The Unapologetic Mathematician

Mathematics for the interested outsider

The Automorphism Group of a Root System

Finally, we’re able to determine the automorphism group of our root systems. That is, given an object in the category of root systems, the morphisms from that root system back to itself (as usual) form a group, and it’s interesting to study the structure of this group.

First of all, right when we first talked about the category of root systems, we saw that the Weyl group \mathcal{W} of \Phi is a normal subgroup of \mathrm{Aut}(\Phi). This will give us most of the structure we need, but there may be automorphisms of \Phi that don’t come from actions of the Weyl group.

So fix a base \Delta of \Phi, and consider the collection \Gamma of automorphisms which send \Delta back to itself. We’ve shown that the action of \mathcal{W} on bases of \Phi is simply transitive, which means that if \tau\in\Gamma comes from the Weyl group, then \tau can only be the identity transformation. That is, \Gamma\cap\mathcal{W}=\{1\} as subgroups of \mathrm{Aut}(\Phi).

On the other hand, given an arbitrary automorphism \tau\in\mathrm{Aut}(\Phi), it sends \Delta to some other base \Delta'. We can find a \sigma\in\mathcal{W} sending \Delta' back to \Delta. And so \sigma\tau\in\Gamma; it’s an automorphism sending \Delta to itself. That is, \tau\in\mathcal{W}\Gamma; any automorphism can be written (not necessarily uniquely) as the composition of one from \Gamma and one from \mathcal{W}. Therefore we can write the automorphism group as the semidirect product:


All that remains, then, is to determine the structure of \Gamma. But each \tau\in\Gamma shuffles around the roots in \Delta, and these roots correspond to the vertices of the Dynkin diagram of the root system. And for \tau to be an automorphism of \Phi, it must preserve the Cartan integers, and thus the numbers of edges between any pair of vertices in the Dynkin diagram. That is, \Gamma must be a transformation of the Dynkin diagram of \Phi back to itself, and the reverse is also true.

So we can determine \Gamma just by looking at the Dynkin diagram! Let’s see what this looks like for the connected diagrams in the classification theorem, since disconnected diagrams just add transformations that shuffle isomorphic pieces.

Any diagram with a multiple edge — G_2, F_4, and the B_n and C_n series — has only the trivial symmetry. Indeed, the multiple edge has a direction, and it must be sent back to itself with the same direction. It’s easy to see that this specifies where every other part of the diagram must go.

The diagram A_1 is a single vertex, and has no nontrivial symmetries either. But the diagram A_n for n\geq2 can be flipped end-over-end. We thus find that \Gamma=\mathbb{Z}_2 for all these diagrams. The diagram E_6 can also be flipped end-over-end, leaving the one “side” vertex fixed, and we again find \Gamma=\mathbb{Z}_2, but E_7 and E_8 have no nontrivial symmetries.

There is a symmetry of the D_n diagram that swaps the two “tails”, so \Gamma=\mathbb{Z}_2 for n\geq5. For n=4, something entirely more interesting happens. Now the “body” of the diagram also has length {1}, and we can shuffle it around just like the “tails”. And so for D_4 we find \Gamma=S_3 — the group of permutations of these three vertices. This “triality” shows up in all sorts of interesting applications that connect back to Dynkin diagrams and root systems.

March 11, 2010 Posted by | Geometry, Root Systems | 1 Comment