User dave penneys - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:56:07Z http://mathoverflow.net/feeds/user/351 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras Examples of noncommutative analogs outside operator algebras? Dave Penneys 2009-11-30T07:46:14Z 2012-06-18T14:45:24Z <p>Theo's <a href="http://mathoverflow.net/questions/7095" rel="nofollow">question</a> made me wonder if there are other "noncommutative analogs" outside of operator algebras. Some noncommutative analogs from operator algebras include:</p> <ul> <li>A $C^\ast$-<a href="http://en.wikipedia.org/wiki/C%2A-algebra" rel="nofollow">algebra</a> is a noncommutative <a href="http://en.wikipedia.org/wiki/Topological%5Fspace" rel="nofollow">topological space</a> (cf. the <a href="http://en.wikipedia.org/wiki/Gelfand%5Ftransform" rel="nofollow">Gelfand transform</a>).</li> <li>The <a href="http://en.wikipedia.org/wiki/Multiplier%5Falgebra" rel="nofollow">multiplier algebra</a> of a nonunital $C^\ast$-algebra is the noncommutative <a href="http://en.wikipedia.org/wiki/Stone%2DCech%5Fcompactification" rel="nofollow">Stone-Cech compactification</a>.</li> <li>A <a href="http://en.wikipedia.org/wiki/Spectral%5Ftriple" rel="nofollow">spectral triple</a> is a noncommutative <a href="http://en.wikipedia.org/wiki/Manifold" rel="nofollow">manifold</a> (add some extra data to the spectral triple to get a noncommutative <a href="http://en.wikipedia.org/wiki/Riemannian%5Fmanifold" rel="nofollow">Riemannian manifold</a> cf. arXiv:0810.2088).</li> <li>A <a href="http://en.wikipedia.org/wiki/Von%5FNeumann%5Falgebra" rel="nofollow">von Neumann algebra</a> is a noncommutative <a href="http://en.wikipedia.org/wiki/Measure%5Fspace" rel="nofollow">measure space</a>.</li> </ul> <p>Are there any other good examples? If you know more in operator algebras, that's great too.</p> <p>EDIT: these algebras should be considered as various functions spaces for noncommutative spaces as per @Yemon's answer. I'm going to leave the above text as is unless there are requests for another edit.</p> http://mathoverflow.net/questions/93265/does-such-an-infinite-index-subgroup-exist Does such an infinite index subgroup exist? Dave Penneys 2012-04-05T23:18:17Z 2012-04-10T21:55:10Z <p>Notation: If $G$ is a countable group and $H$ is a subgroup, for $g\in G$, let $|\mathcal{O}_{gH}|$ be the size of the $H$-orbit of $gH$ in the $H$-set $G/H$. </p> <p>Does there exist a countable group $G$ and a subgroup $H$ with $[G\colon H]=\infty$ such that:</p> <ol> <li>There is a $g\in G$ with $|\mathcal{O} _{gH}|\neq |\mathcal{O} _{g^{-1}H}|$,</li> <li>$|\mathcal{O} _{gH}|=\infty$ if and only if $|\mathcal{O} _{g^{-1}H}|=\infty$, and</li> <li>There is a constant $M>0$ such that $\displaystyle \frac{|\mathcal{O} _{gH}|}{|\mathcal{O} _{g^{-1}H}|} \leq M$ for all $g\in G$ with $|\mathcal{O} _{gH}|,|\mathcal{O} _{g^{-1}H}|\neq \infty$.</li> </ol> <p>For instance, 2 and 3 above would be satisfied if:</p> <ul> <li>If $|\mathcal{O} _{gH}|\neq |\mathcal{O} _{g^{-1}H}|$, then $0&lt; |\mathcal{O} _{gH}|,|\mathcal{O} _{g^{-1}H}|\leq M$ for some $M>1$ independent of $g$.</li> </ul> http://mathoverflow.net/questions/93265/does-such-an-infinite-index-subgroup-exist/93684#93684 Answer by Dave Penneys for Does such an infinite index subgroup exist? Dave Penneys 2012-04-10T17:07:53Z 2012-04-10T21:55:10Z <p>I believe (1 and 2) and (3) are mutually exclusive. Here is a proof:</p> <p>First, the commensurator $$Comm_G(H) = \{g\in G : |\mathcal{O} _{gH}|, |\mathcal{O} _{g^{-1}H}|&lt;\infty\}$$ is a group. We will show:</p> <blockquote> <p>Lemma: $\varphi\colon Comm_G(H)\to \mathbb{Q}_{>0}$ by $g\mapsto \displaystyle\frac{[H\colon H\cap gHg^{-1}]}{[H\colon H\cap g^{-1}Hg]}$ is a homomorphism.</p> </blockquote> <p>From the lemma, if we assume there is a $g\in Comm_G(H)$ such that $\varphi(g)=x>1$ (criteria 1 and 2), then the order of $g$ must be infinite, since $x>1$ implies $x^n>1$ for all $n\geq 1$. Since the order of $g$ is infinite, criterion 3 cannot hold since eventually $\varphi(g^n)=\varphi(g)^n=x^n>M$ for any $M>0$.</p> <p>Proof of the lemma:</p> <p>We must show $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$. Define the following constants:</p> <ul> <li>For $i=1,2$, $a_i = [H\colon H\cap g_iHg_i^{-1}]$ and $b_i = [H\colon H\cap g_i^{-1}Hg_i]=[g_iHg_i^{-1}\colon H\cap g_iHg_i^{-1}]$</li> <li>$a=[H\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b=[(g_1g_2)H(g_1g_2)^{-1}\colon H\cap (g_1g_2)H(g_1g_2)^{-1}]$</li> </ul> <p>Note that since $x\mapsto g_1xg_1^{-1}$ is an automorphism of $G$, we have:</p> <ul> <li>$a_2=[g_1Hg_1^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$ and $b_2=[(g_1g_2)H(g_1g_2)^{-1}\colon g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}]$</li> </ul> <p>Now look at the subgroup $K=H\cap g_1 Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}$, and define</p> <ul> <li>$a_1'=[H\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$</li> <li>$a_2'=[H\cap g_1Hg_1^{-1}\colon K]$</li> <li>$b_1'=[g_1Hg_1^{-1}\cap (g_1g_2)H(g_1g_2)^{-1}\colon K]$</li> </ul> <p>which are all finite, since if we have a quadrilateral of groups $L_1\cap L_2\subset L_1,L_2\subset G$, we must have $[L_1\colon L_1\cap L_2] \leq [G\colon L_2]$. Now since index is multiplicative, we have</p> <ul> <li>$a a_1'=a_1a_2'$</li> <li>$ba_1' = b_2 b_1'$</li> <li>$a_2b_1'=b_1a_2'$</li> </ul> <p>Solving for $a$ and $b$, we get $$\frac{a}{b} = \frac{a_1a_2'}{a_1'}\frac{a_1'}{b_2b_1'}=\frac{a_1a_2'}{b_2b_1'}.$$ Now note that $\displaystyle \frac{a_2'}{b_1'}=\frac{a_2}{b_1}$, so we have $$\varphi(g_1g_2)=\frac{a}{b} = \frac{a_1}{b_2}\frac{a_2}{b_1}= \frac{a_1}{b_1}\frac{a_2}{b_2}=\varphi(g_1)\varphi(g_2).$$</p> http://mathoverflow.net/questions/69115/restriction-on-the-coefficients-for-an-operator-in-the-free-group-factor-l-mat/69134#69134 Answer by Dave Penneys for Restriction on the coefficients for an operator in the free group factor $L(\mathbb{F}_2)$ Dave Penneys 2011-06-29T18:47:28Z 2011-06-29T18:47:28Z <p>For convenience, let's identify $L(G)$ with its image in $\ell^2(G)$ as per @Matthew Daws' answer. For $f=\sum_{g\in G} \mu_g L_g\in\ell^2(G)$, we have $f\in L(G)$ if and only if $f* \xi\in \ell^2(G)$ for all $\xi\in \ell^2(G)$, where $*$ is convolution. Another way of saying this is that $L(G)$ is all $\ell^2$-sums which define bounded operators on $\ell^2(G)$ by convolution.</p> <p>A good reference for this is Vaughan Jones' <a href="http://www.math.berkeley.edu/~vfr/MATH20909/VonNeumann2009.pdf" rel="nofollow">course notes/book</a> on von Neumann algebras.</p> http://mathoverflow.net/questions/67590/representing-tensor-c-categories-in-bim/67669#67669 Answer by Dave Penneys for representing tensor-C*-categories in BIM Dave Penneys 2011-06-13T15:08:10Z 2011-06-13T15:08:10Z <p>I don't know about necessary conditions, but here are some results concerning sufficient conditions:</p> <ul> <li>In MR1749868, Hayashi and Yamagami realize amenable $C^*$-tensor categories in the category of bifinite (Jones index) bimodules of the hyperfinite $II_1$-factor.-</li> <li>In arXiv:0811.1764v4, Stefaan Vaes and Sébastien Falguières showed that "the representation category of any compact group is the [bifinite] bimodule category of a $II_1$-factor," i.e., given a compact group $G$ with representation category $C$, there is a $II_1$-factor $M$ whose category of bifinite bimodules is exactly $C$.-</li> <li>Recently, Sven Raum and Sébastien Falguières showed that "all finite $C^*$-tensor categories are [bifinite] bimodule categories of a $II_1$-factor," i.e., given a finite $C^*$-tensor category $C$, there is a $II_1$-factor $M$ whose category of bifinite bimodules is exactly $C$. This paper has yet to appear on the arXiv. Here is the <a href="http://genco-2011.institut.math.jussieu.fr/program.html" rel="nofollow">link</a> to the conference at which the talk was given.-</li> </ul> <p>I don't know of such results for type $III$ factors.</p> http://mathoverflow.net/questions/66087/is-there-a-fusion-category-with-an-object-which-does-not-commute-with-its-dual Is there a fusion category with an object which does not commute with its dual? Dave Penneys 2011-05-26T18:17:02Z 2011-05-26T19:24:11Z <p>Does there exist a fusion category with an object $X$ such that $XX^*\ncong X^*X$ (where the isomorphism need not be natural in any way)?</p> <p>Feel free to add adjectives such as pivotal, spherical, unitary, etc.</p> http://mathoverflow.net/questions/58656/what-is-the-subfactor-planar-algebra-of-type-tildea-n-of-index-4/58665#58665 Answer by Dave Penneys for What is the subfactor planar algebra of type $\tilde{A}_n$, of index 4? Dave Penneys 2011-03-16T17:28:50Z 2011-03-16T17:28:50Z <p>Yes, it is two dimensional, and this is allowed. It just means the planar algebra is not irreducible. I don't know of anyone that has thought about a presentation by generators and relations of this planar algebra yet. </p> <p>One issue here is that since $d=[M\colon N]^{1/2}=2$ is not generic ($>2$), one has to be careful about the annular multiplicities of the subfactor (arXiv:math/0105071). So I don't know if the planar algebra qualifies as "annular multiplicities $*10$" like (extended) Haagerup.</p> http://mathoverflow.net/questions/52674/invertibility-of-the-planar-algebra-subfactor-correspondence/52872#52872 Answer by Dave Penneys for Invertibility of the planar algebra-subfactor correspondence Dave Penneys 2011-01-22T21:34:51Z 2011-01-23T01:06:07Z <p>Yes, strongly amenable subfactors of the hyperfinite $II_1$-factor are completely classified by their standard invariant. The finite depth case was done by Popa's Classification of subfactors: the reduction to commuting squares (MR1055708), and the infinite depth case was finished by Popa's Classification of amenable subfactors of type $II$ (MR1278111). The reconstruction theorem in this case reproduces a hyperfinite $II_1$-subfactor.</p> <p>The Guionnet-Jones-Shlyakhtenko construction reproduces an inclusion of interpolated free-group factors (arXiv:0911.4728), and there is a specific formula for which factors you get. So you need to start with the right factors (modulo the free-group factor isomorphism problem...).</p> <p>EDIT: Noah's answer makes a really important point. I should point out that at index 6, Bisch, Nicoara, and Popa constructed an uncountable family of (non-amenable) subfactors of the hyperfinite $II_1$-factor with the same standard invariant with property (T) (MR2314611). As they say in the abstract:</p> <blockquote> <p>We exploit the fact that property (T) groups have uncountably many non-cocycle conjugate cocycle actions on the hyperfinite $II_1$ factor.</p> </blockquote> <p>For a discrete group, if you're amenable and you have property (T), then you're finite. For a subfactor, if you're amenable and you have property (T), then you're finite depth. So once you're in the infinite depth property (T) setting, there's no hope for a bijective correspondence between subfactors and planar algebras.</p> http://mathoverflow.net/questions/43683/subfactor-summer-reading-list/43744#43744 Answer by Dave Penneys for Subfactor summer reading list Dave Penneys 2010-10-27T00:52:39Z 2010-10-27T00:52:39Z <p>So it really depends on why you want to learn about subfactors. I'll try and give different reading lists based on different motivation.</p> <p>The basics of $II_1$-subfactors:</p> <p>If you're familiar with $II_1$-factors, then "Introduction to subfactors" (MR1473221) is a good place to start. Of course, Jones' original "Index for subfactors" (MR696688) is an enjoyable read as well. A more advanced and comprehensive treatment is Evans and Kawahigashi's "Quantum symmetries on operator algebras" (MR1642584). </p> <p>Classification:</p> <p>When mathematicians encounter a family of mathematical objects, we feel the need to classify them. Subfactors were first (and still!) classified by their principal graphs, which is the principal part of the Bratteli diagram of the <strong>standard invariant</strong>, or tower of relative commutants (see JS, EK, or Bisch's "Bimodules and higher relative commutants" MR1424954). The first classifications (index $\leq 4$) were completed by Jones, Ocneanu (MR996454) (very hard to find this source), and Popa (MR1278111). Popa showed that "amenable" subfactors of the hyperfinite $II_1$-factor (the index is the norm squared of the principal graph) are completely classified by their principal graphs.</p> <p>Axiomatization:</p> <p>There are several axiomatizations for the standard invariant of a subfactor: Ocneanu's <strong>paragroups</strong> (see EK), Popa's $\lambda$-<strong>lattices</strong> (MR1334479), and Jones' <strong>planar algebras</strong> (see section below on planar algebras). These three different axiomatizations play together nicely, and it's good to have an overview on what's really going on here. Unfortunately, there is no good unified source for this. Yet. However, you should think of it this way:</p> <p>The standard invariant (or "representation theory") of a finite index $II_1$-subfactor $N\subset M$ is a unitary $2$-category with $2$ $0$-morphisms called $N$ and $M$, $1$-morphisms given by various bimodule summands of the basic constructions of $N\subset M$, and $2$-morphisms given by various intertwiner spaces. This is what Ocneanu calls a "paragroup" because it resembles the tensor category of representations of a finite group. This $2$-category is unitary, has nice duals, and satisfies Frobenius reciprocity, and other cool stuff as well. In particular, we can draw planar diagrams to represent different $2$-morphisms in the spaces. So this $2$-category has the structure of a planar algebra. For this category theory stuff, see Mueger's "From subfactors to categories and topology I" (MR1966524). In many cases, we can recover the special $2$-category from <strong>connections</strong> on a bipartite graph or on a <strong>commuting square</strong> (see EK, JS, Popa).</p> <p>Reconstruction:</p> <p>Popa proved that one can start with he standard invariant of a subfactor and reconstruct a subfactor with the same standard invariant (MR1334479). In the "strongly amenable" case (a bit more technical than "amenable"), you get a subfactor of the hyperfinite $II_1$-factor. You can also do the reconstruction completely planar algebraically. This result is due to Guionnet-Jones-Shlyakhtenko (arXiv:0712.2904), and a really easy version to understand was given by Jones-Shlyakhtenko-Walker (arXiv:0807.4146) (Kodiyalam and Sunder also have a version of this).</p> <p>Planar Algebras:</p> <p>If you want to know what a planar algebra is, see Peters' construction of the Haagerup subfactor planar algebra (arXiv:0902.1294), or Morrison-Peters-Snyder "Skein-theory for the $D_{2n}$ planar algebras" (MR2559686). If you want to know how a subfactor actually gives a planar algebra, see the first section of Jones-Penneys (arXiv:1007.3173), which relies on some proofs in Jones' "Planar Algebras I" (arXiv:math/9909027). </p> <p>Examples:</p> <p>A great class of examples is the Bisch-Haagerup subfactors (MR1386923) which are just slightly harder than group-subgroup examples (see JS). Some of the most important examples rely on the above reconstruction theorems. For example, there are the exotic Haagerup and Asaeda-Haagerup subfactors (MR1686551) and the composite Fuss-Catalan subfactors (MR1437496).</p> <p>I'll keep updating this post. Right now I have office hours. Sections to come include:</p> <p>Type III, Recent results</p> http://mathoverflow.net/questions/39968/can-the-minimal-index-of-a-subfactor-take-all-values-in-4cos2pi-nn3-4-5/39982#39982 Answer by Dave Penneys for Can the minimal index of a subfactor take all values in {4cos^2(pi/n);n=3,4,5,...} u [4,infinity]? Dave Penneys 2010-09-25T22:52:44Z 2010-09-25T22:52:44Z <p>There is an irreducible Temperley-Lieb subfactor at every allowed index. For $n\geq 3$, it has index $4\cos^2(\pi/n)$ and principal graph $A_{n-1}$ (in fact all subfactors of index less than $4$ are irreducible), and for every $r\geq 4$, it has index $r$ and principal graph $A_\infty$. Doesn't that do the job by your remark?</p> http://mathoverflow.net/questions/786/hochschild-cyclic-homology-of-von-neumann-algebras-useless Hochschild/Cyclic Homology of von Neumann Algebras: Useless? Dave Penneys 2009-10-16T19:28:36Z 2010-08-08T10:59:12Z <p>Hochschild homology gives invariants of (unital) $k$-algebras for $k$ a unital, commutative ring. If we let our algebra $A$ be the group ring $k[G]$ for $G$ a finite group, we get group homology. There are plenty of other connections to homological algebra. If we use cyclic homology, there are connections to geometry and topology involving the Chern character.</p> <p>Von Neumann algebras are complex algebras, so we can take their Hochschild and cyclic homologies. When I have asked experts in the fields of von Neumann algebras and non-commutative geometry about what you get, I usually hear some approximation of the following: "There's also analysis in von Neumann algebras, so I wouldn't expect an algebraic invariant like Hochschild or cyclic homology to tell you anything useful."</p> <p>Although this answer makes some sense, I find it very displeasing and cryptic. Why shouldn't it tell you something? Is there some way to make "it doesn't tell you anything" quantitative? Is there an example of a von Neumann algebra with nontrivial Hochschild or cyclic homology (different from that of the complex numbers)?</p> <p>EDIT: After reading the responses so far, I should specify that I really want to know if there is a $II_1$-factor with nontrivial Hochschild or cyclic (co)homology.</p> http://mathoverflow.net/questions/31513/when-does-a-matrix-define-a-convolution-operator-on-a-hypergroup When does a matrix define a convolution operator on a hypergroup? Dave Penneys 2010-07-12T07:53:41Z 2010-07-14T01:22:59Z <p>Let $H$ be a discrete hypergroup. Suppose I have a matrix $A=(A_{x,y})$ indexed over $H$ with nonnegative entries which defines a bounded operator on $\ell^2(H)$. When does there exist $f\in\ell^1(H)$ such that $A_{x,y}=\langle f*\delta_x,\delta_y\rangle$, i.e., $A$ is the matrix of transition probabilities for a random walk given by convolution with $f$?</p> <p>A necessary condition is that $A$ commutes with $\ell^1(H)$ convolution on the right. Is this sufficient?</p> http://mathoverflow.net/questions/946/operator-valued-weights Operator Valued Weights Dave Penneys 2009-10-17T21:54:29Z 2010-05-31T22:23:13Z <p>One of the basic tools in subfactors is the conditional expectation. If $N\subset M$ is a $II_1$-subfactor (or an inclusion of finite factors), then there is a unique trace-preserving conditional expectation of $M$ onto $N$. This should be thought of as a (Banach space) projection of norm 1. In fact, it is the restriction of the Jones projection $e_N$ on $L^2(M)$ to $M$. In the finite index case, we get another conditional expectation (Jones projection...) from the basic construction $M_1=\langle M, e_N\rangle$ onto $M$.</p> <p>In his thesis, Michael Burns showed that if we iterate the basic construction in the infinite index case, we only get half the conditional expectations (we only get the odd Jones projections). The other half of the time, we get a generalization of the conditional expectation called an operator valued weight, originally defined by Haagerup.</p> <p>Given an inclusion of semifinite von Neumann algebras $(N, tr_N)\subset (M, tr_M)$, there is a unique normal, faithful, semi-finite trace-preserving operator valued weight $T\colon M_+\to \widehat{N_+}$, where we must take the "extended part" of the positive cone $N_+$ of $N$. </p> <p>Edit as per @Dmitri's answer: Let $$n_T=\{x\in M| T(x^\ast x)\in N_+\}$$ and set $$m_T=n_T^\ast n_T=span\{x^\ast y| x,y\in n_T\}.$$ There is a natural extension of $T$ to $m_T$. Is there an example of a normal, faithful, semifinite operator valued weight such that </p> <ul> <li>$N$ is not contained in $T(m_T)$, and/or</li> <li>$1\notin T(M_+)$?</li> </ul> <p>What about when $M$ and $N$ are factors ($M$ is $II_\infty$)?</p> http://mathoverflow.net/questions/605/ideals-in-factors Ideals in Factors Dave Penneys 2009-10-15T15:58:51Z 2010-04-18T18:47:55Z <p>One can easily prove that factors have no nontrivial ultraweakly closed 2-sided ideals as these are equivalent to nontrivial central projections. One can also show type $I_n$, type $II_1$, and type $III$ factors are algebraically simple (any 2-sided ideal must contain a projection. All projections are comparable in a factor, so you can show 1 is in the ideal). Ideals in $B(H)$ ($\dim(H)=\infty$, $H$ separable) have been studied extensively. What about ideals in $II_\infty$ factors?</p> <p>One might think, since every $II_\infty$ factor $M$ can be written as $N\overline{\otimes} B(H)$ for $N$ a $II_1$ factor, if $I\subset B(H)$ is an ideal, then $N\otimes I$ is a 2-sided ideal. This is false. One needs to take the ideal generated by $N\otimes I$. What does that mean from a von Neumann algebra viewpoint? Is it the same as taking the norm closure? </p> <p>We can also describe some ideals in terms of the trace. One has the equivalent of the Hilbert-Schmidt operators: $$I_2=\{x\in M | tr(x^\ast x)&lt;\infty\}$$ and the trace class operators: $$I_1=\{x\in M | tr(|x|)&lt;\infty\}=I_2^\ast I_2 =\left\{\sum^n_{i=1} x_i^\ast y_i | x_i, y_i\in I_2\right\}.$$ What is the relation of $I_j$ to $N\otimes L^j(H)$ for $j=1,2$ (where $L^2(H)$ is the Hilbert-Schmidt operators and $L^1(H)$ is the trace class operators in $B(H)$)? Is $I_j$ the norm closure of $N\otimes L^j(H)$?</p> http://mathoverflow.net/questions/21401/how-do-you-make-a-good-math-research-poster-for-a-non-mathematical-audience How do you make a good math research poster for a non-mathematical audience? Dave Penneys 2010-04-14T23:05:49Z 2010-04-15T13:02:31Z <p>I have the opportunity to prepare a research poster for a non-mathematical, yet scientifically savvy audience, and I want to do it well. I have asked a few mathematicians, and I have heard the following sound advice:</p> <ul> <li>Use interesting graphics.</li> <li>Elaborate on possible applications to other scientific fields.</li> </ul> <p>Although easier for applied mathematicians, this will be ok as I study subfactors, which have connections to quantum physics and statistical mechanics, and planar algebras, which provide great graphics. But there are practical questions as well:</p> <ul> <li>How can I use LaTeX to make a poster?</li> <li>How can I avoid using mathematical symbols and technical language?</li> </ul> <p>What makes a good math research poster? What are some good ways to target a non-mathematical audience? Does anyone have examples and/or templates using LaTeX? What other advice would you give a mathematician who has never made a research poster before?</p> <p>A few asides:</p> <ul> <li>I'm not sure if this is community wiki. I'm more than happy to click the box if requested.</li> <li>Feel free to retag this question as you see fit.</li> <li>I will answer my own question after the poster presentation, and I will have all of my materials available online.</li> </ul> http://mathoverflow.net/questions/3150/non-commutative-geometry-from-von-neumann-algebras Non-commutative geometry from von Neumann algebras? Dave Penneys 2009-10-28T21:48:34Z 2010-03-08T13:29:03Z <p>The Gelfand transform gives an equivalence of categories from the category of unital, commutative $C^*$-algebras with unital $*$-homomorphisms to the category of compact Hausdorff spaces with continuous maps. Hence the study of $C^*$-algebras is sometimes referred to as non-commutative topology.</p> <p>All diffuse commutative von Neumann algebras acting on separable Hilbert space are isomorphic to $L^\infty[0,1]$. Hence the study of von Neumann algebras is sometimes referred to as non-commutative measure theory.</p> <p>Connes proposed that the definition of a non-commutative manifold is a spectral triple $(A,H,D)$. From a $C^*$-algebra, we can recover the "differentiable elements" as those elements of the $C^*$-algebra $A$ that have bounded commutator with the Dirac operator $D$. </p> <p>What happens if we start with a von Neumann algebra? Does the same definition give a "differentiable" structure? Is there a way of recovering a $C^*$-algebra from a von Neumann algebra that contains the "differentiable" structure on our non commutative measure space? This would be akin to our von Neumann algebra being $L^\infty(M)$ for $M$ a compact, orientable manifold (so we have a volume form). Or are von Neumann algebras just "too big" for this?</p> <p>One of the reasons I am asking this question is Connes' spectral characterization of manifolds (<a href="http://arxiv.org/abs/0810.2088v1" rel="nofollow">arXiv:0810.2088v1</a>) which shows we get a "Gelfand theory" for Riemannian manifolds if the spectral triples satisfy certain axioms. Connes starts with the von Neumann algebra $L^\infty(M)$ instead of the $C^*$-algebra $C(M)$.</p> http://mathoverflow.net/questions/14888/compact-hausdorff-and-c-algebra-objects-in-a-category/14951#14951 Answer by Dave Penneys for Compact Hausdorff and C^*-algebra "objects" in a category. Dave Penneys 2010-02-10T22:03:37Z 2010-02-10T22:03:37Z <p>The "Bohrification" paper arXiv:0905.2275 may be relevant to Question 4. As I understand, they discuss the notion of $C^\ast$-algebra objects in a given topos.</p> http://mathoverflow.net/questions/10480/gelfand-duality-in-ncg/12998#12998 Answer by Dave Penneys for Gelfand duality in NCG Dave Penneys 2010-01-26T00:47:03Z 2010-01-27T00:19:32Z <p>@Meyer has provided a very good answer, so I just want to add something from my notes from Professor Rieffel's $C^\ast$-algebra class about the primitive ideal space, which is a canonical topological space associated to a $C^\ast$-algebra (in fact, it is associated to any normed $*$-algebra, but let's keep it "easy"...).</p> <p>Let $A$ be a $C^\ast$-algebra.</p> <blockquote> <p>Definition: An ideal $I\subset A$ is called primitive if it is the kernel of an irreducible representation. Denote the set of all primitive ideals by $Prim(A)$.</p> </blockquote> <p>Note that $Prim(A)$ is always a set as it is a subset of $P(A)$, the power set of $A$. An important theorem:</p> <blockquote> <p>If $A$ is GCR, then the map $\widehat{A}\to Prim(A)$ by $(\pi, H)\mapsto \ker(\pi)$ is a bijection (where $\widehat{A}$ is the "set" of equivalence classes of irreducible representations). </p> </blockquote> <p>This is far from the case if $A$ is NCR (there is some set theory here, such as the term "unclassifiable," that I don't want to get into as I am not a set theorist), but $Prim(A)$ is still a set, so we still get a topological space using the following fact:</p> <blockquote> <p>All primitive ideals are prime. </p> </blockquote> <p>The space of prime ideals of $A$ comes with the Jacobson, or "hull-kernel" topology, so we get the relative topology on $Prim(A)$.</p> <p>A few facts:</p> <ul> <li>In general, $Prim(A)$ is not Hausdorff, but it is $T_0$ (see @Meyer's comment for a counterexample).</li> <li>$Prim(A)$ is locally compact.</li> <li>If $A$ is separable, $Prim(A)$ has the Baire Category Property.</li> </ul> <p>Once again, this is all from a course I took from Professor Rieffel. I hope it helps!</p> http://mathoverflow.net/questions/3468/do-torsion-free-groups-give-projectionless-group-c-ast-algebras Do torsion-free groups give projectionless group ($C^\ast$) algebras? Dave Penneys 2009-10-30T20:07:07Z 2010-01-21T19:14:49Z <p>One of the reasons I study von Neumann algebras is that they always have plenty of projections. There are many projectionless $C^\ast$-algebras ($0$ and possibly $1$ are the only projections), but the von Neumann algebras they generate must have nontrivial projections (unless it's just the complex numbers, of course). A good example of this is the reduced group $C^\ast$-algebra of any free group $F_n$. If $n=1$, then $C_r^\ast(Z)\cong C(S^1)$ via the Gelfand transform, which is clearly projectionless. If $n\geq 2$, the proof is fairly complicated. See <a href="http://www.amazon.com/Algebras-Example-Fields-Institute-Monographs/dp/0821805991" rel="nofollow">Davidson's book</a> for a proof when $n=2$.</p> <p>If $G$ is a torsion-free group, is the reduced group $C^\ast$-algebra of $G$ projectionless? This $C^\ast$-algebra always contains the group algebra $C[G]$, so a simpler question is whether $C[G]$ is projectionless if $G$ is torsion-free.</p> <p>Note that torsion-free is a necessary condition as one gets a projection from summing up the elements in the cyclic group generated by a torsion element and dividing by the order of the element.</p> <p>EDIT: changed typestting. still some bugs... help please?</p> http://mathoverflow.net/questions/10186/what-are-some-interesting-sequences-of-functions-for-thinking-about-types-of-conv/10188#10188 Answer by Dave Penneys for What are some interesting sequences of functions for thinking about types of convergence? Dave Penneys 2009-12-30T22:38:19Z 2009-12-30T22:38:19Z <p>The sequence $(f_n)$ given by $f_n=n^{-1}\chi_{[-n,n]}$ converges uniformly, but not in $L^1$.</p> http://mathoverflow.net/questions/532/pimsner-popa-bases Pimsner-Popa Bases Dave Penneys 2009-10-15T00:16:56Z 2009-12-30T07:15:30Z <p>Let $N\subset M$ be a finite index $II_1$-subfactor. Let $B=\{b_i\}$ be a finite orthonormal (Pimsner-Popa) basis for $M$ over $N$. Let $d=[M\colon N]^{1/2}$. It is well known that $B_1=\{d b_{i_1} e_1 b_{i_2}\}$ is a (not necessarily orthonormal) basis for $M_1$ over $N$, where $M_1=\langle M, e_1\rangle$ is the basic construction of $N\subset M$ (for example, see <a href="http://www.ams.org/mathscinet-getitem?mr=1424954" rel="nofollow">Bisch, Bimodules and higher relative commutants</a>, page 31). Under what conditions are we assured $B_1$ is orthonormal? orthogonal?</p> <p>Integer index always works (see the Pimsner-Popa paper), and a variation always works in infinite index (with a definition of orthonormal basis due to Burns).</p> http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras/7260#7260 Answer by Dave Penneys for Examples of noncommutative analogs outside operator algebras? Dave Penneys 2009-11-30T11:37:24Z 2009-11-30T11:37:24Z <p>I remembered a few more from operator algebras:</p> <p>a trace is like a noncommutative integral. This has lead to:</p> <ul> <li>The Schatten $p$-ideals</li> <li>If $M$ is a $II_1$-factor, then one forms the noncommutative $L^P$-spaces (see this <a href="http://en.wikipedia.org/wiki/Lp%5Fspace" rel="nofollow">link</a>) $L^p(M)$ as the completion of $M$ with the norm $$tr(|x|^p)^{1/p}$$ for $1\leq p &lt;\infty$ and as $M$ with the operator norm for $p=\infty$. Then we get the duality $L^1(M)^\ast=L^\infty(M)$, and since the trace is finite, we have, as in the finite measure space case, that $$M=L^\infty(M)\subset L^2(M)\subset L^1(M).$$ I believe we have $L^p(M)\subset L^q(M)$ if $p>q$, but I haven't checked it. Is it also true that $L^p(M)^*=L^q(M)$ if $p^{-1}+q^{-1}=1$? I have never checked this either.</li> </ul> http://mathoverflow.net/questions/7095/which-is-the-correct-ring-of-functions-for-a-topological-space/7109#7109 Answer by Dave Penneys for Which is the correct ring of functions for a topological space? Dave Penneys 2009-11-29T04:17:17Z 2009-11-29T06:41:10Z <p>As an operator algebraist, I think the space of continuous functions to $\mathbb{C}$ which vanish at infinity is my preferred choice. Let me tell you why.</p> <p>One of the basic ideas of noncommutative topology/geometry (and probably algebraic geometry, but I don't know much about that) is that we can trade the space for algebras of functions on that space. This is afforded by the <a href="http://en.wikipedia.org/wiki/Gelfand%5Frepresentation" rel="nofollow">Gelfand transform</a>. The spectrum of a commutative $C^\ast$-algebra is the space of characters, i.e., $\ast$-algebra homomorphisms to $\mathbb{C}$.</p> <ul> <li>If $X$ is compact Hausdorff, then the spectrum of $C(X)$ is $X$. </li> <li>If $X$ is locally compact Hausdorff, but not compact, the spectrum of the non-unital $C^\ast$-algebra $C_0(X)$ is $X$. The spectrum of the unitalization of $C_0(X)$ ($C_0(X)\oplus \mathbb{C}$) is the one point compactification of $X$. The spectrum of the unital $C^\ast$-algebra $C_b(X)$ is $\beta X$, the Stone-Cech compactification of $X$. One should note that $C_b(X)$ is also the <a href="http://en.wikipedia.org/wiki/Multiplier%5Falgebra" rel="nofollow">multiplier algebra</a> of $C_0(X)$.</li> <li>If $X$ is compact, but not Hausdorff, then $C(X)$ corresponds to some type of "Hausdorffization" of $X$.</li> </ul> <p>Actually $C(X)$ and $C_0(X)$ are the <a href="http://en.wikipedia.org/wiki/Vanish%5Fat%5Finfinity" rel="nofollow">same</a> if $X$ is compact, but you want to denote it $C(X)$ to emphasize the fact that the algebra is already unital. Otherwise, when you add a unit, you take the one point compactification of a compact space which adds an extra point, which is not what you want.</p> <p>Now let's suppose you have some additional structure, like $X$ is a compact manifold. Then you probably want the $C^\infty$-functions on $X$. However, these can be recovered from $C(X)$ as those operators whose iterated commutator with the Dirac operator is bounded. This inspired the notion of a <a href="http://en.wikipedia.org/wiki/Spectral%5Ftriple" rel="nofollow">spectral triple</a>.</p> <p>EDIT: In my haste to answer this question, I made some mistakes in the earlier answer as pointed out by @Jonas.</p> http://mathoverflow.net/questions/6781/ade-type-dynkin-diagrams/6801#6801 Answer by Dave Penneys for ADE type Dynkin diagrams Dave Penneys 2009-11-25T13:05:52Z 2009-11-25T13:05:52Z <p>They classify principal graphs of $II_1$-subfactors with index less than $4$. The principal graph can be $A_n$, $D_{2n}$, $E_6$, or $E_8$, but $D_{odd}$ and $E_7$ do not occur.</p> http://mathoverflow.net/questions/6647/do-subgroups-have-two-sided-bases Do subgroups have "two sided bases"? Dave Penneys 2009-11-24T03:34:32Z 2009-11-24T18:39:27Z <p>Let $H\leq G$ be an inclusion of finite groups. Define a map $E\colon \mathbb{C}[G]\to \mathbb{C}[H]$ to be the $\mathbb{C}$-linear extension of $$E(g)=\begin{cases} g &amp;\text{if } g\in H\\ 0 &amp;\text{else,} \end{cases}$$ i.e., $E$ is the projection onto $\mathbb{C}[H]$. A finite subset $B\subset \mathbb{C}[G]$ will be called a left basis for $G$ over $H$ if $$x=\sum\limits_{b\in B} b E(b^\ast x)$$ for all $x\in \mathbb{C}[G]$, where $\ast$ is the anti-linear extension of the map $g\mapsto g^{-1}$. For an example, take $B$ to be a set of left-coset representatives. Similarly, we can define a right basis to be a finite subset $B\subset \mathbb{C}[G]$ such that $$x=\sum\limits_{b\in B} E(x b^\ast)b$$ for all $x\in\mathbb{C}[G]$.</p> <p>Note that there exist groups for which there is a basis which is both a left and right basis, but $H$ is not a normal subgroup of $G$. One can take the subgroup of the symmetric group $S_n$ ($n\geq 3$) which fixes $1$. Then a set of left and right coset representatives is given by $$\{ (1 j)|j=1,\dots,n\}.$$ Does there always exist a basis which is both a left and right basis, or are there inclusions of groups for which there is no simultaneous left and right basis?</p> <p>The motivation for this question is another question from subfactor theory: if $N\subset M$ is a finite index, extremal $II_1$-subfactor, does there always exist a Pimsner-Popa basis which is both a left and right basis? The subgroup subfactor is an example of such a subfactor, and the question posed above is a watered-down version of the subfactor question, where perhaps an answer is already known or more easily obtainable.</p> http://mathoverflow.net/questions/6180/why-are-fusion-categories-interesting Why are fusion categories interesting? Dave Penneys 2009-11-19T21:52:58Z 2009-11-21T22:11:03Z <p>In the same vein as <a href="http://mathoverflow.net/questions/544/why-are-subfactors-interesting" rel="nofollow">Kate</a> and <a href="http://mathoverflow.net/questions/2046/how-do-i-describe-a-fusion-category-given-a-subfactor" rel="nofollow">Scott</a>'s questions, why are fusion categories interesting? I know that given a "suitably nice" fusion category (which probably means adding adjectives such as "unitary," "spherical," and "pivotal"), we get a subfactor planar algebra which, in turn, gives us a subfactor. Also, I vaguely understand that these categories give us Turaev Viro TQFTs.</p> <p>What else do fusion categories do? What's a good reference for the Turaev Viro stuff?</p> http://mathoverflow.net/questions/6200/what-is-to-quantize-something/6206#6206 Answer by Dave Penneys for What is to Quantize something? Dave Penneys 2009-11-20T02:00:00Z 2009-11-20T02:00:00Z <p>In mathematics, quantization often refers to some kind of deformation of a classical object. The <a href="http://en.wikipedia.org/wiki/Uncertainty%5Fprinciple" rel="nofollow">Heisenberg Uncertainty Principle</a> says that the position and momentum operators do not commute. In fact, $[X,P]=i\hbar$. In the limit as $\hbar\to 0$, these operators commute once again. Technically speaking, this is nonsense as $\hbar$ is a universal constant, but in mathematics, we are free to play with parameters. A couple of examples include:</p> <ul> <li>the <a href="http://en.wikipedia.org/wiki/Quantum%5Ftorus" rel="nofollow">noncommutative torus</a>, the universal $C^\ast$-algebra generated by two unitaries satisfying $uv=e^{i\theta} vu$. As $\theta\to 0$, we get $C(\mathbb{T}^2)$, the continuous functions on the $2$-torus. We usually think of the deformed algebra as a quantization of the commutative one.</li> <li>some <a href="http://en.wikipedia.org/wiki/Quantum%5Fgroup" rel="nofollow">quantum groups</a> are deformations of universal enveloping algebras, i.e., we get the universal enveloping algebra as $q\to 1$.</li> </ul> http://mathoverflow.net/questions/5299/measurable-functions-and-unbounded-operators-in-von-neumann-algebras/5679#5679 Answer by Dave Penneys for Measurable functions and unbounded operators in von Neumann algebras Dave Penneys 2009-11-16T06:26:57Z 2009-11-16T06:26:57Z <p>I think your question should be as follows:</p> <p>Given a von Neumann algebra $M$, can we define a canonical set $S$ such that</p> <ul> <li>if $M=L^\infty(X)$ acting on $L^2(X)$, then $S$ is (isomorphic to) the set of a.e. defined measurable functions, and</li> <li>if $M=B(H)$ acting on $H$, then $S$ is the <strong>densely defined, closed</strong> unbounded operators?</li> </ul> <p>If we take this slight alteration of the question, then I believe the answer is the closed affiliated operators. Here's a sketch of a proof which I think should work, but you should check the details just to make sure.</p> <p>Let $A=L^\infty(X)$. If $f$ is an a.e. defined measurable function, define as in my other answer $$D(M_f)={ \xi\in L^2(X) | f\xi\in L^2(X)}.$$ Then $M_f\colon D(M_f)\to L^2(X)$ is closed and affiliated with $A$.</p> <p>Now suppose $T$ is a closed, densely defined operator affiliated to $A$ acting in $L^2(X)$ (abbreviated $T\eta A$). Then we can do polar decomposition to get $T=U|T|$ where $U\in A$ and $|T|\eta A$. Hence we have reduced to the case where $T$ is positive and self adjoint. Since $T\eta A$, we must have that $f(T)\in A$ for all bounded Borel functions $f$. In particular, for $n\geq 1$, $T_n=\chi_{[0,n]}(T)\in A$, and $T_n$ increases to $T$. It should be clear how to proceed now to get that $T$ is multiplication by an a.e. defined measurable function.</p> http://mathoverflow.net/questions/5547/ubiquity-importance-of-path-algebras/5552#5552 Answer by Dave Penneys for ubiquity, importance of path algebras Dave Penneys 2009-11-14T19:01:28Z 2009-11-14T19:01:28Z <p>One idea comes to mind for number two: <a href="http://mathoverflow.net/questions/4648/when-to-pick-a-basis/4652#4652" rel="nofollow">Bases</a>! If we have a unital inclusion of finite dimensional semi-simple complex algebras, the isomorphisms with the path algebras fix bases and make our lives easier. </p> http://mathoverflow.net/questions/5299/measurable-functions-and-unbounded-operators-in-von-neumann-algebras/5409#5409 Answer by Dave Penneys for Measurable functions and unbounded operators in von Neumann algebras Dave Penneys 2009-11-13T18:25:59Z 2009-11-14T02:52:27Z <p>It is not always the case that the composite of two operators is nice. In general, the composite of $S$ and $T$ is defined by $$D(ST)={\xi\in D(T) | T\xi\in D(S)}.$$ There is no reason to expect this space to be dense.</p> <p>I agree with @Yemon. You want the affiliated operators. Here's an interesting parallel for finite measure spaces that I heard in a mini-course from Ozawa. Just as $$L^\infty(X,\mu)\subset L^2(X,\mu)$$ which is a subset of all measurable functions, if $(M,tr)$ is a finite von Neumann algebra, $$M=L^\infty(M,tr)\subset L^2(M,tr)\subset \eta(M),$$ the affiliated operators. The inclusion for $L^2(M)$ into $\eta(M)$ is given by $$\xi\mapsto (L_\xi\colon m\mapsto \xi m),$$ (be careful - you need to take the closure of $L_\xi$. Clearly $L_\xi$ commutes with right multiplication by $U(M)$) and the image of $L^2(M)$ is the set of all $T\in \eta(M)$ such that if $$|T|=\int t dE(t)$$ is the usual spectral measure, then $$\int t^2 d tr(E(t)) &lt;\infty.$$</p> <p>About your comment in response to @Yemon: Indeed the trace works. If you have a positive function, use indicator functons to show the trace is semi-finite and faithful (see <a href="http://www.amazon.com/Theory-Operator-Algebras-Non-Commulative-Geometry/dp/354042248X/ref=sr%5F1%5F1?ie=UTF8&amp;s=books&amp;qid=1258133847&amp;sr=1-1" rel="nofollow">Takesaki I</a>). Normal is clear. If you have a measurable function $f$ on $(X,\mu)$, construct a multiplication operator by letting $$D(M_f)={\xi\in L^2(X,\mu)| f\xi \in L^2(X,\mu)}.$$ If $\mu$ is finite, and $f$ is a measurable, real valued, a.e. finite function, then $M_f$ is self-adjoint, and the spectrum is the essential range of $f$.</p> <p><a href="http://www.amazon.com/Functional-Analysis-Methods-Mathematical-Physics/dp/0125850506/ref=sr%5F1%5F1?ie=UTF8&amp;s=books&amp;qid=1258133324&amp;sr=1-1" rel="nofollow">Reed and Simon</a> have a good introduction to unbounded operators, in particular the spectral theorem.</p> <p>Somehow the brackets aren't showing up for the sets above, but I hope it's legible.</p> http://mathoverflow.net/questions/93265/does-such-an-infinite-index-subgroup-exist/93287#93287 Comment by Dave Penneys Dave Penneys 2012-04-09T18:12:25Z 2012-04-09T18:12:25Z And aren't the indices the same as $[H\colon U]=(10)(12)=120=(24)(5)=[H\colon V]$? http://mathoverflow.net/questions/93265/does-such-an-infinite-index-subgroup-exist/93287#93287 Comment by Dave Penneys Dave Penneys 2012-04-09T14:53:09Z 2012-04-09T14:53:09Z Thanks for your answer Mark! I've been working through the example, and 2 and 3 still seem non-trivial (especially 3). For example, $[H\colon H\cap t^2 H t^{-2}]$ seems highly dependent on how $t$ moves $U\cap V$ (and thus depends on the choice of isomorphism $U\cong V$). Are there any good references for the general properties of HNN extensions and why 2 and 3 should hold? Thanks again! http://mathoverflow.net/questions/10186/what-are-some-interesting-sequences-of-functions-for-thinking-about-types-of-conv/10188#10188 Comment by Dave Penneys Dave Penneys 2011-07-01T09:11:31Z 2011-07-01T09:11:31Z It's not like the third one, which only converges almost uniformly, not uniformly. http://mathoverflow.net/questions/42147/name-my-cat-regular-categories-where-inverse-images-also-have-right-adjoint Comment by Dave Penneys Dave Penneys 2010-11-23T19:12:26Z 2010-11-23T19:12:26Z +1 for the title. http://mathoverflow.net/questions/43594/would-a-supersymmetric-theory-of-von-neumann-algebras-be-useful/43648#43648 Comment by Dave Penneys Dave Penneys 2010-10-27T16:37:11Z 2010-10-27T16:37:11Z do you mean &quot;non-commutative&quot; in paragraph 1? http://mathoverflow.net/questions/39968/can-the-minimal-index-of-a-subfactor-take-all-values-in-4cos2pi-nn3-4-5/39982#39982 Comment by Dave Penneys Dave Penneys 2010-09-26T20:16:40Z 2010-09-26T20:16:40Z Every standard invariant that arises from a finite index type $II_1$ subfactor also arises as the standard invariant of a type $III$ subfactor, and vice versa. See Izumi's paper &quot;On type II and type III principal graphs of subfactors.&quot; http://mathoverflow.net/questions/19987/math-paper-authors-order Comment by Dave Penneys Dave Penneys 2010-04-01T17:19:48Z 2010-04-01T17:19:48Z Temperley-Lieb is a pretty famous counterexample. And why is Perron-Frobenius not Frobenius-Perron? http://mathoverflow.net/questions/10480/gelfand-duality-in-ncg/12998#12998 Comment by Dave Penneys Dave Penneys 2010-01-27T00:18:52Z 2010-01-27T00:18:52Z thanks! i will edit the answer to refer to your comment as a counterexample. http://mathoverflow.net/questions/10480/gelfand-duality-in-ncg/12998#12998 Comment by Dave Penneys Dave Penneys 2010-01-26T23:36:31Z 2010-01-26T23:36:31Z Doesn't the primitive ideal space of $B_0(H)$ consist of two points $\\{\\{0\\},B_0(H)\\}$, as the trivial representation is irreducible as well? http://mathoverflow.net/questions/7095/which-is-the-correct-ring-of-functions-for-a-topological-space/7255#7255 Comment by Dave Penneys Dave Penneys 2009-11-30T17:35:48Z 2009-11-30T17:35:48Z @Andrew - most of the non locally compact spaces that I deal with on a day to day basis are infinite dimensional topological vector spaces as you point out. But in this case, wouldn't the &quot;correct&quot; ring of functions be the continuous linear maps? What are some other examples? http://mathoverflow.net/questions/7250/examples-of-noncommutative-analogs-outside-operator-algebras Comment by Dave Penneys Dave Penneys 2009-11-30T07:52:41Z 2009-11-30T07:52:41Z somehow the hyperlinks have a mind of their own... help please? http://mathoverflow.net/questions/7095/which-is-the-correct-ring-of-functions-for-a-topological-space/7109#7109 Comment by Dave Penneys Dave Penneys 2009-11-29T06:45:40Z 2009-11-29T06:45:40Z @Jonas - you have every reason to harp away. somehow my brain is not working clearly right now. i edited again. i think i need to get off of MO right now and get some sleep. hopefully i haven't made any more ridiculous errors. please point them out if you see them. http://mathoverflow.net/questions/7095/which-is-the-correct-ring-of-functions-for-a-topological-space/7109#7109 Comment by Dave Penneys Dave Penneys 2009-11-29T06:13:26Z 2009-11-29T06:13:26Z @Jonas - thank you. don't know what i was thinking... http://mathoverflow.net/questions/7110/monotone-lipschitz-embedding Comment by Dave Penneys Dave Penneys 2009-11-29T05:29:10Z 2009-11-29T05:29:10Z Can you supply a reference? Is Aharoni's proof constructive? Can you give an example of such an embedding? http://mathoverflow.net/questions/6647/do-subgroups-have-two-sided-bases Comment by Dave Penneys Dave Penneys 2009-11-24T18:40:09Z 2009-11-24T18:40:09Z yes, thanks. i fixed it.