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Diagonal Cartan subalgebra and roots of ${s l}_{n}$

Example

For $n \geq 2$ let ${s l}_{n} (C) = {X \in M_{n} (C) : tr X = 0}$ be the Lie algebra of traceless complex $n \times n$ matrices under the commutator, so that $n = 2$ recovers The special linear Lie algebra ${s l}_{2}$ . Let $h = {diag (x_{1}, \dots, x_{n}) : \sum_{i} x_{i} = 0}$ be the diagonal traceless subalgebra, and let $ε_{i} \in h^{*}$ be the restriction of the coordinate functional $H \mapsto x_{i}$ . Then $h$ is a Cartan subalgebra of ${s l}_{n} (C)$ , the roots are the functionals $ε_{i} - ε_{j}$ with $i \neq j$ , and the corresponding root spaces are the lines $g_{ε_{i} - ε_{j}} = C E_{i j},$ so $Φ = {ε_{i} - ε_{j} : i \neq j}$ has $n (n - 1)$ elements.

Facts & Assumptions

Given: The integers $n \geq 2$ , the Lie algebra ${s l}_{n} (C)$ of traceless matrices under the commutator, its diagonal traceless subalgebra $h$ , and the matrix units $E_{i j}$ ; root spaces are those of Root and root space, and nilpotence and normalizers are those of Cartan subalgebra and Normalizer of a Lie subalgebra.

[L1]

The Killing form of ${s l}_{n} (C)$ is $K (X, Y) = 2 n tr (X Y)$ and is nondegenerate for $n \geq 2$ ; a finite-dimensional characteristic-zero Lie algebra is semisimple if and only if its Killing form is nondegenerate (Classical simple Lie algebras and their Killing forms, Cartan's semisimplicity criterion).

Verification

technique · direct

1.1L1

By [L1], ${s l}_{n} (C)$ is semisimple.

1.2givenalgebra

$h$ is a Cartan subalgebra: it is abelian, hence nilpotent, and its normalizer is itself. Indeed if $X = \sum_{a b} x_{a b} E_{a b}$ satisfies $[X, H] \in h$ for every diagonal traceless $H$ , take $H = diag (1, 2, \dots, n) - \frac{n + 1}{2} I$ , whose diagonal entries are pairwise distinct; then $[X, H] = \sum_{a b} (h_{b} - h_{a}) x_{a b} E_{a b}$ has no off-diagonal component, so $(h_{b} - h_{a}) x_{a b} = 0$ and hence $x_{a b} = 0$ whenever $a \neq b$ . Thus $X$ is diagonal, and being in ${s l}_{n}$ it lies in $h$ .

1.3givenalgebra

For $H = diag (x_{1}, \dots, x_{n}) \in h$ and a matrix unit $E_{i j}$ one computes $[H, E_{i j}] = (x_{i} - x_{j}) E_{i j}$ ; note $x_{i} - x_{j}$ depends only on $H$ , so the functional $ε_{i} - ε_{j}$ on $h$ is well defined and $E_{i j}$ is a nonzero eigenvector for the eigenvalue $(ε_{i} - ε_{j}) (H)$ .

2.1givenstep 1.1step 1.2step 1.3algebra∎

By steps 1.1 and 1.2, the root-space decomposition of Root-space decomposition applies. Step 1.3 exhibits, for every pair $i \neq j$ , the nonzero vector $E_{i j} \in g_{ε_{i} - ε_{j}}$ ; conversely every simultaneous $h$ -eigenvector is a linear combination of those $E_{i j}$ whose indices give that functional, and the functionals $ε_{i} - ε_{j}$ for distinct ordered pairs are distinct while the diagonal matrices give the zero weight. Hence the roots are exactly the $n (n - 1)$ functionals $ε_{i} - ε_{j}$ , $i \neq j$ , with one-dimensional root spaces $C E_{i j}$ .

Depends on

Used by

Flag line bundles for SL(2) Example
Regular and singular diagonal elements of slₙ Example
Restricted roots of sl n r Example
Root strings in type A₂ Example
Root-space brackets for matrix units Example
The Killing form identifies roots with coroot directions Example
The root sl₂ triple inside slₙ Example
Two minimal-parabolic projections for SL(3) Example
Vogan diagrams for real forms of sl three c Example
Dynkin diagrams classify real semisimple Lie algebras False statement
The same Dynkin diagram forces isomorphic connected Lie groups False statement

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Sources

Pavel Etingof, MIT 18.745 Lie Groups and Lie Algebras I, Lectures 19–24 (standard reference, not scraped)

Diagonal Cartan subalgebra and roots of slnsln​