PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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The center is the common kernel of the roots inside the Cartan subalgebra

Statement

Assume the Axiom of Choice. Let $h$ be a Cartan subalgebra of a finite-dimensional complex semisimple Lie algebra $g$ , with root set $Φ$ (Root and root space). Then ${H \in h : α (H) = 0 for all α \in Φ} = Z (g) \cap h = 0.$ In particular the roots span $h^{*}$ , and the description of the zero common-root-kernel as the center is an equality inside $h$ .

Facts & Assumptions

Given: The Axiom of Choice and such $g$ and $h$ .

[A1]

The Axiom of Choice is The Axiom of Choice; it licenses the structural suppliers [L1] and [L2].

[L1]

$g = h \oplus ⨁_{α \in Φ} g_{α}$ is a direct sum over the root spaces (Root-space decomposition, Roots of a complex semisimple Lie algebra form a reduced crystallographic root system, Root and root space).

[L2]

$h$ is abelian (Cartan subalgebras are exactly maximal toral subalgebras, Toral and maximal toral subalgebras).

[L3]

$Z (g) = 0$ (Semisimple Lie algebras are centerless and perfect).

Proof

technique · direct

1.1A1L1L2L3algebra

If $H \in h$ has $α (H) = 0$ for every $α \in Φ$ , then $[H, g_{α}] = α (H) g_{α} = 0$ for every root, and $[H, h] = 0$ by [L2]; by [L1] $[H, g] = 0$ , so $H \in Z (g)$ and $H = 0$ by [L3].

2.1L1step 1.1algebra∎

Conversely every central element of $h$ is annihilated by all roots, since $α (H) = 0$ is the eigenvalue of ${ad}_{H}$ on $g_{α}$ and ${ad}_{H} = 0$ for central $H$ . Hence the common kernel equals $Z (g) \cap h = 0$ ; and because no nonzero $H$ annihilates all roots, the finite set $Φ$ spans $h^{*}$ .

Depends on

Used by

The roots form a reduced crystallographic Euclidean root system Proposition

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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