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Difference between revisions of "Talk:Arveson spectrum"

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:[[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 13:15, 15 May 2014 (CEST)
 
:[[User:Boris Tsirelson|Boris Tsirelson]] ([[User talk:Boris Tsirelson|talk]]) 13:15, 15 May 2014 (CEST)
  
thanks for the precisions. For 1a, I just used $U_z U_y \cong U_zy $ which doesn't depend on left or right action, so I would maintain the claim.
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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?

Revision as of 13:00, 15 May 2014

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?

How to Cite This Entry:
Arveson spectrum. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Arveson_spectrum&oldid=32166