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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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You are right, it is not unique – Thomas Apr 21 '12 at 18:30
Why it is unique? It seems that $\mathfrak{g}$ is just all diagonal matrices in $\mathfrak{g}$. – LJR Apr 21 '12 at 18:34
Does my example make sense? – Thomas Apr 21 '12 at 19:17
Cartan subalgebra is unique up to the adjoint action of the corresponding group $G$. – user8268 Apr 21 '12 at 19:19
up vote 4 down vote accepted

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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I know this is a bit late, but to see that the above Cartan subalgebras are nilpotent, shouldn't the determinant of each of the matrices equal zero? But they are not. (I'm just checking the details you have suggested above.) – math-visitor Jun 28 '12 at 8:15
@math-visitor: There is something called a nilpotent matrix ( Here indeed you need the determinant to be zero. However, here we are talking about nilpotent Lie algebras: Note for this that all the commutators are zero. – Thomas Jun 28 '12 at 15:25
Great, thanks! =) – math-visitor Jun 28 '12 at 17:15

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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No, Cartan algebra of a Lie algebra may not be unique. But any two Cartan algebras are conjugate to each other.

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