$\mathfrak{h}_1,\mathfrak{h}_2$ Cartan subalgebras with $\mathfrak{h}_1\cap\mathfrak{h}_2=0$

Question

Let $\mathfrak{g}$ be a finite dimensional simple Lie Algebra over an algebraically closed field $K$. I'm having trouble to show that always exists Cartan subalgebras $\mathfrak{h}_1,\mathfrak{h}_2$ such that $\mathfrak{h}_1\cap\mathfrak{h}_2=0$.

In general, if all (or one) cartan subalgebras of a finite dimensional Lie Algebra $\mathfrak{g}$ are abelian, then $\bigcap_{\mathfrak{h}\text{ cartan}}\mathfrak{h}=\mathfrak{z(g)}$. For this, call $\mathfrak{h}'=\bigcap\mathfrak{h}$. Since $\mathfrak{z(g)}\subset\mathfrak{n(h)}=\mathfrak{h}$, for all $\mathfrak{h}$ cartan subalgebra, we have $\mathfrak{z(g)}\subset\mathfrak{h}'$. Now, let $X\in\mathfrak{h}'$. Since each $\mathfrak{h}$ is abelian, $X$ commutes with all $Y\in\mathfrak{\bar g}=\{\text{regular elements of }\mathfrak{g}\}$, so, $\mathfrak{z}(X)\supset\mathfrak{\bar g}$. Since $\mathfrak{z}(X)$ is an subalgebra (and so a vector subspace), we have $\mathfrak{z}(X)=\mathfrak{g}$, so $X\in\mathfrak{z(g)}$.

From this, since any cartan subalgebra of a simple algebra is abelian, we have that the intersection of all subalgebras is $0$, since in this case $\mathfrak{z(g)}=0$. But I have no idea how to show that only two is sufficient...

Any help will be appreciated!

Answer 1

I think this is an solution.

We can assume $\mathfrak{g}\subset\mathfrak{gl}(V)$, $\dim V=n$, since the adjoint representation has kernel $0$. Also, $\textit tr:g\to\mathbb{K}$ is a null homomorphism. If $\mathfrak{h}\subset\mathfrak{g}$ is a Cartan subalgebra, then, each $X\in\mathfrak{h}$ is semisimple (then diagonalizable) and $\mathfrak{h}$ is abelian, so, there exists a basis $\beta=\{e_1,\cdots,e_n\}$ of $V$ such that each $X\in\mathfrak{h}$ is diagonal.

Let $A_n=\{X\in\mathfrak{gl}(V):[X]_\beta\text{ is diagonal and }\textit{tr}\,(X)=0\}\supset\mathfrak{h}$. Then, if $E_{ij}$ denotes the matrix with $1$ in row $i$ and column $j$ and $H_{ij}=E_{ii}-E_{jj}$, then $\{H_{1i}:2\le i\le n\}$ form a basis for $A_n$. If we define $\beta_0=\{e_1',\cdots,e_n'\}$, where $e_1'=e_1+\cdots+e_n,e_2'=e_1-e_2,\cdots,e_n'=e_1-e_n$, then we have $$ [H_{1i}]_{\beta_0}=\left( \begin{array}{cc}0&\\ \vdots&\\ 0&\\ 1&\quad\ast\quad\\ 0&\\ \vdots&\\ 0& \end{array}\right) $$ Where the $1$ is in the row $i$, and $\ast$ is anything. In particular, in the base $\beta_0$, no element other than $0$ in $A_n$ is diagonal. So there exists a invertible matrix $T\in\mathbb{M}_n(\mathbb{K})$ such that for each $X\in\mathfrak{h}$, $T[X]_\beta T^{-1}$ isn't diagonal. So, if we define $\mathfrak{h}_2=\{Y\in\mathfrak{gl}(V):[Y]_\beta=T[X]_\beta T^{-1},\text{for some }X\in\mathfrak{h}\}$, then $\mathfrak{h}_2\cap\mathfrak{h}=0$ and $\mathfrak{h}_2$ is a Cartan subalgebra, since $T(\,\,\cdot\,\,) T^{-1}$ is an automorphism.

Answer 2

I will propose another approach to solve the question. Since $\mathfrak g$ is semisimple then we can decompose $$\mathfrak g = \mathfrak n^- \oplus \mathfrak h_1 \oplus\mathfrak n^{+}$$ where $$ \mathfrak n^{+} = \sum_{\alpha >0} \mathfrak g_\alpha$$ $$ \mathfrak n^{-} = \sum_{\alpha <0} \mathfrak g_\alpha$$ such that each $\alpha$ is a root.

Now, we can consider the isomorphism of Lie Algebras \begin{align*}\varphi:& \mathfrak g \to \mathfrak g \\ H &\mapsto e^{\text {ad} (X_{\alpha_1})} \cdot \ldots\cdot e^{\text {ad} (X_{\alpha_n})}(H)\end{align*} where $\{X_{\alpha_1},...,X_{\alpha_n}\}$ is a basis of $\mathfrak n^+$, such that, $X_{\alpha_i} \in g_{\alpha_i}$.

Note that $e^{\text{ad}(X_{\alpha_i})}$ is well defined since $\text{ad}(X_{\alpha_i})$ is nilpotent. Moreover it is an isomorphism, because $\text{ad}(X_{\alpha_i})$ is a derivation and $e^{\text{ad}(X_{\alpha_i})}$ is an invertible linear transformation.

Consider $H \in \mathfrak h_1\setminus \{0\}$ since $H \neq 0$, there exists a root $\alpha$, such that $\alpha (H) \neq 0$, otherwise we would conclude that $H \in \mathfrak z (\mathfrak g)$, because $[H,H_1] =0,$ $\forall H_1 \in \mathfrak h_1$ and $[H,X_\alpha] = \alpha(H) X_\alpha =0$ $\forall$ root $\alpha$ which implies that $H \in \mathfrak z (\mathfrak g) \Rightarrow H = 0$ since $\mathfrak g$ is semisimple.

Since there exists a root $\alpha$ such that $\alpha(H) \neq 0$ we conclude that $\varphi (H) \not\in \mathfrak h_1$ for all $H \in \mathfrak h_1 \setminus \{0\}$. Then $\mathfrak h_2 = \varphi (\mathfrak h_1)$ is a Cartan subalgebra that satisfies the required properties.

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Edit: 14/02/2020

In order to check that $\varphi (H) \not\in \mathfrak h_1$. Note that if $\alpha(H)\neq 0$, then $\text{ad}(X_\alpha)(H) = \alpha(H) X_\alpha \in \mathfrak g_\alpha $. Let $\pi_\alpha: \mathfrak g\to \mathfrak g_{\alpha}$ be the natural projection into the subspace $\mathfrak g_{\alpha}$. Since $[\mathfrak g_{\alpha},\mathfrak g_{\beta}] = \mathfrak g_{\alpha + \beta}$ (if $\alpha + \beta$ is not a root the $g_{\alpha + \beta}:= \{0\}$) one can conclude that

$$\pi_\alpha \left(e^{\mathrm{ad}(X_\alpha)}(H)\right) =\pi_\alpha \left(\sum_{n=0}^{m} \frac{1}{n!}\mathrm{ad}(X_\alpha)^{n}(H)\right) =\pi_\alpha \left(H + \alpha(H) X_\alpha + ...\right) = \alpha(H) X_\alpha. $$

Since $\varphi$ is the product of exponencials of $\text{ad}(X_\beta)$ and we are considering only positive roots. Coupled with the fact that the identity matrix appears summed up in the operator $e^{\mathrm{ad}(X_\beta)}$, i.e. $$e^{\mathrm{ad}(X_\beta)} =\color{blue}{\text{Id}} + \text{ad}(X_\beta) + ... + \frac{1}{n!}\text{ad}(X_\beta)^n.$$

This implies that the term $\alpha(H)X_\alpha$ cannot be vanished by the successive application of the operators $e^{\text{ad}(X_\beta)}$, $\beta>0$. Therefore $$\pi_\alpha(\varphi(H)) =\pi_{\alpha}\left( e^{(\text {ad} (X_{\alpha_1})} \cdot \ldots\cdot e^{\text {ad} (X_{\alpha_n})}(H)\right)\neq 0,$$ implying $\varphi(H)\not\in\mathfrak h_1$, because $g_\alpha$ is l.i. with the subspace $\mathfrak h_1$.

Answer 3

If you're comfortable with simple Lie groups and their root subgroups, then here is a more conceptual version of @Yuki's answer. Let $G$ be the adjoint simple Lie group with Lie algebra $\mathfrak g$, and $H_1$ a maximal torus in $G$. Let $u$ be a regular unipotent element of $G$—concretely, one may take $u$ to be the product of a non-trivial element from each simple root subgroup. Then we may take $\mathfrak h_1 = \operatorname{Lie}(H_1)$, and $\mathfrak h_2 = \operatorname{Ad}(u)(\mathfrak h_1)$.