If a Lie algebra $\mathfrak{g}$ with only one Cartan subalgebra, then can we implies that $\mathfrak{g}$ is nilpotent?

Question

We suppose that $\mathfrak{g}$ is a finite dimensional Lie algebra over $\mathbb{C}$. My professor told me that the assertion was clear, but I cannot figure it out. I know that each two Cartan subalgebras are conjugate, but how to deduce that such a Lie algebra with only one Cartan subalgebra is nilopotent?

Answer 1

Let $\mathfrak{h}$ be the unique Cartan subalgebra. Being a Cartan subalgebra, it is nilpotent and its normalizer is $\mathfrak{h}$. By uniqueness, $\mathfrak{h}$ is $\mathrm{Aut}(\mathfrak{g})$-invariant. Hence it is an ideal. So its normalizer is $\mathfrak{g}$.

So $\mathfrak{h}=\mathfrak{g}$ and hence $\mathfrak{g}$ is nilpotent.

Answer 2

Since $\mathfrak{g}$ is a finite dimensional complex vector space, we can find a basis of it and give it a coordinate to view it as $\mathbb{C}^n$. In this way, the unique Cartan subalgebra $\mathfrak{h}$ is a linear subspace that contains all the regular elements (the elements $x$ in $\mathfrak{g}$ such that the dimensional of 0-generalized eigenspace of $\mathrm{ad}x$ is minimal). We try to show the set of regular elements is open in $\mathfrak{g}$ with respect to the Euclidean topology given by the coordination.

To show it, suppose $\dim \mathfrak{g}=n$ and the rank of $\mathfrak{g}$ is r and we can let $$\chi(t,x)=t^r(t^{n-r}+f_1(x)t^{n-r-1}+\cdots+f_{n-r}(x))$$ be the characteristic polynomial of $\mathrm{ad}x$ for any $x\in \mathfrak{g}$ and each $f_i$ is a polynomial with $n$-variables that determines $x$ by the given coordinate.

Observe that $x$ is a regular element if and only if $f_{n-r}(x)\neq 0$. In this way, the set of regular elements is a non-empty open set that is contained in the subspace $\mathfrak{h}$. Note that the subspace is closed and $\mathbb{C}^n$ is connected, so we must have $\mathfrak{h}=\mathfrak{g}$.

This argument is hinted by my professor, who provided me with the key description of regular elements.

Answer 3

This is to generalize your own answer. Until part C of this answer, $k$ is any infinite field and $\mathfrak g$ a finite-dimensional Lie algebra over $k$.

A. In Bourbaki's Lie Groups and Lie Algebras book VII §2 the main theorem (namely, Theorem 1 in no. 3) states, among other things, that in our setting:

The Cartan subalgebras of dimension $rk(\mathfrak g)$ are exactly the generalized (ad-)nilspaces $\mathfrak g^0(x)$ of the regular elements $x$.

(A regular element $x \in \mathfrak g$ is defined exactly as in your answer, or equivalently as one whose generalized nilspace has lowest possible dimension $=: rk(\mathfrak g)$ (which is also equivalent to the definition in your answer).) Particularly helpful for us is Corollary 3 to that theorem which states:

$\mathfrak g$ is the sum of its Cartan subalgebras.

This is proven easily from the theorem once one knows that the regular elements are dense in $\mathfrak g$ with respect to the Zariski topology, which replaces the Euclidean topology in your answer. (Now just note that the sum of Cartan subalgebras is a subspace, hence closed in the Zariski topology.)

Of course this already implies what you want: If there is only one CSA $\mathfrak h$ in $\mathfrak g$, then by that corollary, $\mathfrak g = \mathfrak h$ is nilpotent.

Actually, in Bourbaki's exercise 10 to VII §2 we are asked to show the following generalization of what you want:

The following are equivalent:

1. $rk(\mathfrak g) = \dim(\mathfrak g)$
2. $\mathfrak g$ is nilpotent
3. $\mathfrak g$ contains a finite number of Cartan subalgebras of dimension $rk(\mathfrak g)$
4. $\mathfrak g$ contains only one Cartan subalgebra

(For $3 \implies 4$, I used contraposition and the fact that over an infinite field, a vector space is not a finite union of proper subspaces (cf. this and its many duplicates). As soon as $\mathfrak g$ contains more than one CSA, this allows us to "always find more regular elements" which are not in any finite family of CSA's already constructed.)

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B. What you want also follows immediately from exercise 15 of loc. cit. which states that the following are equivalent:

1. $x$ is in the intersection of all Cartan subalgebras.
2. $x$ is in the intersection of all generalized nilspaces $\mathfrak g^0(y)$ for all $y \in \mathfrak g$.
3. $x \in \bigcup C_n$, where $C_n$ are the steps of the ascending central series (their union sometimes called the "hypercentre").

To prove $1 \implies 2$ one can apply density of regular elements again. $2 \implies 3$ follows from Engel's Theorem, and $3 \implies 2 \implies 1$ are easy.

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C. Finally, it might be helpful to see how all this can fail over finite fields; let's take $k= \mathbb F_2$ for example. Let $\mathfrak h :=k \oplus k$ be the two-dimensional abelian Lie algebra. Denote by $a(h_1,h_2) =h_1, b(h_1,h_2) := h_2, c(h_1,h_2) := h_1+h_2$ the three non-zero elements of $\mathfrak h^*$. Define $\mathfrak g$ as the vector space $$ \mathfrak h \oplus V_a \oplus V_b \oplus V_c$$ where each $V_x$ is one-dimensional, and we define a Lie bracket $[h,v_x]:= x(h)\cdot v_x$ for $x =a,b,c$.

Check that $\mathfrak g$ is solvable, not nilpotent (centre is trivial, lower central series halts after the derived subalgebra $V_a \oplus V_b \oplus V_c$), its only Cartan subalgebra is $\mathfrak h$, all nilspaces (= centralizers) of regular elements in $\mathfrak h$ are strictly bigger than $\mathfrak h$ and not nilpotent, $rk(\mathfrak g) =3$. The first problem here is that regular elements are no longer regular in their nilspaces, which cannot happen over an infinite field due to nonzero polynomials not being identically zero when evaluated on such field. Note that here, as soon as we replace $k =\mathbb F_2$ by a proper field extension, the problem disappears. (For all this, compare exercises 11, 12 and 13 in loc. cit.)