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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.

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    $\begingroup$ You need to sharpen your assumptions and state more clearly what they are, otherwise your question has trivial answers. In particular, there are choices $(a,b,T,\xi)$ for which a pair $\bigl(U(s),w(s)\bigr)$ exists satisfying all of these conditions and there are choices $(a,b,T,\xi)$ such that there exist no $\bigl(U(s),w(s)\bigr)$ satisfying all of these conditions. $\endgroup$
    – Robert Bryant
    Commented Sep 3, 2015 at 11:55
  • $\begingroup$ I see. The important case is typical $a,b$ (ones which generate the entire algebra), I'll add this to the question. $\endgroup$
    – Benjamin
    Commented Sep 3, 2015 at 12:57
  • $\begingroup$ If the existence of any solutions depends on choosing $T$ carefully then this would be very interesting. $\endgroup$
    – Benjamin
    Commented Sep 3, 2015 at 13:04
  • $\begingroup$ Ok, the most important question here is which $(a,b,T,\xi)$ yield no solutions. $\endgroup$
    – Benjamin
    Commented Sep 3, 2015 at 13:10
  • $\begingroup$ Are you also going to require that $B$ be nonzero? $\endgroup$
    – Robert Bryant
    Commented Sep 3, 2015 at 14:25

1 Answer 1

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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.

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  • $\begingroup$ p.s. and f.y.i. The real problem is to work out (using parametric transversality perhaps) if the end point map $V_T$ for the above differential equation is transverse to the foliation of $SU(n)$ into level sets of $|Tr(UG^{\dagger})|^2$ (for any given $G$) for all but a null set of pairs $(a,b)$. I suspect this is true. $\endgroup$
    – Benjamin
    Commented Sep 4, 2015 at 22:47
  • $\begingroup$ If $T$ were close it it's minimum value which made the system controlable and $|w(s)|< k$ was set appropriately, would this all become more easy? My intuition says that this would be enough to ensure no solutions. $\endgroup$
    – Benjamin
    Commented Sep 15, 2015 at 23:34

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