Given two elements $A,B \in \mathfrak{su}(n)$ what is the dimension of the span of the following adjoint orbit: $\{Ad_{e^{sA}}(B) \ | \ s \in [0,t]\}$ for different values of $t$. Does it ever change when $t$ changes or is it the same for all $t>0$?
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The span is the same for all $t>0$. The reason is that the curve $\gamma(s) = \mathrm{Ad}_{e^{sA}}(B)$ is a real-analytic curve in a vector space $V$ (in this case, $V={\frak{su}}(n)$). If $\lambda:V\to\mathbb{R}$ is any linear function, the composition $\lambda\bigl(\gamma(s)\bigr)$ is then a real-analytic function on $\mathbb{R}$ and hence either vanishes identically or only at isolated points. Thus, the annihilator of the span of $\{\gamma(s)\ |\ a<s<b\}$ is the same for all open intervals $(a,b)\subset\mathbb{R}$.
answered May 9, 2015 at 9:48
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$\begingroup$ That's a very nice argument. Does it let one determine the actual dimension though. What's unclear to me is that it will be the same for each $A,B \neq 0$. $\endgroup$– BenjaminCommented May 9, 2015 at 23:30
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1$\begingroup$ Oh, it won't be the same for all $A$ and $B$. For example, if $[A,B]=0$, then the image will just consist of the point $B$, so the span will be the line through $B$. For the generic pair, though, the span will be all of ${\frak{su}}(n)$. $\endgroup$ Commented May 9, 2015 at 23:44
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$\begingroup$ Ah, I should have explicitly excluded the commuting case. It's tricky to remember to say everything relevant when you have an application in mind! The commuting case is not important at all, I should have said. What dimensions can it have? Writing $Ad_{e^{sA}}(B) = e^{s {ad}_{A}}(B)$ suggests that it will be to do with the first value of $k$ for which ${ad}_{A}^{k}(B) = 0$. $\endgroup$– BenjaminCommented May 9, 2015 at 23:48
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1$\begingroup$ Actually, it's just the dimension of the smallest subspace $S\subset{\frak{su}}(n)$ that contains $B$ and is invariant under the action of $\mathrm{ad}(A)$. $\endgroup$ Commented May 10, 2015 at 1:13
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1$\begingroup$ @Benjamin: It's not really the multiplicity (i.e., the sizes of the various eigenspaces of $\mathrm{ad}(A)$) that's important, it's more this: Say that $\mathrm{ad}(A)^2$ (which has real eigenvalues, all nonpositive) has $M$ eigenspaces (one of which is the $0$ eigenspace), and that $m$ of them are orthogonal to $B$. Then the dimension you want is $2(M-m)$ if the $0$-eigenspace is orthogonal to $B$ and $2(M-m)-1$ if the $0$-eigenspace is not orthogonal to $B$. $\endgroup$ Commented May 10, 2015 at 8:27