8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Is the inner product $\langle \cdot, \cdot \rangle$ $g$-invariant as well? $\endgroup$ – Robert Lewis Apr 14 '18 at 4:07
  • 1
    $\begingroup$ @RobertLewis yes! $\endgroup$ – Maffred Apr 14 '18 at 4:26
  • $\begingroup$ The proof that this sum is $G$-invariant is just an expression of the fact that $G$-invariant functions form a ring: Any sum or product of $G$-invariant functions is $G$-invariant. For example $(g \cdot (P \cdot Q))(v)=(P \cdot Q)(g^{-1}v)=P(g^{-1}v)Q(g^{-1}v)=P(v)Q(v)=(P \cdot Q)(v)$ $\endgroup$ – leibnewtz Apr 14 '18 at 8:08
  • $\begingroup$ @leibnewtz $P_i$ and $Q_j$ are derivates! $\endgroup$ – Maffred Apr 15 '18 at 2:04
  • 1
    $\begingroup$ Edited in a line to make it more clear what your question is for improved readability. Please check to ensure I've represented it correctly. $\endgroup$ – Alexander Gruber Apr 16 '18 at 2:44
1
+100
$\begingroup$

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 $\phi$.

$\endgroup$
  • $\begingroup$ I have 3 questions: 1. are you putting an inner product on your $V$ basically? 2. Are all of those maps defined on a basis and extended up to linearity (so they are linear)? 3.Why is $\phi$ $G$-invariant? $\endgroup$ – Maffred Apr 16 '18 at 18:10
  • $\begingroup$ Sorry - I mixed up $V$ and $V^*$ here and have edited to clarify. I've added an explanation of why $\phi$ is a homomorphism. $\endgroup$ – Matthew Towers Apr 16 '18 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.