Consider a smooth curve $U\_s:[0,T] \rightarrow SU(4)$ which solves:


$\frac{d U\_s}{ds} = (a + w(s)b)U\_s$


for some given $a,b \in \mathfrak{su}(4)$ (which generate $\mathfrak{su}(n)$) and a smooth real, bounded function $w: [0,T] \rightarrow \mathbb{R}$ (T>0). Also consider some given $\xi \in \mathfrak{su}(4)$. $\langle \cdot , \cdot \rangle$ is the Killing form in what follows.


Does there exist $w$ such that the following all simultaneously hold:


* for at least one non-zero $B \in \mathfrak{su}(4)$, $\langle B, U\_s \xi U\_s^{\dagger} \rangle = 0 \ \ \forall s \in [0,T]$
* $\int\_0^{T} U\_s \xi U\_s^{\dagger} = 0$
* $\int\_0^{T} w(s) U\_s \xi U\_s^{\dagger} = 0$


These requirements seem extremely demanding on $w$, however, I can't prove that it's impossible.



Simultaneous integral equation on $SU(n)$
Well, I think that, in principle, this is answerable, but it may be hard to come up with an explicit example. I don't have a proof, one way or the other, about what might happen or not, but here are some thoughts that you might find useful:


First, think of this as a problem with a number of parameters: $p = (a,b,\xi,B,T)$, making $4\cdot 15 + 1 = 61$ in all. Now, fix one of these parameters $p$ and consider the following subset of $\mathrm{SU}(4)$:
$$
Z\_p = \{U\in \mathrm{SU}(4) \ | \ \langle B,U\xi U^\dagger\rangle = 0 \}
\subset \mathrm{SU}(4).
$$
Generically, $Z\_p$ will be a (possibly singular in some places) hypersurface in $\mathrm{SU}(4)$, so, generically, it will have dimension $14$. For $U\in Z\_p$, consider the equation
$$
\bigl\langle [B,a + w b],U\xi U^\dagger\bigr\rangle
= \bigl\langle [B,a],U\xi U^\dagger\bigr\rangle + w\bigl\langle [B,b],U\xi U^\dagger\bigr\rangle = 0.
$$
Again, generically, the function $\bigl\langle [B,b],U\xi U^\dagger\bigr\rangle$ on $Z\_p$ will only vanish on a hypersurface $H\_p\subset Z\_p$, and, away from that hypersurface, there will be a unique function $f: Z\_p\setminus H\_p\to\mathbb{R}$ such that 
$$
f(U) = - \frac{\bigl\langle [B,a],U\xi U^\dagger\bigr\rangle}{\bigl\langle [B,b],U\xi U^\dagger\bigr\rangle}.
$$ 
The curves $U(s)$ that you want to study are then the ones that satisfy the ODE
$$
U'(s) = \bigl(a + f\bigl(U(s)\bigr)b\bigr)U(s)
$$
and the initial condition that $U(0)$ lie in $Z\_p$, for these are the ones that satisfy your condition $\bigl\langle B,U(s)\xi U(s)^\dagger\bigr\rangle = 0$ for $0\le s\le T$.
These are the integral curves of a vector field on a (possibly singular) manifold of dimension $14$, so there is a $13$-parameter family of these curves for each value of $p=(a,b,\xi,B,T)$, making a total of $61+31 = 94$ parameters.


Now, your final two conditions are
$$
\int\_0^T U(s)\xi U(s)^\dagger\ \mathrm{d} s = 
\int\_0^T f\bigl(U(s)\bigr)U(s)\xi U(s)^\dagger\ \mathrm{d} s = 0.
$$
and each of these is $14$ equations (since the integrands are perpendicular to $B$ for all values of $s$). 


Thus, *generically* one would expect there to be a solution space of dimension at least
$$
94 - 14 - 14 = 66
$$
if it is not empty. 


I don't, *a priori*, see any reason for the solution space to be empty, particularly, since there are certainly solutions (which you would probably regard as 'degenerate' because, say, $a$ and $b$ for that solution don't generate the entire Lie algebra, and so on). 


To make a real proof out of this, of course, you would have to deal with the singularities of the loci $Z\_p$ and the locations of the hypersurfaces $H\_p$. Probably, it would be more informative to just search around numerically in the parameter space and see whether you can locate a numerical solution that way. 


I suspect that, with patience, this can be done, but, not being sufficiently interested in the problem (because I'm not convinced that it's actually natural or relevant to your real problem, which you haven't described), I don't think I'll spend any more time thinking about it.


