most recent 30 from http://mathoverflow.net 2013-05-26T07:40:24Z http://mathoverflow.net/feeds/user/22260 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93049/countably-complete-filters Douglas Somerset 2012-04-03T22:54:48Z 2012-07-15T23:22:00Z

<p>Is there any description of the set of countably complete filters on the lattice of dense $G_{\delta}$ subsets of a compact, second countable metric space? [I haven't just dreamt this up: it describes a space of ideals that I am looking at in a C*-algebra.]</p>

http://mathoverflow.net/questions/97118/about-subspaces-of-f-spaces/97702#97702 Douglas Somerset 2012-05-22T21:34:59Z 2012-05-22T21:34:59Z

<p>A completely regular space is an $F'$-space if disjoint cozero sets have disjoint closure. Every $F$-space is an $F'$-space and every normal $F'$-space is an $F$-space. There is an example (1.10) in Alan Dow, 'On $F$-spaces and $F'$-spaces', Pac. J. Math., 108 (1983) 275-284 of a locally compact $F'$-space which is not an $F$-space but which is an open subset of a compact $F$-space.</p>

http://mathoverflow.net/questions/96445/iterating-ultrapowers-of-c-algebras-the-calkin-algebra Douglas Somerset 2012-05-09T12:58:10Z 2012-05-15T20:56:16Z

<p><a href="http://mathoverflow.net/questions/95699/iterating-ultrapowers-of-c-algebras" rel="nofollow">Elsewhere</a> I asked about ultrapowers of the C*-algebra $A$ of compact operators on separable infinite-dimensional Hilbert space. My question was whether the process of taking ultrapowers of ultrapowers ever stabilizes. </p> <p>It was pointed out that the definition of an ultrapower of a C*-algebra is slightly different from the usual definition of an ultrapower; that the process never stabilizes in the sense that the canonical embedding is never an isomorphism; but that nevertheless (assuming CH) that if $A^1$ is an ultrapower of $A$ then the ultrapowers of $A^1$ are isomorphic as C*-algebras to $A^1$.</p> <p>This last statement depends on the fact $A$ is separable. Here I would like to ask about the corresponding question for the C*-algebra $B$ of bounded operators on Hilbert space and also for the Calkin algebra $C=B/A$. Does the process of taking ultrapowers of ultrapowers of $B$ ever stabilize in the sense of producing isomorphic C*-algebras? And the same question for $C$?</p>

http://mathoverflow.net/questions/96591/f-spaces-and-points-whose-complements-are-c-embedded Douglas Somerset 2012-05-10T17:07:14Z 2012-05-10T21:22:17Z

<p>Let $X$ be a compact Hausdorff space. If $X$ is extremally disconnected then the complement of every point $x$ in $X$ is C*-embedded in $X$ (i.e every continuous bounded real-valued function on $X\setminus\{x\}$ extends to a continuous function on $X$) because every open subset of $X$ is C*-embedded [Gillman and Jerison, Rings of Continuous Functions; 1H].</p> <p>If $X$ is basically disconnected then the complement of every non-P-point $x$ is C*-embedded in $X$ (because $x$ lies in the boundary of a cozero set and cozero sets in basically disconnected spaces are C*-embedded and have clopen closure [Gillman and Jerison; 1h, 14.25]).</p> <p>My question is whether the complement of every non-P-point in a compact F-space is C*-embedded (an F-space is a Hausdorff space in which disjoint cozero sets are contained in disjoint zero sets).</p> <p>In the opposite direction, I am interested in spaces in which the complement of every non-P-point is C*-embedded. Can this occur in a non-F-space?</p>

http://mathoverflow.net/questions/95699/iterating-ultrapowers-of-c-algebras Douglas Somerset 2012-05-01T21:56:42Z 2012-05-02T03:01:38Z

<p>Let $A$ be something interesting like the C*-algebra of compact operators on a separable infinite-dimensional Hilbert space and let $A^1$ be an ultrapower of $A$. Then $A^1$ is a primitive C*-algebra strictly containing $A$ (Ge and Hadwin). Now let $A^2$ be an ultrapower of $A^1$. Then $A^2$ is a primitive C*-algebra containing $A^1$, presumably strictly. In the same way we may define $A^3$, $A^4$, etc. Since $A^n$ contains $A^{n-1}$ we may define an inductive limit $A^{\omega}$, and then continue transfinitely with <code>$A^{{\omega} +1}$</code> etc.</p> <p>My question is, Does the process ever stabilise? If so does it stabilise at some finite step, or at $\omega$ or at the first uncountable ordinal?</p>

http://mathoverflow.net/questions/93992/closed-meagre-sets Douglas Somerset 2012-04-13T22:04:15Z 2012-04-30T18:04:07Z

<p>A closed meagre subset of $[0,1]$ is either countable or homeomorphic to the Cantor set: either way it is $0$-dimensional.</p> <p>Q.1. Is every closed meagre subset of an $n$-dimensional locally compact Hausdorff space of dimension $\le n-1$ (for $n\ge 1$)?</p> <p>Q.2. If so, is there anything unusual about meagre sets in $0$-dimensional and infinite-dimensional spaces in which closed meagre sets can have dimension equal to the dimension of the whole space (one would think that meagre sets might be 'fatter' in some sense in these cases)?</p> <p>Q.3. Is there a simple example of an uncountable closed meagre subset of the Cantor set? </p>

http://mathoverflow.net/questions/95237/non-p-points-a-baire-space Douglas Somerset 2012-04-26T08:28:52Z 2012-04-26T16:32:23Z

<p>Let $X$ be a compact Hausdorff space. A P-point in $X$ is a point which does not lie in the boundary of the cozero set of a continuous real-valued function on $X$.</p> <p>Question. Suppose that $X$ has no isolated points. Is the set of P-points a meagre subset of $X$? Failing this, is the set of non-P-points at least a Baire space?</p>

http://mathoverflow.net/questions/95266/properties-of-beta-omega-setminus-omega-minus-the-p-points Douglas Somerset 2012-04-26T15:41:02Z 2012-04-26T15:41:02Z

<p>Let $X=\beta\omega\setminus\omega$ and let $Y=X\setminus P$ where $P$ is the set of P-points in $X$. Then $P$ is dense in $X$ if we assume CH but $P$ may be empty otherwise (Shelah). In the one case $Y$ is nowhere locally compact and in the other case $Y$ is a compact Hausdorff space. Elsewhere I have asked whether spaces like $Y$ (i.e. compact Hausdorff spaces with their P-points removed) are Baire spaces. Here I would like to ask about other properties of $Y$.</p> <p>For example, $Y$ is a completely regular Hausdorff space, come what may. Is $Y$ always normal? Is $Y$ always Lindelof/weakly Lindelof? Is the ring of bounded continuous functions on $Y$ independent of the set theory in which one is working (equivalently, does $X$ always equal $\beta Y$)?</p>

http://mathoverflow.net/questions/94757/closed-subset-of-weakly-lindelof Douglas Somerset 2012-04-21T18:05:09Z 2012-04-21T18:18:16Z

<p>A topological space $X$ is weakly Lindelof if every open cover has a countable subfamily $U$ such that $\bigcup \{ V: V\in U\}$ is dense in X.</p> <p>Question: Are closed subsets of weakly Lindelof spaces necessarily weakly Lindelof?</p>

http://mathoverflow.net/questions/51950/derivation-of-von-neumann-algebra-which-is-zero-on-masa/94328#94328 Douglas Somerset 2012-04-17T21:41:36Z 2012-04-17T21:41:36Z

<p>Sorry to be naive but why can't you just represent $M$ on a separable Hilbert space and take the inner derivation induced by any non-central element in the masa? This maps into $M\subset B(H)$, vanishes on the masa, but is non-trivial on $M$ because the element is non-central. </p>

http://mathoverflow.net/questions/11191/nonhausdorff-dimension/94065#94065 Douglas Somerset 2012-04-14T21:34:47Z 2012-04-14T21:34:47Z

<p>I have been interested in a related problem for a long time. For a C*-algebra $A$, the space Prim($A$) of primitive ideals of $A$ with the hull-kernel topology is seldom Hausdorff and one considers its complete regularization Glimm$(A)$. For primitive ideals $P$ and $Q$ one writes $P\sim Q$ if $P$ and $Q$ cannot be separated by disjoint open sets in Prim$(A)$. With this relation, Prim$(A)$ becomes a graph and then Orc$(A)$ is the supremum of the diameters of connected components of this graph. The case when Orc$(A)$ is finite is reasonably well understood (at least when Prim$(A)$ is compact) but little is known about what can happen when Orc$(A)$ is infinite, which would correspond more with the discussion above. </p> <p>I suppose that my question at this stage is whether the work on non-Hausdorff dimension has now been published; and also in what situations would the Hausdorffization coincide with the complete regularization?</p>

http://mathoverflow.net/questions/94052/where-else-do-the-topology-separation-axioms-turn-up/94063#94063 Douglas Somerset 2012-04-14T21:12:50Z 2012-04-14T21:12:50Z

<p>The property of normality turns up in the ideal theory of C*-algebras and related topics. If $A$ is a separable C*-algebra, or more generally if $A$ is $\sigma$-unital, then the complete regularization of the primitive ideal space of $A$ is $\sigma$-compact and hence normal. This fact has been exploited by several authors recently, and in at least two cases properties have been shown to hold when $A$ is $\sigma$-unital which do not hold for general $A$. For example, Aldo Lazar has shown that two natural topologies coincide on this complete regularization space if $A$ is $\sigma$-unital but not in general; see 'Quotient spaces determined by algebras of continuous functions', Israel J. Math., 179 (2010) 145-155.</p> <p>For me, the most useful property of normal (Hausdorff) spaces is that disjoint sets have disjoint closures in the Stone-Cech compactification.</p>

http://mathoverflow.net/questions/92320/union-of-stone-cech-remainders Douglas Somerset 2012-03-26T23:02:16Z 2012-04-12T09:22:01Z

<p>Can anyone point me to a reference or further information on the following construction? Let $X$ be a compact metric space such as $[0,1]$. Let $A$ be the commutative pre-C*-algebra consisting of [equivalence classes of] continuous bounded functions $f$ defined on cofinite subsets of $X$. Thus two such functions are equivalent if they agree on a cofinite subset of $X$; and addition and multiplication are performed by restricting to some cofinite subset on which both functions are defined and then adding or multiplying in the usual way. With the obvious norm, $A$ is a pre-C*-algebra with completion $B$. I am interested in the maximal ideal space of $B$.</p> <p>Here is one way to describe it. Let $Y$ be the absolute of $X$ with the usual map $h: Y\to X$. Then for $x\in X$, $h$ factors through $\beta(Y\setminus{x})$ so there is a canonical map $h_x: Y\to \beta (Y\setminus {x})$. Define an equivalence relation $*$ on $Y$ by $u*v$ if $h(u)=h(v)$ and $h_x(u)=h_x(v)$ (where $x=h(u)$). Then I think that $Y/*$ is homeomorphic to the maximal ideal space of $B$. </p> <p>One can think of this space as the union of the Stone-Cech remainders at all the points of $X$, hence the title of my question.</p>

http://mathoverflow.net/questions/21820/kernel-projections-in-the-universal-representation/93466#93466 Douglas Somerset 2012-04-07T21:18:30Z 2012-04-07T21:18:30Z

<p>I am coming on this problem two years later and trying to remember things I used to know twenty years ago, but my answer (to Jonas' re-stated question) is that $\ker \pi_{\mu}$ is almost never dense in $\ker\pi_{\mu}''$. The universal representation is the direct sum of all the GNS representations and the kernel of $\pi_{\mu}''$ is everything in $A^{**}$ that arises from other GNS representations. So if $\mu$ is a faithful state, $\ker\pi_{\mu}=0$ but $\ker\pi_{\mu}''$ is generally enormous. The projection $p$ is the central cover of the representation $\pi_{\mu}$ and is discussed in the (older?) standard books on C*-algebras.</p>

http://mathoverflow.net/questions/91871/which-spaces-are-characterized-by-functions-with-compact-support/91934#91934 Douglas Somerset 2012-03-22T17:29:38Z 2012-03-22T17:29:38Z

<p>My naive feeling is that the answer is simply the class of locally compact Hausdorff spaces, for the following reasons. First, for a locally compact Hausdorff space $X$, one can recover $C_0(X)$ from $C_c(X)$ by completion in the uniform norm; and the uniform norm is an algebraic feature because it can be derived from the characters on the ring $C_c(X)$. So for $X$ and $Y$ locally compact Hausdorff, if $C_c(X)$ and $C_c(Y)$ are isomorphic then $X$ and $Y$ are homeomorphic.</p> <p>Now suppose that $X$ is any completely regular (Hausdorff) space (and surely it is topological spaces in this class that we are interested in). Then $X$ is the disjoint union of an open locally compact subset $X_0$ consisting of points which have a compact neighbourhood and a closed subset $X_1$ consisting of points which do not. Each $f\in C_c(X)$ vanishes on $X_1$ so we are not going to get any information about $X_1$ from $C_c(X)$. </p> <p>So my feeling is that $C_c(X)$ determines $X_0$ up to isomorphism but gives no information about $X_1$. </p>

http://mathoverflow.net/questions/91869/closed-set-and-z-ultrafilter-on-normal-space Douglas Somerset 2012-03-21T22:51:03Z 2012-03-22T01:57:56Z

<p>Let $X$ be a completely regular, Hausdorff topological space and let $\cal F$ be a $z$-ultrafilter on $X$. Then for each zero set $W$ in $X$, either $W\in \cal F$ or there exists $Z\in \cal F$ such that $Z$ does not meet $W$ (this is the $z$-ultrafilter property). Now suppose that $X$ is additionally normal. Then is it true that for every closed set $W$ in $X$ either $W$ contains an element $Z$ of $\cal F$ or there exists $Z\in \cal F$ such that $Z$ does not meet $W$?</p>

http://mathoverflow.net/questions/84054/a-variant-of-the-stone-weierstrass-theorem/91687#91687 Douglas Somerset 2012-03-20T04:49:15Z 2012-03-21T22:04:59Z

<p>Three thoughts on this. The first is that $A$ probably has to be assumed unital to guarantee that $T$ is compact. </p> <p>Assuming then that $A$ is unital, each point $t\in T$ corresponds to a maximal ideal $M_t$ of $C$ which generates a closed two-sided ideal $G_t$ in $A$. The ideals ${G_t: t\in T}$ are called the Glimm ideals (after James Glimm who used them in the case when $A$ is a von Neumann algebra). For each element $a\in A$, the mapping $t\mapsto \Vert a+G_t\Vert$ is upper semi-continuous but not in general continuous. Indeed these norm funcions are all continuous if and only if the 'complete regularisation' map from the primitive ideal space of $A$ with the hull kernel topology to $T$ is an open map (R-Y Lee, 1970s). The second thought, therefore, is that a necessary condition for the answer to the question to be yes is that the complete regularisation map should be open.</p> <p>Even when the complete regularisation map is open, I expect that one can find examples where the answer to the question is no, although no such example comes to mind just now [in fact, see E. Kirchberg, S.Wassermann, Operations on continuous bundles of C*-algebras, Math. Ann. 303 (1995), 677-697]. The third thought, however, is that Blanchard showed that if $A$ is separable and exact and the complete regularisation map is open then such a $B$ can be found (E. Blanchard, Subtriviality of continuous fields of nuclear C*-algebras, J. Reine Angew. Math. 489 (1997), 133-149).</p>

http://mathoverflow.net/questions/104141/maximal-ideals-of-the-rings-of-baire-one-functions Douglas Somerset 2013-03-22T22:43:59Z 2013-03-22T22:43:59Z

Isn't $Ba_1(X)$ a commutative C*-algebra (in which case it is certainly a PM-ring)?

http://mathoverflow.net/questions/102621/characterizing-posets-by-functions-into-natural-numbers/102695#102695 Douglas Somerset 2012-07-20T23:03:20Z 2012-07-20T23:03:20Z

@Nik. What I had in mind amounts to taking finite subsets of the set of prime numbers whereas you are allowing arbitary subsets. (Actually in $Z$ shouldn't you be including the zero ideal as a prime ideal, so that $P$ isn't an unordered set?) Your approach seems to jettison the topology that is available if $P$ is a poset of prime ideals. But I suppose that not every poset is a poset of prime ideals of a ring?

http://mathoverflow.net/questions/102621/characterizing-posets-by-functions-into-natural-numbers/102695#102695 Douglas Somerset 2012-07-19T21:37:45Z 2012-07-19T21:37:45Z

Presumably in the case when $P$ is the poset of prime ideals of a ring, $P^∗$ is isomorphic to the lattice of semiprime ideals?

http://mathoverflow.net/questions/93049/countably-complete-filters/100965#100965 Douglas Somerset 2012-06-30T19:06:57Z 2012-06-30T19:06:57Z

Thanks. I will have to think about this. Presumably the definition of a Lindelof element is not quite right?

http://mathoverflow.net/questions/96445/iterating-ultrapowers-of-c-algebras-the-calkin-algebra/97050#97050 Douglas Somerset 2012-05-16T21:59:49Z 2012-05-16T21:59:49Z

Many thanks. Fascinating!

http://mathoverflow.net/questions/96591/f-spaces-and-points-whose-complements-are-c-embedded/96616#96616 Douglas Somerset 2012-05-11T08:02:17Z 2012-05-11T08:02:17Z

Thanks. A more complicated problem than I realised! Is it known what the situation is for Shelah's model with no P-points?

http://mathoverflow.net/questions/95699/iterating-ultrapowers-of-c-algebras/95713#95713 Douglas Somerset 2012-05-02T13:26:33Z 2012-05-02T13:26:33Z

I now understand from Nik's answer that the natural containment is strict at each stage, so in this sense the process never stabilizes. On the other hand, as far as constructing new C*-algebras is concerned, the process 'stabilizes' as soon at the ultrapower is isomorphic as a C*-algebra to its predecessor. So I am very interested in this question of whether $A^1$ and $A^2$ are C*-isomorphic (under CH).

http://mathoverflow.net/questions/93992/closed-meagre-sets/95595#95595 Douglas Somerset 2012-05-02T07:57:04Z 2012-05-02T07:57:04Z

Thanks for this: it gives an easy way of visualizing a meagre subset of $C$.

http://mathoverflow.net/questions/95699/iterating-ultrapowers-of-c-algebras Douglas Somerset 2012-05-02T07:51:41Z 2012-05-02T07:51:41Z

@ Joel. Yes, I was just thinking of external isomorphism. So in my question, I am interested in whether $A^1$ and $A^2$ are isomorphic as C*-algebras.

http://mathoverflow.net/questions/95237/non-p-points-a-baire-space/95271#95271 Douglas Somerset 2012-04-26T17:10:34Z 2012-04-26T17:10:34Z

Great! Many thanks.

http://mathoverflow.net/questions/94757/closed-subset-of-weakly-lindelof/94760#94760 Douglas Somerset 2012-04-21T20:04:08Z 2012-04-21T20:04:08Z

Many thanks. Tick to the first answer (by a minute). Looking at the (non-Hausdorff) examples which motivated this question, I now see that one of them has a non-weakly-Lindelof closed set whose complement is a dense open compact set. Thus the whole space is weakly Lindelof. In fact it has the stronger property that every open cover has a finite subfamily with dense union. Presumably this stronger property cannot occur in a non-compact Hausdorff space?

http://mathoverflow.net/questions/93992/closed-meagre-sets/93994#93994 Douglas Somerset 2012-04-14T07:53:04Z 2012-04-14T07:53:04Z

Thanks for these two answers (Andreas and Anton). I am sorry that I have only one tick to allot. Anton's example looks a useful one to remember.

http://mathoverflow.net/questions/92320/union-of-stone-cech-remainders/92542#92542 Douglas Somerset 2012-04-14T07:31:53Z 2012-04-14T07:31:53Z

Since MathOverflow has been kind enough to re-promote this question, let me raise another issue. In the question I mentioned a dense subalgebra of the C*-algebra of continuous bounded function on this space (i.e continuous bounded functions on cofinite subsets of $X$ - see also KPH's answer). What do the rest of the functions in the C*-algebra look like? They must be continuous except on a co-countable subset $D$ of $X$, but what can $D$ look like? Specifically, if $X=[0,1]$ can $D$ be the set of rational numbers in the $X$?

http://mathoverflow.net/questions/93049/countably-complete-filters/93246#93246 Douglas Somerset 2012-04-06T10:34:02Z 2012-04-06T10:34:02Z

In fact, if we take $X=[0,1]$ then for each subset $Y$ there is a largest countably complete filter of dense $G_{\delta}$s with intersection $Y$ (namely all dense $G_{\delta}$s containing $Y$) and a smallest countably complete filter with intersection $Y$ (namely all co-countable dense $G_{\delta}$s containing $Y$). In the case when $Y=\emptyset$, the question is to describe the countably complete filters of dense $G_{\delta}$s which contain the filter of co-countable sets.

http://mathoverflow.net/questions/92515/does-the-hyperfinite-ii-1-factor-admit-two-irreducible-representations-that-are-n Douglas Somerset 2012-04-05T21:01:17Z 2012-04-05T21:01:17Z

@steven: Does the fact that $\pi_p$ is isomorphic to a subrepresentation of $\pi_q^{\oplus I}$ mean that $\pi_p$ and $\pi_q$ can be approximately intertwined, i.e. approximated by unitarily equivalent irreducible representations?