I am to show that for $ G \subset GL(n,\mathbb{C})$ a classical group, $(\cdot, \cdot)$ the usual inner product on $\Phi$ the root system of $G$, and $(\alpha, \alpha) = (\beta, \beta)$, there exists $w \in W$ for which $w \cdot \alpha = \beta$.

What I am confused about is that since the Weyl group permutes the set of roots, isn't this conclusion already true, without having to impose the length equality condition $(\alpha, \alpha) = (\beta, \beta)$? I know that the root systems for type $A$ and $D$ are simply laced so roots have the same length, whereas the roots of type $C$ and $D$ have roots of either length 2 and 4 or 1 and 2 respectively (this is actually the exercise preceding the one of my question), but I don't see how this could be relevant due to I bolded before - I'm not sure where I'm misunderstanding the theory.

The only other approach I can think of is to use Weyl chambers (I have seen the proof of the result that any root can be sent to a unique corresponding element in the Weyl chamber via an element of $W$), but in the assumptions it doesn't seem necessary to choose a set $\Phi^+$ of positive roots.

  • 1
    $\begingroup$ Naively, the sentence "the Weyl group permutes the set of roots" could mean that any root is taken to any other root by the Weyl group, i.e. the action of the Weyl group on the roots is transitive. But this is plainly untrue when roots have unequal length: since the inner product is $W$-invariant, $W$ preserves the length of roots. $\endgroup$ – Joppy Feb 23 at 6:37
  • $\begingroup$ Two followup questions: 1) The inner product I am using is that $(\varepsilon_i, \varepsilon_j) = \delta_{ij}$, but I am struggling to show that it is $W$-invariant - it doesn't seem to be the same inner product as the one on the roots $(\alpha_i, \alpha_j)$. 2) Do you have a hint for how I should show the property using my previous working? $\endgroup$ – mizh Feb 23 at 6:49
  • $\begingroup$ I guess it depends on what definitions you're starting from. If the reflection in $\alpha$ is defined by $s_\alpha(\lambda) = \lambda - 2 \alpha (\lambda, \alpha) / (\alpha, \alpha)$, then you can show directly that $(s_\alpha \lambda, s_\alpha \mu) = (\lambda, \mu)$ and hence $s_\alpha$ is an isometry. $\endgroup$ – Joppy Feb 23 at 10:19
  • $\begingroup$ To be concrete, look at the root system for $B_2$ (symmetry group of a square). The long roots (on the diagonals of the square) belong to one orbit and the short roots (on the bisectors of the sides of the square) belong to another orbit. $\endgroup$ – David Hill Feb 23 at 22:45
  • $\begingroup$ The inner product you are using is correct, but the roots are not the coordinate vectors. In type $A$, the simple roots are of the form $\alpha_i=\epsilon_i-\epsilon_{i+1}$. You can find a table of root systems in an appendix of Bourbaki IV. $\endgroup$ – David Hill Feb 23 at 22:48

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