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Given $A,B \in \mathfrak{su}(n)$ such that $K(A, B)=0$, I am looking for the largest subgroup $H$ of $SU(n)$ for which:

$K \left(A, Ad_{U}(B) \right) = 0, \ \ \forall U \in H$ where $K$ is the Killing form. Finding the Lie algebra of $H$ would be desirable.

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    $\begingroup$ Why do you think that there is one largest subgroup? For example, I imagine that, for the generic pair $A,B\in{\frak{su}}(n)$, there will be two subgroups, $P = \{ U\in\mathrm{SU}(n)\ |\ \mathrm{Ad}_U(A) = A\}$ and $Q = \{ U\in\mathrm{SU}(n)\ |\ \mathrm{Ad}_U(B) = B\}$ that satisfy your condition and, most likely, they will each be maximal, even though they could be of different dimensions. $\endgroup$
    – Robert Bryant
    Commented Apr 21, 2015 at 11:53
  • $\begingroup$ Why do you say most likely they will be maximal? Won't the group of all such $U$ be the largest. By maximal do you mean, one we can't extend to a larger supergroup? Or rather, simply the such largest group. Perhaps I am confused. $\endgroup$
    – Benjamin
    Commented Apr 21, 2015 at 18:29
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    $\begingroup$ Generally, it won't be true that the set $S_{A,B}$ consisting of all $U\in\mathrm{SU}(n)$ that satisfy $K\bigl(A,\mathrm{Ad}_U(B)\bigr)=0$ is a subgroup of $\mathrm{SU}(n)$, so the best you can hope for is to look for the 'largest' (i.e., highest dimension) subset $H\subset S_{A,B}$ that forms a subgroup, i.e., is closed under multiplication and inverse. $\endgroup$
    – Robert Bryant
    Commented Apr 21, 2015 at 19:56
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    $\begingroup$ Well, take a simple example: Let $n=2$ and let $A$ and $B$ be perpendicular unit vectors in ${\frak{su}}(2)$. Then $P$ and $Q$ (as defined in my first comment) are distinct $S^1$-subgroups and their union generates all of $\mathrm{SU}(2)$. It follows that $S_{A,B}$, which is easily seen to be a codimension-$2$ subset of $\mathrm{SU}(2)$, cannot be a subgroup (though it does contain $P$ and $Q$). Instead, it is a $1$-parameter family of cosets of an $S^1$-subgroup. $\endgroup$
    – Robert Bryant
    Commented Apr 22, 2015 at 1:46
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    $\begingroup$ I couldn't figure out how to login in the chat discussion, so I'm continuing here: I didn't assume that $P$ and $Q$ are maximal, I wrote that I thought it was likely. In the particular case $n=2$ above, $\mathrm{SU}(2)$, which is $3$-dimensional, doesn't have any $2$-dimensional subgroups, so $P$ and $Q$ (which are $1$-dimensional) have maximal dimension by default. Is that what you are asking about? $\endgroup$
    – Robert Bryant
    Commented Apr 22, 2015 at 9:21

1 Answer 1

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This is really an extended comment, but, because it's too long to put into a comment box and because it may help answer some of the OP's questions, I'm putting it here.

If one endows $\mathrm{SU}(n)$ with its usual bi-invariant measure $\mathrm{d}\mu$ normalized to have total volume $1$ (aka Haar measure), then one knows that, for any $B\in{\frak{su}}(n)$, $$ \int_{U\in\mathrm{SU}(n)} \mathrm{Ad}_U(B)\ \mathrm{d}\mu = 0. $$ (The integral has to be an element of ${\frak{su}}(n)$ that is $\mathrm{Ad}$-invariant, so it must be zero.)

This implies that, for any $A,B\in {\frak{su}}(n)$ the average value of the function $f:\mathrm{SU}(n)\to\mathbb{R}$ defined by $f(U) = K\bigl(A,\mathrm{Ad}_U(B)\bigr)$ over $\mathrm{SU}(n)$ is also zero. Now, because the linear span of the $\mathrm{Ad}$-orbit of $B$ must be an $\mathrm{Ad}$-invariant subspace, it is either the zero subspace or all of ${\frak{su}}(n)$, so the only way that $f$ could vanish identically would be for either $A$ or $B$ to be zero.

In particular, it follows that, when $A$ and $B$ are both nonzero, the subset $S_{A,B} = f^{-1}(0)\subset \mathrm{SU}(n)$ divides its complement in $\mathrm{SU}(n)$ into two nonempty open sets and hence, since it is defined algebraically, it must be, at most places, a smooth hypersurface, i.e., it must have codimension $1$, at least where it is smooth. (Because $f$ is real-analytic and $\mathrm{SU}(n)$ is connected, its zero locus cannot contain any nonempty open set.)

Now, there are no subgroups of $\mathrm{SU}(n)$ that have codimension $1$, so $S_{A,B}$ cannot ever be a subgroup of $\mathrm{SU}(n)$ when both $A$ and $B$ are nonzero.

Of course, when $K(A,B)=0$, as I already mentioned, $S_{A,B}$ can contain subgroups, in particular, the subgroups $P$ and $Q$, which are the $\mathrm{Ad}$-stabilizers of $A$ and $B$, respectively. Most of the time, though, these will be maximal subgroups of $\mathrm{SU}(n)$ that lie in $S_{A,B}$ because, generically, $P$ and $Q$ will be maximal tori, and hence there will only be a finite number of connected Lie subgroups that lie between, say $P$ and $\mathrm{SU}(n)$ itself. When $B$ is generically chosen with respect to $A$, none of these properly sandwiched subgroups will lie in $S_{A,B}$, so $P$ will be maximal. A similar argument applies to $Q$.

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  • $\begingroup$ Thanks, a great answer as usual. Am I correct in saying that the Lie algebra of $P$ will simply by all the things that commute with $P$ in $\mathfrak{su}(n)$?. I'm trying to understand what else will be in $S_{A,B}$ in the case $K(A,B)=0$ other than $P \cup Q$. $\endgroup$
    – Benjamin
    Commented Apr 22, 2015 at 22:19
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    $\begingroup$ @Benjamin: The Lie algebra of $P$ will be all the things that commute with $A$ in ${\frak{su}}(n)$. It's hard to say what else will be in $S_{A,B}$ since $S_{A,B}$ is of codimension $1$ in $\mathrm{SU}(n)$. It will surely be larger than $P\cup Q$. $\endgroup$
    – Robert Bryant
    Commented Apr 23, 2015 at 1:26
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    $\begingroup$ @Benjamin: Actually, for regular elements, the dimension of $P$ and $Q$ will be $n{-}1$. As long as $[A,B]\not=0$, the set $S_{A,B}$ will be smooth at the identity and its tangent space there will be the hyperplane in ${\frak{su}}(n)$ perpendicular to $[A,B]$. $\endgroup$
    – Robert Bryant
    Commented Apr 23, 2015 at 9:08
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    $\begingroup$ @Benjamin: Correct. For example, when $n=2$, $S_{A,B}$ is a torus embedded in $\mathrm{SU}(2)\simeq S^3$, while the exponentials of linear subspaces of ${\frak{su}}(2)\simeq\mathbb{R}^3$ are either great circles or spheres in $\mathrm{SU}(2)\simeq S^3$. As for characterizing the preimage of $S_{A,B}$ under the exponential map in general, this is probably going to be somewhat messy. $\endgroup$
    – Robert Bryant
    Commented Apr 23, 2015 at 16:37
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    $\begingroup$ @Benjamin: Oh, sorry. That's was a typo; it should have been 'codimension-$1$'. $\endgroup$
    – Robert Bryant
    Commented Apr 23, 2015 at 20:38

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