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awarded  Yearling 
Apr
19 
comment 
Long term behavior of a certain discrete time dynamical system on graphs
Wow, don't know why I didn't realize how easy this was. Thanks! 
Apr
19 
accepted  Long term behavior of a certain discrete time dynamical system on graphs 
Apr
19 
asked  Long term behavior of a certain discrete time dynamical system on graphs 
Mar
11 
awarded  Notable Question 
Feb
24 
awarded  Notable Question 
Apr
29 
awarded  Yearling 
Jul
2 
awarded  Curious 
Apr
29 
awarded  Yearling 
Mar
30 
answered  Amenability of $l^\infty$ 
Dec
13 
awarded  Popular Question 
Sep
22 
awarded  Popular Question 
Sep
15 
awarded  Nice Question 
Sep
1 
comment 
What is the source of this famous Grothendieck quote?
@Yemon: Yes that is what I intended (I only think of normal ones) 
Aug
26 
comment 
Versions of the spectral theorem
@Issam: Yes the spectral theorem does imply that a (noncommutative) vNA is generated by projections. Any vNA is generated by selfadjoint elements, to which you can apply the spectral theorem to get the projections. My perferred book for these topics is "An Introduction to Operator Algebras" by Kehe Zhu. It's quite expensive, though :( 
Aug
24 
comment 
Versions of the spectral theorem
With regards to 3.) In my experience the vN algebra version (ie Borel functional calculas) IS presented as the most general version. For me the way to think about it is not in terms of a the projection valued measure so much but rather in terms of the resulting isomorphism that it gives from the $C^*$ (or vN) algebra generated by $A$ and the algebra of continuous (or measureable) functions on $\sigma(A)$. 
Aug
13 
comment 
von neumann algebras and measurable spaces
@Yemon: Ah ok yes, Now I see. I will add a comment on post. 
Aug
13 
comment 
von neumann algebras and measurable spaces
@Yemon: Doesn't HB do this for you? A bounded linear function on a subspace (which these duals function are when restricted to the predual(original) basis) can be extended to a bounded linear function on the whole space, with the same bound (norm). This is how i've always thought of the nongeometric version of HB, is this not right? 
Aug
13 
comment 
von neumann algebras and measurable spaces
@Yemon: I got your comment, I don't see the issue though, except that maybe you don't want to use axiom of choice to get a basis (since it is weaker than HB no?), but for a finite dimension subspace of $X$ it should be fine. 
Jul
28 
comment 
von neumann algebras and measurable spaces
Maybe I am mistaken, by I thought that they agreed on the unit ball. 