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Given $a,b \in \mathfrak{su}(n)$ and (with $U_0 = I$ taken) the following ODE:

$\frac{d U_t}{dt} = (a + w(t)b)U_t$

consider the "fixed time" endpoint map $V_T: L^2([0,T]) \rightarrow SU(n)$ for which $V[w] := U_T$, the solution to the differential equation at time $T$ if the function $w$ is used in the differential equation.

Is this map a smooth function of $w$? It is acceptable to restrict to only smooth $w$ functions if this helps matters.

Does it also depend smoothly on $a,b$?

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    $\begingroup$ When you write "It is acceptable to restrict to only smooth $w$ functions if this helps matters", do you mean that instead of the Banach space $L^2([0,T])$ you take the usual Fréchet space $C^\infty([0,T])$ as Peter Michor obviously has assumed in his answer and not the space $C^\infty([0,T])$ with the topology induced from $L^2([0,T])$ under the canonical linear embedding? $\endgroup$
    – TaQ
    Commented May 21, 2015 at 20:25

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If you take $w\in C^\infty([0,T])$, then the endpoint map depends smoothly on $w, a, b$. This is most easily seen using convenient calculus: am mapping is smooth if it maps smooth curves to smooth curves, and the space of smooth curves in $C^\infty([0,T])$ is just $C^\infty(\mathbb R\times [0,T])$. So the result follows from smooth depended of solutions of ODE on one further parameter. See this Wikipedia page and literature cited there.

For $L^2([0, T])$ one has to work harder to show smoothness.

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  • $\begingroup$ Thanks, that great. Do you mean that it's harder to prove that $V_T$ is smooth in the $L^2([0,T])$ case or that you need to "work hard" in the sense of adding extra assumptions on $w,a,b$. $\endgroup$
    – Benjamin
    Commented May 20, 2015 at 20:15
  • $\begingroup$ Also, do you think this would work if the set of $w$ was just some smooth (finite dim) vector space of $C^{\infty}$ functions rather than an infinite dim one. $\endgroup$
    – Benjamin
    Commented May 20, 2015 at 20:16
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    $\begingroup$ A finite dimensional vector space is completely fine. Anything that maps smoothly into $C^\infty(\mathbb R, \mathfrak{su}(n)$ just follows. I might add something on $L^2$ when I have time. You might look into mat.univie.ac.at/~michor/convenient-overview.pdf $\endgroup$
    – Peter Michor
    Commented May 21, 2015 at 6:17
  • $\begingroup$ Great. How special is this property of the end point map being smooth? Am I right in saying that no specific property of $SU(n)$ was used? Or in fact even of the specific differential equation? $\endgroup$
    – Benjamin
    Commented May 21, 2015 at 6:26
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    $\begingroup$ Its true on every Lie group. Even on diffeomorphism groups. Groups with this property are called regular Lie groups. Look into the references. $\endgroup$
    – Peter Michor
    Commented May 21, 2015 at 8:14

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