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I know that the Weyl group of a root system is a subgroup of its isometry group, but (as in the case of $A_2$) it isn't always the whole isometry group. Why isn't the Weyl group defined as the isometry group?

mathlander
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1 Answers1

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Indeed, negation of the roots is not always an element of the Weyl group.

For me the most natural reason to restrict ourselves to the group generated by reflections corresponding to the roots comes from representation theory. The formal characters of finite dimensional simple modules are invariant under the Weyl group, but not always under negation of weights. When you negate the weights of the formal character of a f.d. simple module $V$, you get the formal character of the dual module $V^*$. We don't always have $V\simeq V^*$. Basically because sometimes there exists weights $\lambda$ such that the difference $\lambda-(-\lambda)$ is not in the root lattice, and hence both $\lambda$ and $-\lambda$ cannot appear in the same simple module.

It may have also played a role that according to the standard definition the Weyl group is always a Coxeter group. I'm afraid I don't know which appeared earlier in the development of mathematics. The nice theory of Coxeter groups makes a number of arguments simpler, but I don't know which is an extension/special case of the other historically.

Jyrki Lahtonen
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    Maybe it should also be mentioned that modulo some assumptions and identifications, the difference (or rather quotient) between the full isometry group and the Weyl group "is" the difference (or rather quotient) of the full automorphism group of a Lie Algebra / Group and its inner automorphisms. Or, even more vague but catchy: The Weyl group is something like the "inner" automorphisms of the root system. – Torsten Schoeneberg Nov 29 '22 at 17:59
  • Agree with Torsten. May be the automorphisms coming from graph automorphisms of the Dynkin diagram can similarly be thought of as "outer" automorphisms :-) – Jyrki Lahtonen Nov 30 '22 at 09:16