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Consider a curve $\gamma_s= U_s^{\dagger} b U_s$ in $\mathfrak{su}(n)$ where $U_s$ is a smooth curve on $SU(n)$ (starting at $U_0 = \mathbb{I}$) and nonzero $b\in \mathfrak{su}(n)$ and $s \in[0,T]$ for some $T>0$.

assume that the span of all the points on $\gamma_s$ has dim one less than $\mathfrak{su}(n)$, i.e. a co-rank 1 singular curve.

Is it now true that either of the `surfaces' $Q_2(x,y):= \gamma_x \gamma_y$ or $Q_3(x,y,z):= \gamma_x \gamma_y \gamma_z$, have the property that the span of the points on the surface includes all of $\mathfrak{su}(n)$ for $x,y,(z) \in [0,T]$?

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  • $\begingroup$ No: consider $n = 2$ and $b = 0$. $\endgroup$
    – LSpice
    Mar 3, 2016 at 22:10
  • $\begingroup$ By the way, do you really mean to look at the products $\gamma_x \gamma_y$ (which need not lie in $\mathfrak{su}(n)$)? $\endgroup$
    – LSpice
    Mar 3, 2016 at 22:14
  • $\begingroup$ b is non zero, and yes $\endgroup$
    – Benjamin
    Mar 3, 2016 at 22:18
  • $\begingroup$ Do you have an example of such a curve? (For the record, the condition that $b$ be non-0 was edited in after my comment.) $\endgroup$
    – LSpice
    Mar 3, 2016 at 22:32
  • $\begingroup$ No, I don't have an example. However, I conjecture that it's always true. $\endgroup$
    – Benjamin
    Mar 9, 2016 at 3:32

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