5
$\begingroup$

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.)

$\endgroup$
4
  • 1
    $\begingroup$ @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. $\endgroup$ 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
    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. $\endgroup$ 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. $\endgroup$ Nov 18, 2014 at 2:13

1 Answer 1

6
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Perfect, that was precisely the calculation I was hoping for. Thanks! :) $\endgroup$ Nov 18, 2014 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.