Jtdylktuy

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 $mathfrak{h}_1capmathfrak{h}_2=0$.

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 $mathfrak{h}'=bigcapmathfrak{h}$. Since $mathfrak{z(g)}subsetmathfrak{n(h)}=mathfrak{h}$, for all $mathfrak{h}$ cartan subalgebra, we have $mathfrak{z(g)}subsetmathfrak{h}'$. Now, let $Xinmathfrak{h}'$. Since each $mathfrak{h}$ is abelian, $X$ commutes with all $Yinmathfrak{bar g}={text{regular elements of }mathfrak{g}}$, so, $mathfrak{z}(X)supsetmathfrak{bar g}$. Since $mathfrak{z}(X)$ is an subalgebra (and so a vector subspace), we have $mathfrak{z}(X)=mathfrak{g}$, so $Xinmathfrak{z(g)}$.

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!

I think this is an solution.

We can assume $mathfrak{g}subsetmathfrak{gl}(V)$, $dim V=n$, since the adjoint representation has kernel $0$. Also, $textit tr:gtomathbb{K}$ is a null homomorphism. If $mathfrak{h}subsetmathfrak{g}$ is a Cartan subalgebra, then, each $Xinmathfrak{h}$ 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 $Xinmathfrak{h}$ is diagonal.

Let $A_n={Xinmathfrak{gl}(V):[X]_betatext{ is diagonal and }textit{tr},(X)=0}supsetmathfrak{h}$. Then, if $E_{ij}$ denotes the matrix with $1$ in row $i$ and column $j$ and $H_{ij}=E_{ii}-E_{jj}$, then ${H_{1i}:2le ile n}$ 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
$$
[H_{1i}]_{beta_0}=left(
begin{array}{cc}0&\
vdots&\
0&\
1&quadastquad\
0&\
vdots&\
0&
end{array}right)
$$
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 $Tinmathbb{M}_n(mathbb{K})$ such that for each $Xinmathfrak{h}$, $T[X]_beta T^{-1}$ isn't diagonal. So, if we define $mathfrak{h}_2={Yinmathfrak{gl}(V):[Y]_beta=T[X]_beta T^{-1},text{for some }Xinmathfrak{h}}$, then $mathfrak{h}_2capmathfrak{h}=0$ and $mathfrak{h}_2$ is a Cartan subalgebra, since $T(,,cdot,,) T^{-1}$ is an automorphism.

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 $mathfrak h_2 = operatorname{Ad}(u)(mathfrak h_1)$.

I will propose another approach to solve the question. Sinse $mathfrak g$ is semisimple then we can decompose $$mathfrak g = mathfrak n^- oplus mathfrak h_1 oplusmathfrak n^{+}$$ 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 ldotscdot 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 sinse $text{ad}(X_{alpha_i})$ 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 $H in mathfrak h_1setminus {0}$ once $H neq 0$, 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 $H in mathfrak z (mathfrak g) Rightarrow H = 0$ sinse $mathfrak g$ is semisimple.

Once there exists a root $alpha$ such that $alpha(H) neq 0$ we conclude that $varphi (H) notin mathfrak h_1$ 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.