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Let $\mathfrak{g}$ be a Lie algebra. Is Cartan subalgebra of $\mathfrak{g}$ unique? I see in some places it is written "Let $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$".

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    Cartan subalgebra is unique up to the adjoint action of the corresponding group $G$.2012-04-21

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Here is an example. If $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{R})$, then both of $ \mathfrak{h}_1 = \mathbb{R}\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \quad , \quad \mathfrak{h}_2 = \mathbb{R}\begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix} $ are Cartan subalgebras.

Add: To add a bit more detail: It isn't too difficult to check that both of these subalgebras are (1) nilpotent and (2) they are both equal to their own normalizers in $\mathfrak{g}$.

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    Great, thanks! =)2012-06-28
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You can think in terms of the Lie group, any element is in a maximal torus and any two maximal torus are conjugate to each other. For simple-connected Lie groups the maximal torus correspond to the Cartan subalgebra. So one should expect many different copies of Cartan subalgebra in the Lie algebra. But they are all isomorphic.

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    "any element is in a maximal torus"—only if you have a *compact* Lie group. (Why do you restrict to simply connected to establish a bijection between maximal tori and CSA's? I only know about linear Lie groups, but there neither the Lie algebra, nor the collection of tori, sees the fundamental group—although of course individual tori do.)2016-09-28
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No, Cartan algebra of a Lie algebra may not be unique. But any two Cartan algebras are conjugate to each other.