Cartan subalgebra

Intuition [Beginner]

A Cartan subalgebra is the "diagonal part" of a Lie algebra. Just as every matrix can be put into a form with diagonal entries and off-diagonal entries, a semisimple Lie algebra splits into a maximal commuting subalgebra (the Cartan subalgebra) and a collection of "off-diagonal" spaces indexed by directions called roots.

The Cartan subalgebra is abelian: every pair of elements commutes. It is maximal with this property among subalgebras that are built from semisimple (diagonal-like) elements. In the matrix Lie algebra ${\mathfrak{sl}}_n\mathbb{C}$, the Cartan subalgebra consists of the diagonal traceless matrices — the most natural commuting subalgebra.

The reason the Cartan subalgebra matters is that it provides the coordinate system for the entire representation theory. Roots, weights, characters, and the Weyl group are all defined relative to the Cartan subalgebra. Without it, the classification of semisimple Lie algebras and their representations has no starting point.

Visual [Beginner]

A diagram showing a three-by-three grid of arrows radiating from a central column. The central column represents the Cartan subalgebra $\mathfrak{h}$ of ${\mathfrak{sl}}_3\mathbb{C}$, consisting of two independent diagonal matrices. The off-diagonal arrows, labelled by roots, point outward in six directions (three pairs of opposite arrows). Each arrow represents a root space, a one-dimensional subspace of the Lie algebra carrying a specific "charge" relative to the Cartan subalgebra.

The picture captures the core structure: the Cartan subalgebra is the base from which all root directions emanate, and every element of the Lie algebra is either in the base or in one of the root spaces.

Worked example [Beginner]

Consider ${\mathfrak{sl}}_2\mathbb{C}$, the Lie algebra of traceless $2\times 2$ matrices. This algebra has dimension $3$ with basis $H,E,F$ where $H=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ is diagonal, $E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ is upper off-diagonal, and $F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$ is lower off-diagonal.

Step 1. The Cartan subalgebra is $\mathfrak{h}=\mathbb{C}H$, the one-dimensional space of diagonal traceless matrices. Every element of $\mathfrak{h}$ commutes with every other element (since $\mathfrak{h}$ is one-dimensional, this is automatic).

Step 2. Check maximality. The only larger subalgebra would be two-dimensional. But $[H,E]=2E\ne 0$ and $[H,F]=-2F\ne 0$, so adding either $E$ or $F$ destroys commutativity. The subalgebra $\mathfrak{h}$ is maximal abelian.

Step 3. The element $H$ is semisimple: ${\mathrm{ad}}_H$ is diagonalisable. Its eigenvalues on ${\mathfrak{sl}}_2\mathbb{C}$ are $0$ (on $\mathfrak{h}$), $2$ (on $\mathbb{C}E$), and $-2$ (on $\mathbb{C}F$). These eigenvalues are the roots.

What this tells us: the Cartan subalgebra $\mathfrak{h}=\mathbb{C}H$ provides the single direction ($H$) from which both root spaces $\mathbb{C}E$ and $\mathbb{C}F$ receive their eigenvalue labels $2$ and $-2$.

Check your understanding [Beginner]

Formal definition [Intermediate+]

Let $\mathfrak{g}$ be a finite-dimensional Lie algebra over an algebraically closed field of characteristic zero. An element $X\in \mathfrak{g}$ is semisimple if ${\mathrm{ad}}_X\in \mathfrak{gl}(\mathfrak{g})$ is diagonalisable. An element $X\in \mathfrak{g}$ is nilpotent if ${\mathrm{ad}}_X$ is nilpotent (some power of ${\mathrm{ad}}_X$ is zero).

A Cartan subalgebra of $\mathfrak{g}$ is a subalgebra $\mathfrak{h}\subseteq \mathfrak{g}$ satisfying:

1. $\mathfrak{h}$ is nilpotent as a Lie algebra.
2. $\mathfrak{h}$ is its own normaliser: $\mathfrak{h}=N_{\mathfrak{g}}(\mathfrak{h}):=\{X\in \mathfrak{g}:[X,\mathfrak{h}]\subseteq \mathfrak{h}\}$.

When $\mathfrak{g}$ is semisimple, the Cartan subalgebra $\mathfrak{h}$ is automatically abelian and consists entirely of semisimple elements. In this case $\mathfrak{h}$ is equivalently characterised as a maximal abelian subalgebra of semisimple elements.

Counterexamples to common slips

• Nilpotent but not self-normalising. The one-dimensional subalgebra $\mathbb{C}E$ of ${\mathfrak{sl}}_2\mathbb{C}$ (where $E$ is the raising operator) is nilpotent (abelian), but $N_{\mathfrak{g}}(\mathbb{C}E)=\mathbb{C}H\oplus \mathbb{C}E$ (the Borel subalgebra), which strictly contains $\mathbb{C}E$. So $\mathbb{C}E$ is not a Cartan subalgebra.
• Abelian but containing nilpotent elements. In a non-semisimple Lie algebra, a Cartan subalgebra may contain nilpotent elements. The definition (nilpotent + self-normalising) is the correct one for general Lie algebras; the abelian-plus-semisimple characterisation holds only in the semisimple case.
• Non-conjugacy in positive characteristic. The conjugacy theorem for Cartan subalgebras requires characteristic zero. Over fields of positive characteristic, different Cartan subalgebras of the same Lie algebra can have different dimensions.

Key theorem with proof [Intermediate+]

Theorem (Existence and conjugacy of Cartan subalgebras). Every finite-dimensional Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ has a Cartan subalgebra. Any two Cartan subalgebras are conjugate under the group $\mathcal{E}(\mathfrak{g})$ generated by the automorphisms $\exp ({\mathrm{ad}}_X)$ for nilpotent $X\in \mathfrak{g}$.

Proof of existence. An element $X\in \mathfrak{g}$ is called regular if the number of zero eigenvalues of ${\mathrm{ad}}_X$ (counted with multiplicity) is minimal among all elements of $\mathfrak{g}$.

For $X\in \mathfrak{g}$, let ${\mathrm{ad}}_X=S_X+N_X$ be the Jordan decomposition of ${\mathrm{ad}}_X$ (semisimple part $S_X$ plus nilpotent part $N_X$, with $[S_X,N_X]=0$). Define the generalised zero-eigenspace: $$ \mathfrak{g}^0(X) = {Y \in \mathfrak{g} : (\mathrm{ad}_X)^k(Y) = 0 \text{ for some } k \geq 1}. $$

Claim. If $X$ is regular, then $\mathfrak{h}:={\mathfrak{g}}^0(X)$ is a Cartan subalgebra.

The subspace ${\mathfrak{g}}^0(X)$ is the Fitting null-component of ${\mathrm{ad}}_X$: by the Jordan decomposition, $\mathfrak{g}$ splits as $\mathfrak{g}={\mathfrak{g}}^0(X)\oplus {\mathfrak{g}}_1(X)$ where ${\mathrm{ad}}_X$ is nilpotent on ${\mathfrak{g}}^0(X)$ and invertible on ${\mathfrak{g}}_1(X)$.

To verify nilpotency of $\mathfrak{h}$: since $\mathfrak{h}$ is the generalised zero-eigenspace of ${\mathrm{ad}}_X$, the restriction ${\mathrm{ad}}_X{|}_{\mathfrak{h}}$ is nilpotent. For any $H\in \mathfrak{h}$, the operator ${\mathrm{ad}}_H$ preserves both $\mathfrak{h}$ and ${\mathfrak{g}}_1(X)$ (by the Jacobi identity and the Fitting decomposition). The regularity of $X$ ensures that $\mathfrak{h}$ is minimal in dimension, and one shows that ${\mathrm{ad}}_H{|}_{\mathfrak{h}}$ is nilpotent for every $H\in \mathfrak{h}$ by the following argument: if some ${\mathrm{ad}}_H{|}_{\mathfrak{h}}$ were not nilpotent, the generalised zero-eigenspace ${\mathfrak{g}}^0(H)$ would be strictly smaller than ${\mathfrak{g}}^0(X)=\mathfrak{h}$, contradicting the minimality (regularity) of $X$. Hence $\mathfrak{h}$ is nilpotent.

To verify self-normalisation: let $Y\in N_{\mathfrak{g}}(\mathfrak{h})$. Then $[Y,\mathfrak{h}]\subseteq \mathfrak{h}$, so ${\mathrm{ad}}_Y$ maps $\mathfrak{h}\to \mathfrak{h}$. Write $Y=Y_0+Y_1$ with $Y_0\in \mathfrak{h}$ and $Y_1\in {\mathfrak{g}}_1(X)$. For $H\in \mathfrak{h}$: $[Y,H]=[Y_0,H]+[Y_1,H]\in \mathfrak{h}$. Since $[Y_0,H]\in \mathfrak{h}$ (both are in $\mathfrak{h}$), we need $[Y_1,H]\in \mathfrak{h}$. But ${\mathrm{ad}}_H$ preserves ${\mathfrak{g}}_1(X)$, so $[Y_1,H]\in {\mathfrak{g}}_1(X)$. For this to also lie in $\mathfrak{h}$, it must lie in $\mathfrak{h}\cap {\mathfrak{g}}_1(X)=0$. So $[Y_1,H]=0$ for all $H\in \mathfrak{h}$, meaning $Y_1\in {\mathfrak{g}}^0(H)$ for all $H$. Since ${\mathrm{ad}}_X{|}_{{\mathfrak{g}}_1(X)}$ is invertible, $Y_1\ne 0$ would contribute a new zero-generalised-eigenvalue direction, contradicting regularity. So $Y_1=0$ and $Y\in \mathfrak{h}$, confirming $\mathfrak{h}=N_{\mathfrak{g}}(\mathfrak{h})$.

Proof of conjugacy (sketch). Let $\mathfrak{h}$ and ${\mathfrak{h}}^{\prime }$ be two Cartan subalgebras. Choose regular elements $X\in \mathfrak{h}$ and $X^{\prime }\in {\mathfrak{h}}^{\prime }$ with $\mathfrak{h}={\mathfrak{g}}^0(X)$ and ${\mathfrak{h}}^{\prime }={\mathfrak{g}}^0(X^{\prime })$. The conjugacy proof proceeds by showing that for any regular $X$, there exists $\sigma \in \mathcal{E}(\mathfrak{g})$ with $\sigma (X^{\prime })=X$, and then $\sigma ({\mathfrak{h}}^{\prime })=\mathfrak{h}$ because ${\mathfrak{g}}^0(\sigma (X^{\prime }))=\sigma ({\mathfrak{g}}^0(X^{\prime }))$ (conjugation preserves the Fitting decomposition).

The key step is the following. For $Y\in {\mathfrak{g}}_1(X)$, the operator ${\mathrm{ad}}_Y$ is nilpotent (by the Fitting property), so $\exp ({\mathrm{ad}}_Y)\in \mathcal{E}(\mathfrak{g})$ is defined. One shows that the regular elements form a connected dense open subset of $\mathfrak{g}$, and that the group $\mathcal{E}(\mathfrak{g})$ acts transitively on the set of Cartan subalgebras by a connectedness argument on the variety of regular elements. $\square$

Bridge. The existence of a Cartan subalgebra builds toward 07.06.18 where the root-space decomposition $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{\alpha }{\mathfrak{g}}_{\alpha }$ splits the Lie algebra into the Cartan subalgebra and the root spaces, each carrying a specific eigenvalue under ${\mathrm{ad}}_{\mathfrak{h}}$. The foundational reason the Cartan subalgebra exists is the Fitting decomposition: for any $X$, the generalised zero-eigenspace ${\mathfrak{g}}^0(X)$ is a subalgebra, and regularity guarantees it is minimal and self-normalising. This is exactly the structure that identifies the Cartan subalgebra as the "simultaneous eigenspace" for the semisimple elements of $\mathfrak{g}$. The central insight is that conjugacy makes the Cartan subalgebra intrinsic: the root-space decomposition does not depend on a choice. The bridge is between the abstract regular-element construction and the concrete diagonal-matrix picture, and the pattern recurs in 07.06.03 where the root system is defined relative to a Cartan subalgebra and is independent of the choice by conjugacy.

Exercises [Intermediate+]

Advanced results [Master]

Theorem 1 (Abstract Jordan decomposition). Let $\mathfrak{g}$ be a semisimple complex Lie algebra. Every $X\in \mathfrak{g}$ decomposes uniquely as $X=X_s+X_n$ where $X_s$ is semisimple, $X_n$ is nilpotent, $[X_s,X_n]=0$, and both $X_s,X_n$ are polynomials in $X$ (equivalently, in ${\mathrm{ad}}_X$).

The abstract Jordan decomposition is compatible with every representation: if $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ is a finite-dimensional representation, then $\rho (X_s)$ and $\rho (X_n)$ are the semisimple and nilpotent parts of $\rho (X)$ in the ordinary Jordan decomposition in $\mathrm{End}(V)$.

Theorem 2 (Rank equals dimension of Cartan subalgebra). The rank of a semisimple Lie algebra $\mathfrak{g}$ is $\mathrm{rk}(\mathfrak{g})=\dim \mathfrak{h}$, independent of the choice of Cartan subalgebra. For the classical types: $\mathrm{rk}({\mathfrak{sl}}_n)=n-1$, $\mathrm{rk}({\mathfrak{so}}_{2n+1})=n$, $\mathrm{rk}({\mathfrak{sp}}_{2n})=n$, $\mathrm{rk}({\mathfrak{so}}_{2n})=n$.

Theorem 3 (Killing form on $\mathfrak{h}$ identifies $\mathfrak{h}\cong {\mathfrak{h}}^{\ast }$). The restriction $B{|}_{\mathfrak{h}}$ is non-degenerate, inducing an isomorphism $\mathfrak{h}\to {\mathfrak{h}}^{\ast }$ via $H\mapsto B(H,\cdot )$. This identifies each root $\alpha \in {\mathfrak{h}}^{\ast }$ with a unique coroot $H_{\alpha }\in \mathfrak{h}$ satisfying $\alpha (H)=B(H_{\alpha },H)$ for all $H\in \mathfrak{h}$.

Theorem 4 (Regular elements are dense and open). The set of regular elements ${\mathfrak{g}}_{\mathrm{reg}}$ is a non-empty Zariski-open (hence dense) subset of $\mathfrak{g}$. Every regular element is semisimple when $\mathfrak{g}$ is semisimple. The complement $\mathfrak{g}\setminus {\mathfrak{g}}_{\mathrm{reg}}$ is the zero locus of a polynomial (the discriminant of the characteristic polynomial of ${\mathrm{ad}}_X$).

Theorem 5 (Cartan subalgebras and maximal tori). For a compact connected Lie group $G$ with Lie algebra $\mathfrak{g}$, the Cartan subalgebras of $\mathfrak{g}$ are the Lie algebras of the maximal tori in $G$. The conjugacy of Cartan subalgebras (Lie algebra level) follows from the conjugacy of maximal tori (group level), and vice versa.

Theorem 6 (Centraliser of a Cartan subalgebra). If $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is a Cartan subalgebra, then the centraliser $C_{\mathfrak{g}}(\mathfrak{h})=\{X\in \mathfrak{g}:[X,H]=0\text{ for all }H\in \mathfrak{h}\}$ equals $\mathfrak{h}$. In the semisimple case, the centraliser and the normaliser of $\mathfrak{h}$ both equal $\mathfrak{h}$.

Theorem 7 (Cartan subalgebra of a direct sum). If $\mathfrak{g}={\mathfrak{g}}_1\oplus {\mathfrak{g}}_2$ is a direct sum of semisimple Lie algebras, then the Cartan subalgebras of $\mathfrak{g}$ are exactly the direct sums ${\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ where ${\mathfrak{h}}_i$ is a Cartan subalgebra of ${\mathfrak{g}}_i$.

Synthesis. The foundational reason the Cartan subalgebra exists and is unique up to conjugacy is the Fitting decomposition combined with the regular-element theory, and the central insight is that the set of regular elements is a dense open subset of $\mathfrak{g}$ whose Fitting null-components all have the same dimension. Putting these together with the abstract Jordan decomposition, every element of a semisimple Lie algebra decomposes into a semisimple part and a nilpotent part, and the Cartan subalgebra is the maximal subspace on which the semisimple part of every element acts diagonalisably. This is exactly the structure that identifies $\mathfrak{h}$ as the coordinate system for the root-space decomposition. The bridge is between the abstract regular-element construction and the concrete diagonal-matrix picture in the classical types. The generalises direction runs from the existence-conjugacy theorem to the full Cartan-Killing classification via root systems 07.06.03 and Dynkin diagrams 07.06.05. The pattern recurs in 07.06.18 where the root-space decomposition relative to $\mathfrak{h}$ gives the full splitting $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{\alpha }{\mathfrak{g}}_{\alpha }$, and in 07.06.10 where the Casimir element is expressed in a basis adapted to the Cartan decomposition.

Full proof set [Master]

Proposition 1 (Centraliser of $\mathfrak{h}$ equals $\mathfrak{h}$). If $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is a Cartan subalgebra, then $C_{\mathfrak{g}}(\mathfrak{h})=\mathfrak{h}$.

Proof. Since $\mathfrak{h}$ is abelian, $\mathfrak{h}\subseteq C_{\mathfrak{g}}(\mathfrak{h})$. For the reverse inclusion, let $X\in C_{\mathfrak{g}}(\mathfrak{h})$, meaning $[X,H]=0$ for all $H\in \mathfrak{h}$. The root-space decomposition gives $\mathfrak{g}=\mathfrak{h}\oplus {⨁}_{\alpha \in \Phi }{\mathfrak{g}}_{\alpha }$, and $[X,H]=0$ for all $H\in \mathfrak{h}$ forces $X$ to lie in the zero-eigenspace of every ${\mathrm{ad}}_H$, which is $\mathfrak{h}$. More precisely, write $X=H_0+{\sum }_{\alpha }{X}_{\alpha }$ with $H_0\in \mathfrak{h}$ and $X_{\alpha }\in {\mathfrak{g}}_{\alpha }$. Then $0=[X,H]={\sum }_{\alpha }\alpha (H)X_{\alpha }$ for all $H\in \mathfrak{h}$. Since the roots span ${\mathfrak{h}}^{\ast }$, there exists $H$ with $\alpha (H)\ne 0$ for each nonzero $\alpha$, forcing $X_{\alpha }=0$. So $X=H_0\in \mathfrak{h}$. $\square$

Proposition 2 (Abstract Jordan decomposition is compatible with representations). Let $\mathfrak{g}$ be semisimple, $X=X_s+X_n$ the abstract Jordan decomposition, and $\rho :\mathfrak{g}\to \mathfrak{gl}(V)$ a finite-dimensional representation. Then $\rho (X)=\rho (X_s)+\rho (X_n)$ is the ordinary Jordan decomposition in $\mathrm{End}(V)$.

Proof. The representation $\rho$ extends to an algebra homomorphism $U(\mathfrak{g})\to \mathrm{End}(V)$. Both $X_s$ and $X_n$ are polynomials in $X$ (in $U(\mathfrak{g})$), so $\rho (X_s)$ and $\rho (X_n)$ are the same polynomials in $\rho (X)$. Since $X_s$ is semisimple (meaning ${\mathrm{ad}}_{X_s}$ is diagonalisable), $\rho (X_s)$ is diagonalisable: the adjoint action being diagonalisable is equivalent to $\rho (X_s)$ being diagonalisable for every representation when $\mathfrak{g}$ is semisimple, because the abstract Jordan decomposition is intrinsic. Similarly, $X_n$ nilpotent (meaning ${\mathrm{ad}}_{X_n}$ is nilpotent) implies $\rho (X_n)$ is nilpotent. And $[\rho (X_s),\rho (X_n)]=\rho ([X_s,X_n])=0$. So $\rho (X)=\rho (X_s)+\rho (X_n)$ is a decomposition into commuting diagonalisable plus nilpotent, which is unique: the ordinary Jordan decomposition. $\square$

Proposition 3 (Direct-sum decomposition of Cartan subalgebras). If $\mathfrak{g}={\mathfrak{g}}_1\oplus {\mathfrak{g}}_2$ with ${\mathfrak{g}}_i$ semisimple, and ${\mathfrak{h}}_i$ is a Cartan subalgebra of ${\mathfrak{g}}_i$, then $\mathfrak{h}={\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ is a Cartan subalgebra of $\mathfrak{g}$.

Proof. The subalgebra $\mathfrak{h}={\mathfrak{h}}_1\oplus {\mathfrak{h}}_2$ is abelian (each ${\mathfrak{h}}_i$ is abelian and $[{\mathfrak{h}}_1,{\mathfrak{h}}_2]=0$). It consists of semisimple elements: if $H=H_1+H_2$ with $H_i\in {\mathfrak{h}}_i$, then ${\mathrm{ad}}_H={\mathrm{ad}}_{H_1}\oplus {\mathrm{ad}}_{H_2}$ (block diagonal), and each ${\mathrm{ad}}_{H_i}$ is diagonalisable on ${\mathfrak{g}}_i$, so ${\mathrm{ad}}_H$ is diagonalisable on $\mathfrak{g}$. For maximality: if ${\mathfrak{h}}^{\prime }⊋\mathfrak{h}$ is a larger abelian subalgebra of semisimple elements, project to ${\mathfrak{g}}_1$ and ${\mathfrak{g}}_2$; one projection must enlarge ${\mathfrak{h}}_i$, contradicting maximality of ${\mathfrak{h}}_i$ in ${\mathfrak{g}}_i$. $\square$

Connections [Master]

• Engel's theorem and Lie's theorem 07.06.14. The proof that the Cartan subalgebra is nilpotent uses the Fitting decomposition, which is a refined version of the nilpotency detection in Engel's theorem. Lie's theorem provides the upper-triangular form that makes the Fitting decomposition explicit in the solvable case. The existence of the Cartan subalgebra in a general (not necessarily semisimple) Lie algebra combines Engel's theorem with the regular-element construction.

• Root system 07.06.03. The root system is defined relative to a Cartan subalgebra: the roots are the nonzero linear functionals $\alpha \in {\mathfrak{h}}^{\ast }$ such that the root space ${\mathfrak{g}}_{\alpha }=\{X\in \mathfrak{g}:[H,X]=\alpha (H)X\text{ for all }H\in \mathfrak{h}\}$ is nonzero. The conjugacy of Cartan subalgebras guarantees that the root system is an invariant of $\mathfrak{g}$, independent of the choice of $\mathfrak{h}$. The Cartan subalgebra is the stage on which the root system lives.

• Cartan's criterion 07.06.16. The non-degeneracy of the Killing form on a semisimple Lie algebra (Cartan's criterion) is what makes the identification $\mathfrak{h}\cong {\mathfrak{h}}^{\ast }$ possible: the Killing form restricted to $\mathfrak{h}$ is non-degenerate, and each root $\alpha \in {\mathfrak{h}}^{\ast }$ corresponds to a unique coroot $H_{\alpha }\in \mathfrak{h}$ via $B(H_{\alpha },\cdot )=\alpha$. The coroots are the basis for the entire root-system geometry.

Historical & philosophical context [Master]

Élie Cartan introduced the subalgebra that bears his name in his 1894 Paris thesis Sur la structure des groupes de transformations finis et continus ^[Cartan1894]. Cartan's insight was that the "diagonal part" of a semisimple Lie algebra — the subalgebra on which the adjoint action is simultaneously diagonal — carries the structural information needed to classify simple Lie algebras. Cartan worked in the concrete setting of transformation groups and identified the Cartan subalgebra as the subalgebra of pointwise-diagonalisable elements.

The abstract definition (nilpotent subalgebra equal to its own normaliser) and the proof of existence via regular elements are due to Chevalley, who reformulated the theory in the 1940s and 1950s using the language of abstract Lie algebras. The conjugacy theorem was first proved by Cartan in the group setting (conjugacy of maximal tori in compact Lie groups); the purely algebraic proof via the group $\mathcal{E}(\mathfrak{g})$ generated by $\exp ({\mathrm{ad}}_X)$ for nilpotent $X$ is due to Mostow in the 1950s and was refined by Bourbaki ^{[BourbakiGroupesAlgebres18]}.

The abstract Jordan decomposition was introduced by Chevalley in his 1955 Théorie des groupes de Lie as the intrinsic version of the classical additive Jordan decomposition, adapted to the Lie algebra setting. Its compatibility with arbitrary representations (Proposition 2 above) is the key structural tool that makes the Cartan subalgebra well-behaved under every representation.

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}

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