User owen sizemore - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T10:33:08Z http://mathoverflow.net/feeds/user/5732 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85236/examples-of-monster-groups Examples of "Monster" groups Owen Sizemore 2012-01-09T07:38:08Z 2013-03-05T05:08:37Z <p>I am planning a talk for a general graduate student audience. The topic is exotic examples of countable discrete groups ("monsters"). Some examples of properties that I'm interested in are:</p> <p>1.) Non-amenable groups without free subgroups</p> <p>2.) Groups such that $x^n=e \hspace10pt\forall x$</p> <p>3.) Groups with all proper subgroups cyclic</p> <p>4.) Groups such that every proper subgroup is finite and cyclic of a given order</p> <p>5.) Groups such that every elements has roots of all orders</p> <p>The main source I have been using so far is a survey by Mark Sapir <a href="http://arxiv.org/abs/0704.2899" rel="nofollow">http://arxiv.org/abs/0704.2899</a>.</p> <p>I would like additional sources. Additional properties to the ones above would also be great. Also examples that arise "naturally" (say as a group of symmetries of some nice space rather than a combinatorial construction would be great.) </p> http://mathoverflow.net/questions/119910/how-similar-different-are-dense-subgroups-of-a-compact-group How similar/different are dense subgroups of a compact group. Owen Sizemore 2013-01-26T04:07:41Z 2013-01-28T10:44:29Z <p>Let $\Gamma, \Lambda\subset G$ be countably infinite subgroups of a common compact subgroup G. I am interested in properties that one would have to inherit from the other (ie if $\Gamma$ has this property then so does $\Lambda$.)</p> <p>One initial thing that I thought was that you have a left-right action of $\Gamma\times\Lambda$ on $G$ with the Haar measure. Thus from this you might be able to induce unitary representation from $\Gamma$ to $\Lambda$ much like you would do with measure equivalence and hopefully from this follows some things that we know to be measure equivalence invariants (amenability, property T, etc.) Though I haven't actually verified that this works. Beyond this I can only think of abelian. I would be particularly interested in the case that G is the profinite completion of both $\Gamma$ and $\Lambda$. </p> <p>Alternatively, are there examples where $\Gamma$ and $\Lambda$ can be quite different. </p> http://mathoverflow.net/questions/108093/topics-for-an-undergraduate-expository-paper-in-number-theory Topics for an Undergraduate Expository Paper in Number Theory Owen Sizemore 2012-09-25T19:46:59Z 2012-12-28T01:57:29Z <p>I am teaching an undergraduate course in number theory and am looking for topics that students could take on to write an expository paper (~10 pages). No new results are expected of them. Many of the students have an undergraduate course in abstract algebra and a course in real analysis (but few have any complex analysis background).</p> <p>The only topics that I have come up with are</p> <p>1.) Elliptic curves</p> <p>2.) Cryptography</p> <p>Of course these are related but I think these could be two projects. Other topics (such as the prime number theorem) seem too difficult to me. What other good projects are there? In particular are there good projects based on analysis? References would be greatly appreciated, including references for the two projects above.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/100830#100830 Answer by Owen Sizemore for Blackbox Theorems Owen Sizemore 2012-06-28T02:04:50Z 2012-06-28T02:04:50Z <p>The fact that every von Neumann algebra is the direct integral of factors. Every operator algebraist knows this, and could probably more or less explain the proof, but the details are tedious (and kind of useless in practice.) There are similar facts about decomposition of non-singular actions into ergodic ones and representation into irreducible representations.</p> http://mathoverflow.net/questions/100361/a-question-about-automorphisms-of-ii-1-factors/100396#100396 Answer by Owen Sizemore for A question about automorphisms of $II_1$ factors Owen Sizemore 2012-06-22T21:17:31Z 2012-06-22T21:17:31Z <p>In regards to your original question. It is known that this is not always true. For example, Connes proof that a property (T) factor has countable fundamental group and Out(M)=Aut(M)/Inn(M), is done by taking the quotient as topological groups with both have the pointwise $\|\cdot\|_2$ topology. </p> http://mathoverflow.net/questions/97041/what-is-the-probability-that-two-numbers-are-relatively-prime What is the probability that two numbers are relatively prime? Owen Sizemore 2012-05-15T19:57:36Z 2012-05-15T20:19:47Z <p>The basic question that I have is in the title, but let us make it more rigorous below.</p> <p>Let $N={1, 2, ..., n}$, and put the (normalized) counting measure, $\mu_n$, on $N\times N$. </p> <p>Let $\mathcal{S}_n= { (a, b)\in N\times N: gcd(a, b)=1}$</p> <p>and $x_n=\mu_n(\mathcal{S}_n).$</p> <p>Then what is the assymptotic behavior of $x_n$ as $n\rightarrow\infty$.</p> http://mathoverflow.net/questions/57493/is-there-an-infinite-group-whose-elements-all-have-finite-order/57496#57496 Answer by Owen Sizemore for Is there an infinite group whose elements all have finite order? Owen Sizemore 2011-03-05T20:16:29Z 2011-03-05T20:48:32Z <p>Yes, as per Ryan's comment you can just take an infinite direct sum of finite groups. However the more interesting problem is: are there (infinite) $\textit{finitely generated}$ groups with all elements of finite order? </p> <p>The answer to this was open for a long time, but it is indeed yes. In fact this was known as Burnside's problem</p> <p>The first examples were given by Golod &amp; Shafarevich.</p> <p>There is a lot of info on the wikipedia page</p> <p><a href="http://en.wikipedia.org/wiki/Burnside%27s_problem" rel="nofollow">http://en.wikipedia.org/wiki/Burnside%27s_problem</a> </p> http://mathoverflow.net/questions/50194/irreducible-unitary-representations-of-locally-compact-groups/50210#50210 Answer by Owen Sizemore for Irreducible unitary representations of locally compact groups Owen Sizemore 2010-12-23T00:01:02Z 2010-12-23T00:01:02Z <p>I'm not quite sure if this is the answer that you looking for but anyway he we go. For a locally compact group you are going to generally want to look at strongly continuous representation. By this is mean endow $B(H)$, the bounded operators on a hilbert space $H$ with the topology of point-wise norm convergence. And only consider reps $\pi:G\rightarrow B(H)$ that are continuous with this topology. Now such a rep is unitary if, for every $g\in G$, $\pi(g)$ is a unitary operator.</p> <p>Now the notion of "occur in" that you mention seems to be the notion of strong containment. We say that $\rho:G\rightarrow B(K)$ is strongly contained in $\pi:G\rightarrow B(H)$ if there is a $G$-equivarient unitary operator from $K$ to a closed subspace of $H$. </p> <p>So it now seems that you are asking when does the left regular rep ($L^2(G)$) strongly contain all irreducibles. So yes for compact Lie groups this follows from Peter-weyl this is true.</p> <p>However as soon as you go to something non-compact this might not be true. </p> <p>In fact, there is a much weaker notion known as weak containment of representation. and it is known that $L^2(G)$ weakly contains all irreducible reps if and only if $G$ is amenable. </p> <p>Non-compact Lie groups are in general not amenable, (any groups which contains $\mathbb{F}_2$ the free group on 2 generators is non-amenable) </p> <p>There is much more to be said about this but I think that this should suffice for now</p> http://mathoverflow.net/questions/43537/property-t-and-subgroups-of-finite-index/43540#43540 Answer by Owen Sizemore for Property (T) and subgroups of finite index Owen Sizemore 2010-10-25T16:31:20Z 2010-10-25T16:31:20Z <p>Yes. For groups $H\subset G$, with H a lattice, H has (T) iff G has (T). When both groups are discrete being a lattice is the same as being finite index.</p> <p>Almost every thing you ever need to know about Property (T) can be found here</p> <p><a href="http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf" rel="nofollow">http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf</a></p> http://mathoverflow.net/questions/42512/awfully-sophisticated-proof-for-simple-facts/42531#42531 Answer by Owen Sizemore for Awfully sophisticated proof for simple facts Owen Sizemore 2010-10-17T19:04:52Z 2010-10-17T19:04:52Z <p>The proof that the reduced $C^*$-algebra of the free group has no projections has the nice corollary that the circle is connected. </p> http://mathoverflow.net/questions/40756/averaging-in-von-neumann-algebras/40766#40766 Answer by Owen Sizemore for "Averaging" in von Neumann algebras Owen Sizemore 2010-10-01T15:37:02Z 2010-10-01T15:37:02Z <p>As far as the Dixmier property, there is a relative version that Popa has shown holds for for certain inclusions. (i don't have the paper in front of me and i don't quite remember exactly what he shoes). He also has one for $C^*-algebras.$</p> <p>The papers are</p> <p>S. Popa, The relative Dixmier property for inclusions of von Neumann algebras of ¯nite index, Ann. Sci. Ec. Norm. Sup., 32 (1999), 743-767.</p> <p>and</p> <p>S. Popa, On the relative Dixmier property for inclusions of C¤-algebras, Journal of Funct. Analysis, 171 (2000), 139-154.</p> <p>Secondly, (and i would say more interestingly for me) is the averaging involved in Popa's intertwining by bi-modules technique. Without going into too much detail, if two subalgebras are close in norm on their unit balls, then by averaging, one can get a partial isometry that intertwines a corner of one into a corner of the other. This is an absolutely crucial element of all the progress in the past 8 years or so in the classification of von Neumann algebras coming from groups, or ergodic group actions.</p> <p>For references prob the best place to start would be Popa's ICM talk</p> <p><a href="http://www.math.ucla.edu/~popa/ICMpopafinal.pdf" rel="nofollow">http://www.math.ucla.edu/~popa/ICMpopafinal.pdf</a></p> <p>or appendix C of Stefaan Vaes's Bourbaki seminar.</p> <p><a href="http://arxiv.org/PS_cache/math/pdf/0603/0603434v2.pdf" rel="nofollow">http://arxiv.org/PS_cache/math/pdf/0603/0603434v2.pdf</a></p> http://mathoverflow.net/questions/27609/an-example-of-a-non-amenable-exact-group-without-free-subgroups An example of a non-amenable exact group without free subgroups. Owen Sizemore 2010-06-09T18:59:28Z 2010-09-12T18:03:40Z <p>A countable discrete group $\Gamma$ is said to be exact if it admits an amenable action on some compact space. </p> <p>So clearly amenable groups are exact, but large familes of non-amenable groups are as well.</p> <p>For many of the families that I know of (ex. linear groups, hyperbolic groups) that are exact, they also satisfy the von Neumann conjecture (i.e. that if they are non-amenable then they have subgroup isomorphic to a free group.)</p> <p>So my questions is:</p> <p>Are there examples of exact groups that are non-amenable and do not contain free subgroups?</p> http://mathoverflow.net/questions/35970/construction-of-a-proper-uncountable-subgroup-of-mathbbr-without-choice Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice. Owen Sizemore 2010-08-18T14:03:40Z 2010-08-19T19:02:44Z <p>It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the group generated by the basis or the vector subspace generated by some proper uncountable set of the basis). </p> <p>However, the first step (constructing the basis) requires the axiom of choice.</p> <p>So does anyone know of any proper uncountable subgroup of $\mathbb{R}$ that does not require choice to construct?</p> <p>or is this not possible.</p> <p>Meaning are there models not involving choice where every uncountable subgroup of $\mathbb{R}$ is equal to $\mathbb{R}$. </p> http://mathoverflow.net/questions/35840/the-role-of-completeness-in-hilbert-spaces/35848#35848 Answer by Owen Sizemore for The role of completeness in Hilbert Spaces Owen Sizemore 2010-08-17T06:56:00Z 2010-08-17T06:56:00Z <p>This isn't as much about Hilbert Spaces per se. But generally, the Hilbert spaces that are interesting (to me) and arise naturally (say in quantum mechanics) are infinite dimensional. </p> <p>Now once one tries to do linear algebra on infinite dimensional normed spaces, analysis becomes crucial. It is just a general principle that when doing analysis on some space, you want it to be complete. </p> <p>The point is that you want to use analysis to approximate things, meaning that in order to study things, you instead study approximations, (i.e. sequences that converge to them). So to do this properly you want to know that the sequences which you know should converge (Cauchy seq), actually do converge (Completeness).</p> <p>Note: The requirement that it be complete isn't really a restriction, since any inner product space can be embedded densely into a complete one. </p> http://mathoverflow.net/questions/35778/what-does-the-representation-theory-of-the-reduced-c-algebra-correspond-to/35779#35779 Answer by Owen Sizemore for What does the representation theory of the reduced C*-algebra correspond to? Owen Sizemore 2010-08-16T17:16:30Z 2010-08-17T02:00:36Z <p>So there is a similar property. </p> <p>Now $C^*_r(G)$ is the $C^\star$-algebra generated by the left-regular rep. It a general theorem that if you have a unitary rep $\pi:G\rightarrow \mathcal{U} (H)$, and if $\rho: G\rightarrow \mathcal{U}(K)$ is another unitary rep that is weakly contained ($\rho\prec\pi$) in $\pi$, then there is a surjective map from the reduced $C^\star$-algebra to the algebra generated by $\rho(G)\subset B(K)$</p> <p>So $C^\star_r(G)$ surjects onto all reps that weakly contain the left-regular. </p> <p>Note: $C^\star_r(G)\simeq C^\star(G)$ iff G is amenable. </p> <p>A good source for most of this</p> <p><a href="http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf" rel="nofollow">http://perso.univ-rennes1.fr/bachir.bekka/KazhdanTotal.pdf</a></p> <p>This is the pdf of a book about Property (T). Appendix F.4 is about the above questions but the whole book is of interest for people in operator algebras, representation theory, geometric group theory, and many other fields. </p> <p>EDIT: Another good source, which is directed to Yemon's comment is <a href="http://arxiv.org/PS_cache/math/pdf/0509/0509450v1.pdf" rel="nofollow">http://arxiv.org/PS_cache/math/pdf/0509/0509450v1.pdf</a></p> <p>This is a survey, by Pierre de la Harpe, of groups whose reduced $C^\star$-algebra is simple. </p> http://mathoverflow.net/questions/29087/commutative-subalgebras-of-m-n/30849#30849 Answer by Owen Sizemore for Commutative subalgebras of M_n Owen Sizemore 2010-07-07T03:13:58Z 2010-07-07T03:13:58Z <p>If you are only concerned about commutative subalgebras of $M_n(\mathbb{C})$ then there is a fairly easy characterization. So any abelian algebra is generated by a single self adjoint element (spectral theorem). Call this element T. Then T is diagonalizable and so the algebra it form will be the algebra of polynomials over it. Since it is diagonalizable that is a unitarty $u$ with $uTu^*$ diagonal. And the algebra has dimension $k$ exactly when T has $k$ distinct non-zero eigenvalues. </p> <p>Note: This is assuming that T is invertible. If T is not invertible then the polynomial algebra since it contains the constants will have dimension $k+1$. </p> <p>So then we can view the algebra generated by T as an algebra of the form $u^*Au$ where A is an algebra of diagonal matrices.</p> http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/27242#27242 Answer by Owen Sizemore for Is there an i.c.c. nonamenable simple group that is inner amenable? Owen Sizemore 2010-06-06T14:37:13Z 2010-06-06T14:37:13Z <p>Hey Jon</p> <p>So my initial thought would be no. </p> <p>First, in full generality every group is virtually inner-amenable. Meaning that for any group $G$, the group $G \times \mathbb{Z}/2\mathbb{Z}$ is inner amenable. In fact, any non-icc group is inner amenable just by taking the mean to be the counting measure on a finite conjugacy class, and 0 elsewhere.</p> <p>Even if we restrict to icc groups then, for any icc group $G$, $G\times S_\infty$ (or just choose the second group to be anything inner amenable) is still inner amenable. </p> <p>And because the group is formed as a direct product there is not any way for the generators of $S_\infty$ to sort of "slow down" the growth in the $G$ factor.</p> <p>Now a final way to maybe make something out of this is to ask </p> <p>"If $G$ is inner amenable and along with all of its quotients, then is there a growth contsraint."</p> <p>This will get rid of the examples above. Amenable groups fall into this class, and I would be willing to bet that there are others as well (if anyone knows examples that would be nice) but I can't think of any on the spot.</p> <p>AS for this class.... I have no idea. </p> http://mathoverflow.net/questions/26919/real-analysis-has-no-applications/26934#26934 Answer by Owen Sizemore for Real analysis has no applications? Owen Sizemore 2010-06-03T16:09:56Z 2010-06-03T16:09:56Z <p>While I agree that the paragraph is largely a sales pitch, I think it does hit on something else. It says that real analysis doesn't involve applications to other science. I take this to mean that when you are doing (or studying in a first course) real analysis you don't look at applications to science. This is in contrast to calculas, whereas many of the problems in calculus books are focused on all kinds of problems from classical mechanics and other areas.</p> <p>Just a final note. I thing that Pugh's book is amazing, the best undergrad analysis text out there. Mainly because of the HUGE number of very good problems.</p> http://mathoverflow.net/questions/26758/hausdorff-dimension-of-cayley-graphs-of-groups Hausdorff Dimension of Cayley Graphs of Groups Owen Sizemore 2010-06-01T20:41:49Z 2010-06-02T02:13:17Z <p>I was wondering what has been done concerning the Hausdorff measure of the Cayley graphs of finitely generated countable groups. There are number of issues that would need to be dealt with:</p> <p>1.) By the Cayley graph should we mean just the vertices or the entire 1-skeleton?</p> <p>2.) How could one embed such a graph into $\mathbb{R}^n$?</p> <p>3.) How does the Hausdorff dimension change with a change of generating set?</p> <p>I nice example of somewhere to start would be the Cayley graph of $\mathbb{F}_2$, the free group on 2 generators, with the two generators as the generating set. </p> <p>A nice picture of this, with a possible embedding into Euclidean space can be found at <a href="http://en.wikipedia.org/wiki/File:Cayley_graph_of_F2.svg" rel="nofollow">http://en.wikipedia.org/wiki/File:Cayley_graph_of_F2.svg</a></p> <p>Any comments on this would be very much appreciated.</p> http://mathoverflow.net/questions/23759/a-reference-for-schuberts-theorem A Reference for Schubert's Theorem Owen Sizemore 2010-05-06T18:22:07Z 2010-05-08T17:07:50Z <p>Schubert's Theorem in Knot Theory says that any knot can be uniquely decomposed as the connected sum of prime knots. </p> <p>Unfortunately the original paper is in German. </p> <p>Does anyone know a good english reference for this. Or just the special case of the unknot. (i.e. that the unknot can't be written as the connected sum of two knots which aren't the unknot.) </p> http://mathoverflow.net/questions/19014/finding-questions-between-functional-analysis-and-set-theory/23880#23880 Answer by Owen Sizemore for Finding questions between functional analysis and set theory Owen Sizemore 2010-05-07T17:45:09Z 2010-05-07T17:45:09Z <p>Though this is a little more advanced, there is actually some very exciting research right now at the intersection of descriptive set theory, ergodic theory, and von Neumann algebras. It is quite striking that the three areas have powerful tools for looking at similar problems, and yet tend to be applicable in different cases. For a nice introduction to some of these ideas from a more set-theoretical point of view I would say check out "Topics in Orbit Equivalence" by Kechris and Miller. </p> <p><a href="http://www.springerlink.com/content/0pwfmbrandag/" rel="nofollow">http://www.springerlink.com/content/0pwfmbrandag/</a></p> <p>Here is a link where you can download it (though you might need a subscription but many universities will have it so it should work on a department computer.) It is actually quite elementary, you need some basic descriptive set theory and measure theory, but arrives at quite deep theorems. </p> http://mathoverflow.net/questions/23719/lifting-surjective-von-neumann-algebra-homomorphisms/23751#23751 Answer by Owen Sizemore for Lifting surjective von Neumann algebra homomorphisms Owen Sizemore 2010-05-06T17:19:30Z 2010-05-06T17:19:30Z <p>Yes you can get a $\phi$ that is a homomorphism. Here is a quick sketch. </p> <p>First let $p=sup$ {$p_\alpha,$ projections in $Ker \theta$}. So $p\in Ker \theta$. Furthermore $p\in Z(M)$, the center of M. </p> <p>To see this note that if this were not true then we could find a unitary $u\in M$ with $p\neq upu^\star$. So then $p\wedge upu^\star$ would be a projection in $Ker \theta$ bigger than $p$. </p> <p>From here you can get that $Ker \theta=pMp$, and so we can decompose $M=Ker \theta \oplus M_1$ and $\theta|_{M_1}$ is injective and thus an isomorphism, thus $\phi$ can be chosen to just be the inverse of $\theta$ on $M_1$. </p> <p>Note that if we demand that $\phi$ be unital, this doesn't work and I don't think it can be done in general. I will have to give it more thought.</p> http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23595#23595 Answer by Owen Sizemore for Examples of common false beliefs in mathematics. Owen Sizemore 2010-05-05T15:03:29Z 2010-05-05T15:03:29Z <p>I remember from my first analysis class thinking that if $\mathbb{Q}\subset E\subset\mathbb{R}$ with $E$ open, then $E$ would have to be all of $\mathbb{R}$ (at least more or less, maybe up to countably many points). And once we started measure theory I remember arguing with a friend over it for a good two hours. </p> http://mathoverflow.net/questions/22748/normalizer-of-algebras-and-groups/23050#23050 Answer by Owen Sizemore for normalizer of algebras and groups Owen Sizemore 2010-04-29T22:55:19Z 2010-04-29T22:55:19Z <p>So I don't think it is known in full generallity but there are some partial results. For example if additionally we assume that for any $c,d\in G\setminus H$ the stabilizer subgroups are either equal of noncommensurable then it is true. Much of the results rely on the Pukanzki invarient for $L(H)$ (and if $L(H)$ is cartan then the invarient is {1}), which in some cases you can calculate by the number of left-right cosets. </p> <p>This is mostly from memory, but Sinclair and Smith have a book "Finite von Neumann algebras and MASAS", and there is a chapter about the pukanzki invarient and masas coming from groups. So check that as well as references therein </p> http://mathoverflow.net/questions/129647/unbounded-metrics-on-groups Comment by Owen Sizemore Owen Sizemore 2013-05-04T18:02:19Z 2013-05-04T18:02:19Z Do you mean $\textit{countable}$ group? Otherwise just take the uncountable product of a discrete group. The result is reasonably nice (ie it's a compact topological group). However, the underlying space is not metrizable. http://mathoverflow.net/questions/119910/how-similar-different-are-dense-subgroups-of-a-compact-group Comment by Owen Sizemore Owen Sizemore 2013-01-26T14:55:00Z 2013-01-26T14:55:00Z @Misha: This was basically my motivation. I am familiar with the case of lattices and I was wondering if there was anything similar in this sort of &quot;opposite&quot; situation. http://mathoverflow.net/questions/119910/how-similar-different-are-dense-subgroups-of-a-compact-group Comment by Owen Sizemore Owen Sizemore 2013-01-26T05:56:55Z 2013-01-26T05:56:55Z mmm...seems like my initial thoughts are totally wrong http://mathoverflow.net/questions/101321/injective-von-neumann-algebra Comment by Owen Sizemore Owen Sizemore 2012-07-04T17:36:16Z 2012-07-04T17:36:16Z There is no reason that this subalgebra will be a direct summand http://mathoverflow.net/questions/101260/number-of-projections-in-a-von-neumann-algebra Comment by Owen Sizemore Owen Sizemore 2012-07-04T06:59:05Z 2012-07-04T06:59:05Z @MTS. Yes this was what I was trying to say in my first comment. As soon as there is any non-commutativity there are going to be lots (uncountably many...?) projections. So the least of amount of projections would occur in an abelian algebra. My statement about $I+2$ was stupid. Now that I think it should be that the least amount of projections occurs in $l^\infty(I)$, which should have number of projections equal to the cardinality of the power set of $I$. http://mathoverflow.net/questions/101260/number-of-projections-in-a-von-neumann-algebra Comment by Owen Sizemore Owen Sizemore 2012-07-04T01:59:27Z 2012-07-04T01:59:27Z @MTS. Yes I was not interpreting the questions that way but now that you mention it I think that is what is meant by &quot;$I$ has the minimal cardinality for which it holds&quot;. In that case the cardinality of Proj(A) should be at least cardinality of $I + 2$. I haven't thought through the details, but I'm pretty sure that should be true http://mathoverflow.net/questions/101260/number-of-projections-in-a-von-neumann-algebra Comment by Owen Sizemore Owen Sizemore 2012-07-04T00:06:25Z 2012-07-04T00:06:25Z You should think about this more. Just think of the example where $|I|=2$. So you are asking about unital subalgebras of $M_2(\mathbb{C})$. There depending on which algebra A is there are either the minimal possible for a von Neumann algebra, 2, or as many as uncountably many (which is the maximum if $|I|$ is at most countable. http://mathoverflow.net/questions/100361/a-question-about-automorphisms-of-ii-1-factors/100396#100396 Comment by Owen Sizemore Owen Sizemore 2012-06-23T00:27:43Z 2012-06-23T00:27:43Z Hmm, it's been a while since I looked at that. It is certainly true that you can also do it with the $\|\cdot\|_2$ norm, since that is the appropriate norm for also considering the convergence of the bimodules associated to the automorphisms. http://mathoverflow.net/questions/100361/a-question-about-automorphisms-of-ii-1-factors Comment by Owen Sizemore Owen Sizemore 2012-06-22T15:35:35Z 2012-06-22T15:35:35Z Actually the point wise $\|\cdot\|_2$ topology is really the appropriate topology for studying Aut(M). Because in this topology you can use the Hilbert space structure. For example, Popa's notion of malleable deformation is a one-parameter family that converges to the identity in point wise $\|\cdot\|_2$ http://mathoverflow.net/questions/98911/subfactors-of-lf-infty Comment by Owen Sizemore Owen Sizemore 2012-06-06T02:15:49Z 2012-06-06T02:15:49Z Hi Mike, I don't think this is known. In fact, I don't even think it is known if any finite index subfactor is an interpolated free group factor. The problem is that there isn't alot of ways to prove something is a free group factor. One possibility, which might not be too far out of reach, is to show that if a factor has lots of &quot;free malleable deformations&quot; in the sense of Popa (which is something like a one parameter group of automorphisms that takes a large subalgebra to a free copy of itself) then it is isomorphic to a free product. This still seems somewhat out of reach though. http://mathoverflow.net/questions/91673/is-it-true-that-c-0x-ell-1x Comment by Owen Sizemore Owen Sizemore 2012-03-20T00:35:54Z 2012-03-20T00:35:54Z Do you mean just $l_1(x)$ not $l_1(X^*)$? http://mathoverflow.net/questions/86550/positive-but-not-completely-positive/88309#88309 Comment by Owen Sizemore Owen Sizemore 2012-02-13T01:42:34Z 2012-02-13T01:42:34Z @Yemon: I don't see how your WLOG works. It seems to be saying that the algebra generated by any finite dimensional operator system is finite dimensional. Right? Surely this is not true, just take something in a free product. http://mathoverflow.net/questions/27233/is-there-an-i-c-c-nonamenable-simple-group-that-is-inner-amenable/48106#48106 Comment by Owen Sizemore Owen Sizemore 2011-03-07T20:34:09Z 2011-03-07T20:34:09Z What is $(A_1)^{x_1}$? I have not seen this notation before. http://mathoverflow.net/questions/56213/self-taught-undergrad-math-ordering-of-topics Comment by Owen Sizemore Owen Sizemore 2011-02-21T21:07:27Z 2011-02-21T21:07:27Z Just to comment on 2). It depends on what you mean by topology. If you mean point-set topology (as is often the first seen) then it is much closer to analysis. While algebraic topology is of course closer to algebra. Also manifold theory (differentiable top) you should definitely have familiarity with multi variable calc and linear algebra http://mathoverflow.net/questions/48572/weakly-solid-factors Comment by Owen Sizemore Owen Sizemore 2011-01-12T16:23:42Z 2011-01-12T16:23:42Z Hey Jon, There are weaker forms, such as semi-solidity, or even say prime. Of course these are weakenings of a different form. For instance, semi-solid you restrict which algebras you are taking relative commutants of, while $\Gamma$-solidity eases the requirement of the relative commutant itself. Also I was just wondering, is this notion of $\Gamma$-solid in the literature, i haven't heard of it before.