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)