**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. 



Calculation with weights of $E_6$
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.


