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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 |
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Versions of the spectral theorem
@Issam: Yes the spectral theorem does imply that a (non-commutative) vNA is generated by projections. Any vNA is generated by self-adjoint 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 |
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von neumann algebras and measurable spaces
@Yemon: Ah ok yes, Now I see. I will add a comment on post. |
Aug
13 |
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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 non-geometric version of HB, is this not right? |
Aug
13 |
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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 |
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von neumann algebras and measurable spaces
Maybe I am mistaken, by I thought that they agreed on the unit ball. |
Jul
24 |
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von neumann algebras and measurable spaces
@Yemon: Yes, I did gloss over this. Though to be honest, ultrapowers are a convenient language to talk about sequences of separable algebras, and every use of ultrapowers that I know of fits into this role. |
Jul
24 |
answered | von neumann algebras and measurable spaces |
Jul
23 |
comment |
What's a noncommutative set?
With regards to your last questions it is equivalent to the free group factor isomorphism problem. Specifically $L(\mathbb{F}_n)\sim L(\mathbb{F}_\infty)\Leftrightarrow L(\mathbb{F}_n)\simeq L(\mathbb{F}_\infty)$. This is because in the non-isomorphic case you need $P\otimes N\simeq P$ and these factors are prime. More generally we have that if $M, N$ are prime $II_1$ factors then $M\sim N\Leftrightarrow M\simeq N$. |
Jul
13 |
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Is the fundamental group of a maximal subfactor always $\mathbb{R}_{+}^{*}$?
Stable means the that $P\subset Q\simeq P\otimes\mathcal{R}\subset Q\otimes \mathcal{R}$. |
Jul
12 |
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What's the natural equivalence of subfactors in general?
$P^t\subset Q^t=pM_n(P)p\subset pM_n(Q)p$ for an appropriate projection $p\in P$. Similarly for $P^t\otimes\mathcal{R}\subset Q^t\otimes\mathcal{R}$, but $p\in P\otimes\mathcal{R}$ in the last case. The isomorphism type only depends on the trace of $p$ and since $P\otimes\mathcal{R}$ is a factor we get a projection with correct trace in $\mathcal{R}$. So we can assume $p\in 1\otimes\mathcal{R}$. Then we just view $M_n(P)\otimes\mathcal{R}$ as $M_n(\mathbb{C})\otimes P \otimes\mathcal{R}= P\otimes M_n(\mathcal{R})$. Since $p\in P'$ we get the amplification is happening on $\mathcal{R}$. |
Jul
12 |
comment |
What's the natural equivalence of subfactors in general?
However, this notion kills any information from the fundamental group. Specifically, consider the BNP examples $P\subset Q$. Then $P$ and $Q$ are the hyperfinite $II_1$, which we call $\mathcal{R}$, which is absorbing for itself. Then we have $P^t\otimes\mathcal{R} \subset Q^t\otimes\mathcal{R}\simeq P\otimes\mathcal{R}^t\subset Q\otimes\mathcal{R}^t\simeq P\otimes\mathcal{R}\subset Q\otimes \mathcal{R}$. This will always happen if the absorbing factor ($M$ above) has full fundamental group, and anything that absorbs $\mathcal{R}$ is a McDuff factor and has full fundamental group. |