Timeline for About the category of von neumann algebras
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25 events
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| Mar 17, 2012 at 15:51 | comment | added | YCor | [...], $g(1_N)=1$ and $g$ vanishes on finitely supported functions. (I forgot to say I assume $k$ is the real field, although I haven't used it yet). Let $u\in k^N$ be a function growing to infinity. Then for every positive integer d we can write $u=2^d+v^2+w$ where $w$ is finitely supported (define $v=\sqrt{\max(0,u−2^d)}$. So $g(u)=2^d+g(u)^2\ge 2^d$ for all $d$ . This is a contradiction. On the other hand you have plenty of nontrivial homomorphisms $k^S\to k$ when $k$ is the complex field $C$ ,as soon as $S$ is infinite, because any countable ultraproduct of $C$ embeds into $C$. | |
| Mar 17, 2012 at 15:39 | comment | added | YCor | @Theo: that's correct. If $S$ is measurable, you obviously have nonstandard morphisms. Conversely if you've got a ring morphism $f$ from $k^S$ to $k$ vanishing on the finitely supported functions, let $U$ be the set of $I$ such that $f(1_I)=1$. This is an ultrafilter and you need to check that it's stable under intersections. If not, you easily find a sequence of disjoint sets $I_n$ with union $J$ with $I_n\notin U$ and $J\in U$. Define, if $N$ is the set of integers and $u\in k^N$, define $g(u)=f(\sum_n u(n)1_{I_n})$. Then $g$ is a ring homomorphism $k^N\to k$ [...] | |
| Mar 16, 2012 at 4:05 | comment | added | Theo Johnson-Freyd | @Dmitri: The issue is much more subtle than either of us thought. In short, it seems that I need $S$ to be a measurable cardinal in order for my claim to be correct. Existence of measurable cardinals is not provable in ZFC if ZFC is consistent (it would imply that ZFC proves its own consistency, which it cannot if it is consistent by Godel). I learned the argument from Will A. Johnson, one of the grad students at Berkeley. There's one step I don't understand; when I do, I will post a new "answer" with the correct story. | |
| Feb 13, 2012 at 2:48 | comment | added | Theo Johnson-Freyd | @Dmitri: I haven't forgotten that I still owe you a proof of my claim. Unfortunately, I haven't had a lot of time to think about it, and in the time I have had I haven't been able to fill in the proof. So this comment is just to mention that I hope still to get back to you about it. I think I can prove the weaker claim that there exists some set $S$ such that the map $S \hookrightarrow \hom_k(k^S,k)$ to the set of $k$-algebra homomorphisms is not an iso. But this is of course far from my claim that it is never an iso if $|S| > |k|$. So I will keep it in the back of my mind. | |
| Jan 28, 2012 at 17:39 | history | edited | Dmitri Pavlov | CC BY-SA 3.0 |
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| Jan 28, 2012 at 17:39 | comment | added | Dmitri Pavlov | @Jesse: You are certainly right, the two notions are equivalent only in the abelian case and in the nonabelian case a σ-finite von Neumann algebra only admits a representation with a countable separating set, which does not imply that there is a representation with countable dimension. | |
| Jan 28, 2012 at 17:10 | comment | added | Jesse Peterson | @Dmitri: A von Neumann algebra is $\sigma$-finite if it admits at most countably many orthogonal projections. This is the same as admitting a faithful normal state (which the trace provides for all II$_1$ factors). It is not the same as admitting a faithful representation on a separable Hilbert space (there are many non-separable II$_1$ factors, for instance the free group factor on uncountably many generators). See Definition 3.18 and Proposition 3.19 in Takesaki volume 1. | |
| Jan 28, 2012 at 11:49 | comment | added | Dmitri Pavlov | @Jesse: By definition, a von Neumann algebra is separable if it admits a faithful representation on a separable Hilbert space. This is equivalent to σ-finiteness. Takesaki never defines separability, but uses the above definition, as you can see from the fourth paragraph on the page 374 of his Theory of Operator Algebras I. | |
| Jan 28, 2012 at 5:59 | comment | added | Jesse Peterson | $\sigma$-finite is not the same as separable. For instance, all II$_1$ factors are $\sigma$-finite. | |
| Jan 27, 2012 at 10:22 | comment | added | Dmitri Pavlov | @Theo: You are certainly right that my argument is incomplete. Furthermore, I have no idea how to fix it, so your claim seems plausible to me now, even though its purported proof is unclear. | |
| Jan 27, 2012 at 6:32 | comment | added | Theo Johnson-Freyd | On the other hand, I hope it's at least clear that your proposed argument presenting the bijection fails: you just can't a priori guarantee to check on the finitely-supported functions, because it's at least a priori possible that on all finitely-supported functions the homomorphism vanishes. | |
| Jan 27, 2012 at 6:28 | comment | added | Theo Johnson-Freyd | My memory is that the application of Zorn's lemma is essentially the same as the usual case. But you're right, there's something else subtle that I'm forgetting. | |
| Jan 27, 2012 at 6:03 | comment | added | Theo Johnson-Freyd | "Then the filter of co-finite subsets of M is a k-filter." Sorry, that was a misprint. I was trying to skip a step. I should have said: consider the filter of subsets whose complement is strictly smaller than k. This is a k-filter: the union of strictly-less-than-k strictly-smaller-than-k things is strictly-smaller-than-k. | |
| Jan 26, 2012 at 17:47 | comment | added | Dmitri Pavlov | I also cannot see how to apply the axiom of choice to construct a maximal k-filter that contains the given k-filter. Presumably one would use Zorn's lemma, but unless k is countable I don't see how to verify the chain condition. | |
| Jan 26, 2012 at 13:03 | comment | added | Dmitri Pavlov | "Then the filter of co-finite subsets of M is a k-filter." I'm afraid I don't understand this statement. If you intersect an infinite number of pairwise different cofinite subsets you will get a noncofinite subset. So the filter of cofinite subsets of M is not a k-filter unless k is aleph-0. | |
| Jan 26, 2012 at 4:04 | comment | added | Theo Johnson-Freyd | I should mention, I was utterly surprised by the fact I quoted above. I worked it out in conversation with Nick Rozenblyum while on a hike organized by an operator-algebras workshop in Eugene a few years ago. | |
| Jan 26, 2012 at 2:16 | comment | added | Theo Johnson-Freyd | ... maximal k-filter. Thus I will build the homomorphism C(M)→R which sends f to the value of its (unique) level set that is in the chosen maximal k-filter. This is my claimed "extra" point. Thus I have shown that if M is a (paracompact!) zero-dimensional manifold of cardinality strictly greater than R, and assuming Choice, there exist nonprincipal ultrafilters with residue field R. Of course, similar constructions work when M is higher-dimensional but still similarly huge. And conversely, as soon as you ask the homomorphism C(M)→R to be continuous, these extra points evaporate. | |
| Jan 26, 2012 at 2:13 | comment | added | Theo Johnson-Freyd | ... The point is the following. Let k be a cardinality, and call a filter on M a "k-filter" if it has the property that any intersection of strictly fewer than k things in the filter is again in the filter. So a usual filter is an "aleph-null-filter". Let M has cardinality at least k. Then the filter of co-finite subsets of M is a k-filter. By axiom of choice, any k-filter is contained in a maximal k-filter. So there exists a non-principal maximal k-filter; choose one. Now let k be strictly larger than the cardinality of R. Then any function f:M→R has a level-set which is in the chosen ... | |
| Jan 26, 2012 at 1:56 | comment | added | Theo Johnson-Freyd | @Dmitri: It is possible that a homomorphism function C(M)→R vanishes on all functions that are non-zero only only finitely many components. Let's consider the zero-dimensional case, where M is a discrete topological space. Then the maximal ideals of C(M) are precisely the ultrafilters on M, and of course the nonprincipal ultrafilters are distinguished by the property that evaluating a finitely-supported function at any gives 0. Now, usually the residue field at a nonprincipal ultrafilter is huge. It is always a model of the first-order theory of R, but usually it is a nonstandard model. ... | |
| Jan 23, 2012 at 20:44 | vote | accept | Oliver | ||
| Jan 23, 2012 at 20:29 | comment | added | Dmitri Pavlov | @Theo: Here is an argument that establishes a bijection for an arbitrary paracompact manifold. Suppose M: C(M)→R is a morphism of rings. Restricting to functions that are nonvanishing only on a finite number of connected components, we immediately find a unique point p∈M such that the evaluation at p coincides with the restriction of M. Now if f∈C(M) is an arbitrary function and g is the characteristic function of the connected component of p, then M(f)=M(f)·1=M(f)·g(p)=M(f)·M(g)=M(f·g)=(f·g)(p)=f(p)·g(p)=f(p)·1=f(p). Hence M is given by the evaluation at p on the entire ring C(M). | |
| Jan 23, 2012 at 20:16 | comment | added | Dmitri Pavlov | @Theo: I don't think that your claim is actually true. Would you mind elaborating on the construction of “extra” points? | |
| Jan 23, 2012 at 18:09 | comment | added | Theo Johnson-Freyd | That said, it's easy to fix the "problem" with too-big manifolds. Let $C(M,A)$ denote the ring of $A$-valued smooth functions on $M$, where $A$ is a topological $\mathbb R$-algebra. Then for any manifold $M$, one can always find a "large enough" $A$ so that the canonical inclusion $M \hookrightarrow \hom_{A{\rm -alg}}(C(M,A),A)$ is an isomorphism. (Unless I think hard about it, to be safe let me mumble some words about "topological $A$-algebras", and ask the hom be taken in that category.) | |
| Jan 23, 2012 at 18:06 | comment | added | Theo Johnson-Freyd | Incidentally, here's a possible reason to ask manifolds to be second countable. Let $M$ be a manifold. One has an inclusion of sets $M \hookrightarrow \hom_{\rm ring}(C(M),\mathbb R)$, where $C(M)$ is the ring of real-valued smooth functions. When $M$ is second countable, this inclusion is an isomorphism of sets. When $M$ is too big, the latter set is "larger" than the former (constructing "extra" points requires a week form of Choice). | |
| Jan 23, 2012 at 12:00 | history | answered | Dmitri Pavlov | CC BY-SA 3.0 |