Root system of a semi-simple Lie algebra

Associating a root system to a semisimple Lie algebra

Let be a complex semisimple Lie algebra. Let further be a Cartan subalgebra of . Then acts on via simultaneously diagonalizable linear maps in the adjoint representation. For λ in define the subspace by

We say that is a root if and the subspace is nonzero. In this case is called the root space of λ. For each root , the root space is one-dimensional.[1] Meanwhile, the definition of Cartan subalgebra guarantees that .

Let R be the set of all roots. Since the elements of are simultaneously diagonalizable, we have

The Cartan subalgebra inherits a nondegenerate bilinear form from the Killing form on . This form induces a form on and the restriction of that form to the real span of the roots is an inner product. One can show that with respect to this inner product R is a reduced crystallographic root system.[2]

Associating a semisimple Lie algebra to a root system

The relations

Let E be a Euclidean space and R a reduced crystallographic root system in E. Let moreover Δ be a choice of simple roots. We define a complex Lie algebra over the generators

with the Chevalley–Serre relations[3]

Here is the coefficient of the Cartan matrix, given by


Note that if and are in with , then , so that is a non-positive integer and is a positive integer.

Finding generators with these relations

Now, if we are given a semisimple Lie algebra with root system , it is not particularly difficult to find a set of generators for satisfying the above relations.[4] For each simple root , we can find in the root space , in the root space and in the Cartan subalgebra satisfying the standard relations: , , and . These will be our generators.

Now, the element is the coroot associated to , which means that after we identify with its dual, we have [5] Then we have, for example,

This sort of reasoning verifies the first four relations above.

The last two relations hold because belongs to the root space , and more generally, belongs to . But, as we shall see momentarily, if and are simple roots, then is not a root if , so that must be zero. To see that is not a root, note that if and are distinct elements of , then cannot be a root, for this would violate one of the defining properties of a base—that the expansion of a root in terms of the base cannot have both positive and negative coefficients. But then if is the reflection associated to , we can easily calculate that


Then since is not a root, neither is .

Note that the elements do not span as a vector space, because does not range over all the positive roots, but only over the base. Nevertheless, these elements generate as a Lie algebra.[6]

Serre's theorem

Serre's theorem asserts much more than just the existence of such generators for a given semisimple Lie algebra . First, the claim is that the above relations completely determine ; that is, there are no other relations in besides ones that follow from these. Even more, however, Serre's theorem asserts that starting from an arbitrary root system—not assumed to come from a semisimple Lie algebra—we can use the above relations to define a Lie algebra, the Lie algebra is finite-dimensional and semisimple, and the root system of that Lie algebra is the root system we started from.[7]

A consequence of Serre's theorem is this:

  • Every (reduced, crystallographic) root system comes from a semisimple Lie algebra.

Application to the classification of semisimple Lie algebras

The preceding results help in the process of reducing the classification of semisimple Lie algebras to the task of classifying reduced crystallographic root systems, which is then done in terms of Dynkin diagrams. Although the above describes how to construct a root system from a Lie algebra and vice versa, there are still two things to prove before we can obtain a one-to-one correspondence.

  • First, we need to know that each Lie algebra gives only one root system (up to isomorphism). This is shown by proving that the Cartan subalgebra of a semisimple Lie algebra is unique up to automorphism.[8] It follows that all Cartan subalgebras of give isomorphic root systems.
  • Second, we must prove that each root system gives only one Lie algebra (up to isomorphism). That is, we must show that if two Lie algebras have isomorphic root systems, the Lie algebras are also isomorphic.[9]

See also


  1. Hall 2015 Theorem 7.23
  2. Hall 2015 Theorem 7.30
  3. Humphreys 1973 Section 18.1
  4. Humphreys 1973 Proposition 18.1
  5. Hall 2015 Equation (7.9)
  6. Humphreys 1973 Section 18.3
  7. Humphreys 1973 Section 18.3
  8. Humphreys 1973 Corollary 16.4
  9. Humphreys 1973 Theorem 14.2


This article incorporates material from Root system underlying a semi-simple Lie algebra on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.

  • Hall, Brian C. (2015), Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Graduate Texts in Mathematics, 222 (2nd ed.), Springer, ISBN 978-3319134666
  • Humphreys, James (1973), Introduction to Lie Algebras and Representation Theory, Graduate Texts in Mathematics, 9, Springer
  • V.S. Varadarajan, Lie groups, Lie algebras, and their representations, GTM, Springer 1984.
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