Does all semi-simple Lie algebra has at least one pair of Cartan subalgebras with greater than {0} intersection?

[mpham26uchicago]

Let $\mathfrak{g}$ be a semi-simple Lie algebra with the Cartan decomposition $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{m}$. Does $\mathfrak{m}$ always contain two Cartan subalgebras $\mathfrak{h}, \mathfrak{h}^\prime$ such that $\{0\} \subsetneq \mathfrak{h} \cap \mathfrak{h}^\prime$?

[mpham26uchicago]

If proving this state for general semi-simple $\mathfrak{g}$ is too hard, then I could just do $\mathfrak{g} = \mathfrak{su}(2^n)$. I could even restrict this to just product operator basis.

[mpham26uchicago]

If proving this state for general semi-simple $\mathfrak{g}$ is too hard, then I could just do $\mathfrak{g} = \mathfrak{su}(2^n)$. I could even restrict this to just product operator basis.

Here's a non-constructive proof.

Definition (Commutative set): Let $a \in \mathfrak{m}$ be non-zero. The commutative set of $a$, $\mathcal{C}_a$, is defined as

$$\mathcal{C}_a = \{b \in \mathfrak{m} : [a, b] = 0\}$$

Observe that $\mathcal{C}_a$ forms a subspace of $\mathfrak{m}$.

Proof. To show that $\mathfrak{m}$ contain two Cartan subalgebras $\mathfrak{h}, \mathfrak{h}^\prime$ such that $\{0\} \subsetneq \mathfrak{h} \cap \mathfrak{h}^\prime$, we just need to show that there exists a non-zero $a \in \mathfrak{m}$ such that

$$\dim(\mathcal{C}_a) > \dim(\mathfrak{h}) $$

This being true, we can construct two distinct Cartan subalgebras that both contain $a$.

Here's a related question For non-zero $a,b \in \mathfrak{m}$, is $\dim(\mathcal{C}_a) = \dim(\mathcal{C}_b)$?

[mpham26uchicago]

To generalize beyond the product operator basis, I could choose an arbitrary vector in $\mathfrak{m}$ and consider orthogonal vectors (inner product = 0)