So I've read that the Weyl group of the $A_3$ root system is the full symmetry group of a tetrahedron $T_h$, $C_3$ is $O_h$, and $H_3$ is $I_h$. I've read some complicated proofs of this but I don't feel like I really understand why this is the case. I am hoping that the symmetric points of one of those 3d point groups can be produced by integer linear combinations of the root system (of a different sort for edge, vertex and face points), but I'm not sure if that's the case. Something like that would help connect the ideas visually for me.

The reason I'm interested is that I want to know how many symmetric points are fixed by a reflection, and I know how to find out how many roots are fixed by a reflection; that is, how many roots lie in the plane perpendicular to another root.

Thanks in advance!

  • $\begingroup$ @MarianoSuárez-Álvarez The symmetric points are the edge-centres, face-centres, and vertices. I’m wanting to know why it is that (particularly 3d) point groups are the Weyl groups of these root systems. $\endgroup$ Commented Dec 31, 2022 at 20:43
  • 1
    $\begingroup$ $H_3$ is not a Weyl group. $\endgroup$
    – JBL
    Commented Jan 1, 2023 at 3:47

1 Answer 1


I think you just mixes up root systems and Coxeter grous.

Within wikipedia you'll find the first one here: root systems, then providing esp. this graphic of associated Dynkin diagrams enter image description here which then are bound to the crystallographic restriction, i.e. allowing for angles of 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180° only (simply because $(2\cos{\theta})^2\in\mathbb{Z}$).

The second one then is found here: Coxeter groups, then esp. providing this graphic of associated Coxeter graphs enter image description here then no longer distinguishing between the directedness of the 4-fold or 6-fold link and esp. additionally allowing for a 5-fold link as well.

Now adding the latter with the according kaleidoscopic Wythoff construction of polytopes (within your case: of polyhedra) you'd derive

enter image description here

thus representing directly the required for polyhedral symmetries.

--- rk

  • $\begingroup$ Thanks Dr Klitzing. I’m not just mixing them up though, it’s explained here jstor.org/stable/26583576 that the 3 root systems in question have as Weyl groups the 3 symmetry groups in question. $\endgroup$ Commented Dec 31, 2022 at 20:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .