Can the Intersection of two Cartan subalgebras be bigger than {0}?

[mpham26uchicago]

If $\mathfrak{h}, \mathfrak{h}^\prime$ are two maximal abelian subalgebra of $\mathfrak{m}$, can the intersection $\mathfrak{h} \cap \mathfrak{h}^\prime$ be bigger than {0}?

[mpham26uchicago]

Suppose this is true there is an $h \neq 0 \in \mathfrak{h}$, then by the adjoint orbit theorem, there exist $k \in K$ such that

$$k\mathfrak{h} k^\dagger = \mathfrak{h}^\prime$$

I was thinking khk = h but this doesn't have to be true.

[mpham26uchicago]

The answer is YES, the intersection can be bigger than ${0}$. For example, take $\mathfrak{so}(4)$ under type AI Cartan involution. We have the $-1$ eigenspace.

$$\mathfrak{m} = {\text{IX, IY, IZ}, \mathbf{XI}, \mathbf{XX}, \mathbf{XZ}, \text{ZI, ZX, ZZ}, \mathbf{YY}}$$

Example: $\{\text{IZ, ZI, ZZ} \} \cap \{\text{XX, YY ,ZZ} \} = \{ZZ\}$

Observe that both sets on the left-hand side are maximal abelian subalgebra.

In general, if there exists non-zero and distinct $a, b, c \in \mathfrak{m}$ such that $ab = ba$ and $ac = ca$ but $bc \neq cb$, then we can construct two distinct maximal abelian subalgebra with bigger than zero intersection.