Question: Consider the complex simple Lie group $E_6$. Let $\lambda_1$ and $\lambda_6$ be the fundamental weights defining the $27$-dimensional representation $V$ and $V^*$, resp. Consider the complex span $\mathfrak a^*=\langle \lambda_1,\lambda_6\rangle$. The subquotient of the Weyl group $W=W(E_6)$ that acts on $\mathfrak a^*$ is $\Gamma:=N_W(\mathfrak a^*)/Z_W(\mathfrak a^*)=A_2$.

If two elements in $\mathfrak a^*$ are $W$-conjugate, are they $\Gamma$-conjugate? Equivalently, does $$W\cdot\lambda\cap\mathfrak a^*=\Gamma\cdot\lambda$$ hold for every $\lambda\in\mathfrak a^*$?

Motivation: If true, it would imply that the all invariants of $V\oplus V^*$ as a representation of $\mathbb C^\times\times E_6$ would be pull-backs of coadjoint invariants under the moment map.

Some details: The roots of $E_6$ are (we use Bourbaki conventions) $$ \pm\epsilon_i\pm\epsilon_j\quad1\leq i<j\leq5\quad\mbox{(for $Spin(10)$)} $$ and $$ \frac12(\epsilon_8-\epsilon_7-\epsilon_6+\sum_{i=1}^5\nu_i\epsilon_i)\quad\Pi_{i=1}^5\nu_i=1\quad\mbox{(for the half-spin representation)}. $$ The $27$-dimensional irreducible representation $V$ of $E_6$ and its contragredient $V^*$ have respectively highest weights $$\lambda_1 = \frac23\Omega\quad\mbox{and}\quad\lambda_6=\frac13\Omega+\epsilon_5 $$ where $\Omega=\epsilon_8-\epsilon_7-\epsilon_6$. These are minuscule weights, which means that the Weyl group $W$ acts transitively on the weights of $V$ (resp. $V^*$); in particular they all have multiplicity one. The weights of $V$ can be listed as $$ \frac23\Omega,\ \frac16\Omega-\frac12\sum_{i=1}^5\nu_i\epsilon_i\ (\Pi_{i=1}^5\nu_i=1), -\frac13\Omega\pm\epsilon_i\ (1\leq i\leq5).$$ In particular, $$ \lambda_1,\ -\lambda_6,\ -\lambda_1+\lambda_6 $$ are weights of $V$; the longest element in $W$ maps the highest weight $\lambda_1$ and the lowest weight $-\lambda_6$ one to the other and thus fixes $-\lambda_1+\lambda_6$. One can indeed find elements in $W$ permuting $\lambda_1$, $-\lambda_6$, $-\lambda_1+\lambda_6$. The angle between $\lambda_1$ and $\lambda_6$ is $60^o$. Now it is not difficult to see that $\Gamma$ is the Coxeter group of type $A_2$.

Edit: Here is a naive argument which appears to give an answer. Suppose $w\xi_1=\xi_2$ where $w\in W$ and $\xi_1$, $\xi_2\in\mathfrak a^*$. Since the action of $W$ preserves the real span $\mathfrak a^*_{\mathbb R}$ of $\lambda_1$, $\lambda_6$, we may assume $\xi_1$, $\xi_2\in\mathfrak a^*_{\mathbb R}$. Since $\Gamma$ is a subquotient of $W$ and the action of $\Gamma$ on $\mathfrak a^*$ is generated by reflections on the real lines through $\lambda_1$, $\lambda_6$, we may replace $\xi_1$ and $\xi_2$ by suitable $\Gamma$-conjugates and assume both are linear combinations of $\lambda_1$, $\lambda_6$ with non-negative coefficients. Now $\lambda_1$, $\lambda_6$ are fundamental weights of $E_6$, so this implies that $\xi_1$, $\xi_2$ belong to the positive Weyl chamber of $E_6$ and thus $w=1$.

Update: The argument in the first edit seems OK, but about the above motivation, it is not true that it implies that the invariants are all pull-backs of coadjoint invariants.

  • $\begingroup$ The dot sometimes denotes the affine action, i.e. $w\cdot \alpha = w(\alpha+\rho)-\rho$. I guess this is not the case here. $\endgroup$ – Vít Tuček Jul 11 '14 at 6:31
  • $\begingroup$ @Vit: No, it is not. $\endgroup$ – Claudio Gorodski Jul 11 '14 at 10:36
  • $\begingroup$ @Claudio: Can you clarify your notation $A_2$? $\endgroup$ – Jim Humphreys Jul 11 '14 at 13:34
  • $\begingroup$ @Jim: I meant the Coxeter group of type $A_2$. $\endgroup$ – Claudio Gorodski Jul 11 '14 at 15:27
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    $\begingroup$ @Claudio: In your added paragraph, what do you mean by saying $\Gamma \subset W$? Some points are still a little out of focus, though what you are aiming at is probably OK. $\endgroup$ – Jim Humphreys Jul 19 '14 at 21:36

The answer is yes. Let $\lambda,\mu\in{\mathfrak a}^*$ and $w\in W$ such that $w(\lambda)=\mu$. Then we want to show that there exists $w'\in Z_W(\mu)$ such that $w'w({\mathfrak a}^*)={\mathfrak a}^*$.

We can think of ${\mathfrak a}^*$ as a subalgebra of ${\mathfrak g}$ (via the Killing form restricted to the Cartan subalgebra). Then the centralizer of ${\mathfrak a}^*$ in ${\mathfrak g}$ is just the standard type $D_4$ Levi subalgebra ${\mathfrak l}$ with simple roots $\alpha_2,\alpha_3,\alpha_4,\alpha_5$. Thus the centralizer of ${\mathfrak a}^*$ in $W$ contains the corresponding type $D_4$ parabolic subgroup. Let $\Phi_0$ be the root subsystem of $\Phi$ spanned by $\alpha_2,\alpha_3,\alpha_4,\alpha_5$. Now the normalizer of ${\mathfrak a}^*$ in $W$ is clearly equal to the normalizer of $\Phi_0$ in $W$.

The set $\Phi^\mu$ of roots $\alpha\in\Phi$ such that $\langle \mu,\alpha\rangle=0$ is a subsystem of $\Phi$ which contains $\Phi_0$, so is either all of $\Phi$, is $\Phi_0$, or is a subsystem of type $D_5$ containing $\Phi_0$. In the first and the last case the existence of a $w'$ as above is clear. But any two subsystems of type $D_4$ of a root system of type $D_5$ are $W(D_5)$-conjugate, so in the last case we can also find $w'\in W(\Phi^\mu)$ such that $w'(w(\Phi_0))=\Phi_0$, and we are done.

I'm sure there's a much neater way of showing this.

I should point out that the representation $V\oplus V^*$ of the group ${\mathbb C}^\times\times E_6$ is well-understood, as it arises from an involution on an $E_7$ Lie algebra. In particular, the invariants form a polynomial ring with generators of degree 2, 4 and 6.


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