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Problem

Given a C*-algebra $\mathcal{A}$.

Consider dynamics $\tau:\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$ and $\tau':\mathbb{R}\to\mathrm{Aut}(\mathcal{A})$.
(More precisely, strongly continuous one-parameter groups.)

Denote their derivations by $\delta:\mathcal{D}\to\mathcal{A}$ and $\delta':\mathcal{D}'\to\mathcal{A}$.

Then one has: $$\delta=\delta'\implies\tau=\tau'$$ (Here, equality is meant in terms of operators resp. maps.)

How do I check this?

For dynamics over Hilbert spaces I would proceed by: $$i\frac{\mathrm{d}}{\mathrm{d}t}\|\varphi(t)\|^2=\langle H\varphi(t),\varphi(t)\rangle-\langle\varphi(t),H\varphi(t)\rangle=0$$ But for the C*-algebra case this path is not directly available.

Disclaimer

I hope to get a hint from here.
(I haven't got any respond yet from stack exchange.)

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    $\begingroup$ @JonBannon; No, I mean infinitesimal generator but there's no Hilbert space and so neither a concept of selfadjointness nor unitarity. All I have is a C*-algebra and a strongly continuous one-parameter group of automorphisms and that's the problem as I pointed out. $\endgroup$
    – C-star-W-star
    Commented Nov 18, 2014 at 1:53
  • $\begingroup$ I guess the hint would be that exponentiating the generators should recover the one-parameter groups. The key technical point is that the generators have dense domain, which you can prove using a mollifier. $\endgroup$
    – Nik Weaver
    Commented Nov 18, 2014 at 2:07
  • $\begingroup$ Yep, I mean $t_n\to t$ implies $\tau^{t_n}(A)\to\tau^t(A)$ for every $A\in\mathcal{A}$. In principle that is nothing but continuity w.r.t. sort of pointwise topology. $\endgroup$
    – C-star-W-star
    Commented Nov 18, 2014 at 2:08
  • $\begingroup$ @NikWeaver: Denseness I checked but I'm still hanging at the point that showing that exponentiating really recovers the group at least on a dense domain. Then the rest would be just a result by uniform extension. $\endgroup$
    – C-star-W-star
    Commented Nov 18, 2014 at 2:13

1 Answer 1

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As was discussed in the comments, it suffices to see that $\delta$ determines $\tau$ uniquely on $\mathrm{dom}(\delta)$ which is dense in $\mathcal{A}$. Suppose that $x_0 \in \mathrm{dom}(\delta)$. Check that $t \mapsto \tau^t(x_0)$ is a solution to the initial value problem \begin{align*} \frac{d}{dt} x(t) = \delta( x(t)) && x(0) = x_0. \end{align*} If $x$ is any solution then $$\frac{d}{dt} \left( \tau^{-t}(x(t)) \right) = -\tau^{-t}\delta(x(t)) + \tau^{-t}\left(\delta(x(t))\right) = 0$$ and it follows that $\tau^{-t}(x(t)) \equiv x_0$ so that $x(t) = \tau^t (x_0)$ is the unique solution.

Basically this all works for Banach space flows too. See the answers to my own question here.

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  • $\begingroup$ Perfect, that was precisely the calculation I was hoping for. Thanks! :) $\endgroup$
    – C-star-W-star
    Commented Nov 18, 2014 at 12:48

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