Given $\xi \in \mathfrak{su}(4)$ and positive $T \in \mathbb{R}$, is it possible to find all smooth curves $U_s \in SU(4)$ with $U_0 = I$ such that
$$\int_0^T U_s \xi U_s^{\dagger} ds =0\; ?$$
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It is not clear what you mean by 'find' (give an explicit parametrization?) and 'curve' (continuous? differentiable? smooth? closed?). Certainly, it is possible to 'write down' many curves satisfying your requirements, but 'find'? I don't know what that means.
The first thing to recognize is that the orbit $$ M_\xi = \mathrm{SU}(4)\cdot \xi = \left\{U\xi U^\dagger\ | U\in\mathrm{SU}(4)\right\} \subset {\frak{su}}(4)\simeq \mathbb{R}^{15} $$ is a smooth submanifold endowed with a natural $\mathrm{SU}(4)$-invariant measure with respect to which the average of the position vector in ${\frak{su}}(4)$ is zero. This means, in particular, that the convex hull of $M_\xi$ contains $0$. Consequently, there are many curves $\eta:[0,T]:M_\xi$ with $\eta(0)=\xi$ such that $$ \int_0^T \eta(s)\ \mathrm{d} s = 0 $$ and any such curve $\eta$ can be lifted to a curve in $\mathrm{SU}(4)$ since the natural mapping $\mathrm{SU}(4)\to M_\xi$ is a smooth fiber bundle with fibers being the left cosets of the subgroup $G_\xi$ consisting of those $g\in \mathrm{SU}(4)$ that satisfy $g\xi g^\dagger = \xi$. It doesn't matter how you lift such an $\eta$, so the problem really is to describe the 'balanced' curves in $M_\xi$ that start at $\xi$.
Now, the balancing condition is only $15$ equations on $\eta$ and the set of (continuous, differentiable, smooth, or real-analytic; pick your flavor) curves $\eta:[0,T]\to M_\xi$ with $\eta(0)=\xi$ is infinite dimensional when you put any reasonable topology on it, so you are asking for a description of an infinite dimensional space. This raises the question of how you are going to 'parametrize' this space in such a way that you could pick out the 15 parameters that need to vanish in order to lie in the 'balanced' set. I don't see any particularly good way to do this, though I can see some ideas about computing the 'tangent space' if the space of curves is sufficiently nice and is given the right topology.
Here is a simpler problem; if you can tell us what you would take for an answer for it, maybe we can give you a similar answer for your question: Consider the set of smooth curves $\gamma:[0,1]\to S^2\subset\mathbb{R}^3$ with $\gamma(0)=(0,0,1)$. Describe the codimension $3$ subset that consists of those $\gamma$ satisfying $$ \int_0^1 \gamma(s)\ \mathrm{d} s = 0\in\mathbb{R}^3. $$
answered Sep 1, 2015 at 11:18
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$\begingroup$ This was pretty much what I meant when I said "find all Xs", I really just meant, "please can someone tell me about the nature of all Xs". As there are an infinite dim family (which seems obvious now you say it), I see that I need to add more conditions for my application to work at all, so there's no point in "finding them all" in any strong sense than your answer. Thanks for your help. $\endgroup$– BenjaminCommented Sep 1, 2015 at 13:31
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$\begingroup$ So, to avoid asking another similar question, what if we now insist that $\frac{d U_s}{ds}= (a+w(s)b)U_s$ for some $a,b \in \mathfrak{su}(n)$ and some smooth, real $w$. This seems that there should be far fewer curves with both properties. $\endgroup$– BenjaminCommented Sep 1, 2015 at 13:40
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$\begingroup$ Yes, there will be far fewer such curves. Indeed, there could be none. For example, if $a$ and $b$ happen to commute with $\xi\not=0$, then there will be none since, for all such curves $U_s\xi U_s^\dagger$ will be constant and equal to $\xi$. $\endgroup$ Commented Sep 1, 2015 at 15:42
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1$\begingroup$ I believe that, if $a$, $b$, and $\xi$ are generically chosen in ${\frak{su}}(4)$, then there will be such curves and, in fact, an infinite dimensional family of them, though they might not be easy to construct. $\endgroup$ Commented Sep 1, 2015 at 16:39
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2$\begingroup$ It would not be easy to explain in a short note, but the basic point is that the system you are describing above is a right-invariant control system with drift on the compact manifold $\mathrm{SU}(4)$, and, for generic choice of $a$ and $b$, it is controllable via the single control ($w(s)$), as is not difficult to show using standard control theory. Thus, as long as $T$ is sufficiently large, you have enough time to steer all over $\mathrm{SU}(4)$ and to equidistribute the resulting curve with respect to the position map defined by $\xi$. $\endgroup$ Commented Sep 1, 2015 at 21:34