# 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.Noix07 (talk) 19:31, 15 May 2014 (CEST)

- 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)

**How to Cite This Entry:**

Arveson spectrum.

*Encyclopedia of Mathematics.*URL: http://encyclopediaofmath.org/index.php?title=Arveson_spectrum&oldid=32185