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HandWiki. Root System of a Semi-simple Lie Algebra. Encyclopedia. Available online: https://encyclopedia.pub/entry/35013 (accessed on 21 June 2024).
HandWiki. Root System of a Semi-simple Lie Algebra. Encyclopedia. Available at: https://encyclopedia.pub/entry/35013. Accessed June 21, 2024.
HandWiki. "Root System of a Semi-simple Lie Algebra" Encyclopedia, https://encyclopedia.pub/entry/35013 (accessed June 21, 2024).
HandWiki. (2022, November 17). Root System of a Semi-simple Lie Algebra. In Encyclopedia. https://encyclopedia.pub/entry/35013
HandWiki. "Root System of a Semi-simple Lie Algebra." Encyclopedia. Web. 17 November, 2022.
Root System of a Semi-simple Lie Algebra

In mathematics, there is a one-to-one correspondence between reduced crystallographic root systems and semisimple Lie algebras. Here the construction of a root system of a semisimple Lie algebra – and, conversely, the construction of a semisimple Lie algebra from a reduced crystallographic root system – are shown.

crystallographic root system root systems

## 1. Associating a Root System to a Semisimple Lie Algebra

Let $\displaystyle{ \mathfrak g }$ be a complex semisimple Lie algebra. Let further $\displaystyle{ \mathfrak h }$ be a Cartan subalgebra of $\displaystyle{ \mathfrak g }$. Then $\displaystyle{ \mathfrak h }$ acts on $\displaystyle{ \mathfrak g }$ via simultaneously diagonalizable linear maps in the adjoint representation. For λ in $\displaystyle{ \mathfrak h^*, }$ define the subspace $\displaystyle{ \mathfrak g_\lambda\subset\mathfrak g }$ by

$\displaystyle{ \mathfrak{g}_\lambda := \{X\in\mathfrak{g}: [H,X]=\lambda(H)X\text{ for all }H\in\mathfrak{h}\}. }$

We say that $\displaystyle{ \lambda\in\mathfrak h^* }$ is a root if $\displaystyle{ \lambda\neq 0 }$ and the subspace $\displaystyle{ \mathfrak g_\lambda }$ is nonzero. In this case $\displaystyle{ \mathfrak g_\lambda }$ is called the root space of λ. For each root $\displaystyle{ \lambda }$, the root space $\displaystyle{ \mathfrak g_\lambda }$ is one-dimensional.[1] Meanwhile, the definition of Cartan subalgebra guarantees that $\displaystyle{ \mathfrak g_0=\mathfrak h }$.

Let R be the set of all roots. Since the elements of $\displaystyle{ \mathfrak h }$ are simultaneously diagonalizable, we have

$\displaystyle{ \mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\lambda\in R}\mathfrak{g}_\lambda. }$

The Cartan subalgebra $\displaystyle{ \mathfrak h }$ inherits a nondegenerate bilinear form from the Killing form on $\displaystyle{ \mathfrak g }$. This form induces a form on $\displaystyle{ \mathfrak h^* }$ 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]

## 2. Serre's Relations: Associating a Semisimple Lie Algebra to a Root System

### 2.1. 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

$\displaystyle{ H_\lambda,X_\lambda,Y_\lambda\text{ for }\lambda\in\Delta }$

with the Chevalley–Serre relations[3]

\displaystyle{ \begin{align}[][H_\lambda,H_\mu] &=0 \text{ for all }\lambda,\mu\in\Delta,\\ \left[H_\lambda,X_\mu\right] &= C_{\mu,\lambda}X_\mu,\\ \left[H_\lambda,Y_\mu\right] &= -C_{\mu,\lambda}Y_\mu,\\ \left[X_\mu,Y_\lambda\right] &= \delta_{\mu\lambda}H_\mu,\\ \mathrm{ad}_{X_\lambda}^{1-C_{\mu,\lambda}}(X_\mu) &= 0\text{ for }\lambda\ne\mu,\\ \mathrm{ad}_{Y_\lambda}^{1-C_{\mu,\lambda}}(Y_\mu) &= 0\text{ for }\lambda\ne\mu.\end{align} }

Here $\displaystyle{ C_{\lambda,\mu} }$ is the coefficient of the Cartan matrix, given by

$\displaystyle{ C_{\lambda,\mu}=2\frac{(\lambda,\mu)}{(\mu,\mu)} }$.

Note that if $\displaystyle{ \lambda }$ and $\displaystyle{ \mu }$ are in $\displaystyle{ \Delta }$ with $\displaystyle{ \lambda\neq\mu }$, then $\displaystyle{ (\lambda,\mu)\leq 0 }$, so that $\displaystyle{ C_{\lambda,\mu} }$ is a non-positive integer and $\displaystyle{ 1-C_{\lambda,\mu} }$ is a positive integer.

### 2.2. Finding Generators with These Relations

Now, if we are given a semisimple Lie algebra $\displaystyle{ \mathfrak g }$ with root system $\displaystyle{ R }$, it is not particularly difficult to find a set of generators for $\displaystyle{ \mathfrak g }$ satisfying the above relations.[4] For each simple root $\displaystyle{ \lambda\in\Delta }$, we can find $\displaystyle{ X_\lambda }$ in the root space $\displaystyle{ \mathfrak g_\lambda }$, $\displaystyle{ Y_\lambda }$ in the root space $\displaystyle{ \mathfrak g_{-\lambda} }$ and $\displaystyle{ H_\lambda }$ in the Cartan subalgebra satisfying the standard $\displaystyle{ \mathrm{sl}(2;\mathbb C) }$ relations: $\displaystyle{ [H_\lambda,X_\lambda]=2X_\lambda }$, $\displaystyle{ [H_\lambda,Y_\lambda]=-2Y_\lambda }$, and $\displaystyle{ [X_\lambda,Y_\lambda]=H_\lambda }$. These will be our generators.

Now, the element $\displaystyle{ H_\lambda }$ is the coroot associated to $\displaystyle{ \lambda }$, which means that after we identify $\displaystyle{ \mathfrak h }$ with its dual, we have $\displaystyle{ H_\lambda=2\lambda/(\lambda,\lambda). }$[5] Then we have, for example,

$\displaystyle{ [H_\lambda,X_\mu]=(\mu,H_\lambda)X_\mu=2\frac{(\mu,\lambda)}{(\lambda,\lambda)}X_\mu=C_{\mu,\lambda}X_\mu. }$

This sort of reasoning verifies the first four relations above.

The last two relations hold because $\displaystyle{ [X_\lambda,X_\mu] }$ belongs to the root space $\displaystyle{ \mathfrak g_{\lambda+\mu} }$, and more generally, $\displaystyle{ \mathrm{ad}_{X_\lambda}^{k}(X_\mu) }$ belongs to $\displaystyle{ \mathfrak g_{k\lambda+\mu} }$. But, as we shall see momentarily, if $\displaystyle{ \lambda }$ and $\displaystyle{ \mu }$ are simple roots, then $\displaystyle{ k\lambda+\mu }$ is not a root if $\displaystyle{ k=1-C_{\mu,\lambda} }$, so that $\displaystyle{ \mathrm{ad}_{X_\lambda}^{k}(X_\mu) }$ must be zero. To see that $\displaystyle{ k\lambda+\mu }$ is not a root, note that if $\displaystyle{ \lambda }$ and $\displaystyle{ \mu }$ are distinct elements of $\displaystyle{ \Delta }$, then $\displaystyle{ -\lambda+\mu }$ 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 $\displaystyle{ s_\lambda }$ is the reflection associated to $\displaystyle{ \lambda }$, we can easily calculate that

$\displaystyle{ s_\lambda\cdot(-\lambda+\mu)=\lambda+\mu-C_{\mu,\lambda}\lambda=k\lambda+\mu }$.

Then since $\displaystyle{ -\lambda+\mu }$ is not a root, neither is $\displaystyle{ k\lambda+\mu }$.

Note that the elements $\displaystyle{ \{X_\lambda,Y_\lambda,H_\lambda\},\,\lambda\in\Delta, }$ do not span $\displaystyle{ \mathfrak g }$ as a vector space, because $\displaystyle{ \lambda }$ does not range over all the positive roots, but only over the base. Nevertheless, these elements generate $\displaystyle{ \mathfrak g }$ as a Lie algebra.[6]

### 2.3. Serre's Theorem

Serre's theorem asserts much more than just the existence of such generators for a given semisimple Lie algebra $\displaystyle{ \mathfrak g }$. First, the claim is that the above relations completely determine $\displaystyle{ \mathfrak g }$; that is, there are no other relations in $\displaystyle{ \mathfrak g }$ 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 $\displaystyle{ R }$ we started from.[7]

A consequence of Serre's theorem is this:

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

## 3. 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 $\displaystyle{ \mathfrak g }$ is unique up to automorphism.[8] It follows that all Cartan subalgebras of $\displaystyle{ \mathfrak g }$ 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]

### References

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
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