\text{multiplication}}\to R.$$
Since it is a $G$-homomoprhism, the image of $P\otimes Q$ will be $G$-invariant --- but the image is your 'weird invariant.'
To see why $\phi$ is a $G$-homomorphism you need a bit of multilinear algebra.  For any $G$ and any $G$-module $M$, there's a 'contraction' map $c: S^r(M) \otimes M^* \to S^{r-1}(M)$ which sends $f\otimes m_i^*$ to $\frac{\partial f}{\partial m_i}$ and is a $G$-homomorphism --- you can read about this in Fulton and Harris Representation theory for example.  Here $S^r(M)$ is the $r$th symmetric power, so the polynomial ring on $M$ which I will call $S(M)$ is just the direct sum of the $S^r(M)$ for $r\geq 0$.  Now there is a $G$-map $\iota: S(M) \to S(M)\otimes M^* \otimes M$ sending $f \mapsto f\otimes \sum_i m_i^* \otimes m_i$, where the $m_i$ are a basis of $M$, because $\sum_i m_i^*\otimes m_i$ spans a trivial submodule of $M^*\otimes M$. Then $(c\otimes 1)\circ \iota$ is the map $\phi$.
