Talk:Arveson spectrum

1. A rapid check shows that $\hat{x}(n)$ does not exactly satisfy the indicated equation but rather with an inverse $$U_z \hat{x}(n) = (z)^{-1} \hat{x}(n)$$ Even if there is no ambiguity, it would also be better if we make it clear that $\hat{x}(n)$ is still a function on $T$ (while usually in Fourier transform, the transform is a function on the "Fourier space" which I admit is just a word that doesn't explain anything)

2. A question: "vector-valued Riemann integral", is that the same thing as Bochner integrals?

1a. I guess the result depends on the interpretation of the phrase "take translation for $\{U_z\}$". What is meant by the translation by $z$ of a function $x$? Is it the function $w \mapsto x(zw)$ or $w \mapsto x(z^{-1}w)$?

1b. The "Fourier space" (dual to T) is the group of integers, and $n \mapsto \hat{x}(n)$ is indeed a function on this space. But it is a vector-valued function, and its values are (in the special case considered) functions.

2. As far as I understand, Riemann integrability implies Bochner integrability, but is not equivalent to it.

Boris Tsirelson (talk) 13:15, 15 May 2014 (CEST)

thanks for the precisions. For 1a, I just used $U_z U_y \equiv U_{zy} $ which should hold no matter if it is a left or right action?

Yes, I see, you are right; it should be $U_z \hat{x}(n) = z^{-n} \hat{x}(n)$, indeed. (Though, not $ (z)^{-1} \hat{x}(n)$.)

But there is also another problem: if $dz$ is treated in the complex sence (that is, as $i \E^{i\phi} \rd\phi$ for $z=\E^{i\phi}$) then $U_z \hat{x}(n) = z^{-n-1} \hat{x}(n)$; in the article probably $dz$ is treated in the real sence (that is, as $\rd\phi$ for $z=\E^{i\phi}$), which should be stated explicitly. Boris Tsirelson (talk) 16:10, 15 May 2014 (CEST)

$z^{-n}$ indeed. I actually also thought about the measure, and guessed that it was the Haar measure on the circle considered as a group.

But now I have another question. Is there a name for the theorem (§ after the condition $||x||=\sup_{\rho\in \chi^*} \cdots$):

"if $\lbrace U_t\rbrace_{t\in G} $ is an isometric representation of $G$ that is continuous in the weak topology, then for each finite regular Borel measure $\mu$ on $G$ there is an operator $U_{\mu}$ on $\chi$ such that $\langle U_{\mu}(x),\rho\rangle =\int_G \langle U_t(x),\rho\rangle d\mu(t),\ \rho\in\chi^*$"

I have found this, Dinculeanu, A. and C. Ionescu-Tulcea theorem p.2, but I'm not confident with measure theory;

Coincidentally, I've read from notes by Paul Garrett (quasi-completeness p.2) that the condition that convex hull of compact subset have to be compact is required for the existence of Pettis integrals.

Yes, it is the Haar measure. But the article does not say so.

About that theorem. I do not know. I do not have Arveson's texts. Looking at their MR abstracts (by the way, (a1) should be Vol.15, not 13), I do not know whether Pettis integral was used, or not. I only see in the abstract to (a1): "Define a representation $\pi$ of $L_1 (G)$ into the bounded linear operators on $X$ by $\pi(f )x = \int_G f (t)U_t x \rd t$, $f \in L_1(G)$.

And please sign your messages with four tildas: ~~~~. :-) Boris Tsirelson (talk) 18:53, 15 May 2014 (CEST)