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Let $\mathfrak g$ be a compact simple real Lie algebra and let $x\in\mathfrak g$.

What is the intersection of all maximal abelian subalgebras of $\mathfrak g$ which contain $x$?

For instance, in the simplest case of $\mathfrak{so}(3)$, with $x\neq 0$, the only maximal torus containing $x$ is the 1-dimensional space spanned by $x$. In Lie algebras of higher rank, there are in general multiple maximal tori containing a given $x$.

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    $\begingroup$ Some examples and heuristics? $\endgroup$
    – Andrea Marino
    Commented May 20, 2022 at 0:11
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    $\begingroup$ It's the center of the centralizer. $\endgroup$
    – Friedrich Knop
    Commented May 20, 2022 at 6:19
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    $\begingroup$ In $\mathfrak{su}(n)$ for a diagonal matrix $(a_1,\dots,a_n)$ these are the diagonal matrices $(d_1,\dots,d_n)$ in $\mathfrak{su}(n)$ such that $a_i=a_j$ implies $d_i=d_j$. $\endgroup$
    – YCor
    Commented May 20, 2022 at 7:32
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    $\begingroup$ It depends on the level of regularity of the element. Generally speaking, it is the algebra generated by the lowest dimensional face of a Weyl chamber, the element sits in. $\endgroup$
    – user473423
    Commented May 20, 2022 at 14:28
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    $\begingroup$ In $\mathfrak{so}(n)$, these are if I'm correct, those elements of $\mathfrak{so}(n)$ that are proportional to $x$ on each generalized eigenspace of $x$ (i.e. $\mathrm{Ker}(x)$ or $\mathrm{Ker}(x^2+s)$ for $s>0$), except precisely, when $ K_x=\mathrm{Ker}(x)$ is 2-dimensional (happens only when $n$ is even), in which case the element is allowed to be arbitrary on the 2-plane $K_x$. Here "proportional to 0" means "equal to 0". $\endgroup$
    – YCor
    Commented May 24, 2022 at 12:47

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