most recent 30 from http://mathoverflow.net 2013-06-19T02:49:58Z http://mathoverflow.net/feeds/user/3557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16967/is-there-an-additive-model-of-the-stable-homotopy-category Don Stanley 2010-03-03T11:20:10Z 2011-12-13T20:48:55Z

<p>Is there a model category $C$ on an additive category such that its homotopy category $Ho(C)$ is the stable homotopy category of spectra and the additive structure on $Ho(C)$ is induced from that on $C$.</p> <p>Basically I want to add and subtract maps in $C$ without going to its homotopy category. </p> <p>I'm not asking for $C$ to be a derived category or anything like that. Just that it should have an additive structure. </p> <p>As John Palmieri pointed out I should really say what structure I want the equivalence (between $Ho(C)$ and the stable homotopy category) to preserve. Since I do want it to be a triangulated equivalence, Cisinski indicates why this is not possible. </p>

http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils Don Stanley 2010-04-14T12:20:47Z 2011-07-29T19:18:24Z

<p>Suppose that $M$ is a symplectic manifold with a Hamiltonian circle action. Is there a topological Lefschetz pencil on $M$, $f\colon M-A \rightarrow S^2$, such that the fibers are symplectic submanifolds of $M$ and such that the circle action restricts to a Hamiltonian circle action on the fibers?</p> <p>(After reading Tim's answer.)</p> <p>Are there some cases when there do exist such things?</p>

http://mathoverflow.net/questions/23857/what-does-the-classifying-space-of-a-category-classify Don Stanley 2010-05-07T14:22:01Z 2010-05-07T18:27:43Z

<p>A finite group $G$ can be considered as a category with one object. Taking its nerve $NG$, and then geometrically realizing we get $BG$ the classifying space of $G$, which classifies principle $G$ bundles.</p> <p>Instead starting with any category $C$, what does $NC$ classify? (Either before or after taking realization.) Does it classify something reasonable?</p>

http://mathoverflow.net/questions/21709/what-are-the-advantages-of-phrasing-results-in-terms-of-exact-sequences-and-commu/21831#21831 Don Stanley 2010-04-19T10:51:07Z 2010-04-19T10:51:07Z

<p>My answer is not so different from Emerton's, but it's mainly an answer to the poster rather than just the question. </p> <p>I think often it's often because of taste and experience that people use commutative diagrams. I find in my own research I'm often checking that diagrams commute or checking that sequences are exact, by computing with the actual underlying formulas. However I find it easier to state (and remember) things with the diagrams. In terms of intuition diagrams give a different point of view and so may lead to different ideas and understandings of things (which is a good thing). I have you a bigger commutative diagram (say like a cube) saying the diagram is commutative, is easier to understand and phrase. Rewording this in terms of equations would probably make it less clear. A very simple diagram is $A\rightarrow B$, do you feel this is a useful diagram, or would you prefer to have maps only described by equations? That might be hard if you were dealing with abelian categories. </p> <p>In terms of the example of tensor products, I also can understand the actual object better as a more concrete thing, but when you want to construct maps out of it, sometimes it's easier to use its "universal property" (ie that bilinear maps out of the product factor uniquely through it). Once I was siting in on a course on representation theory by a colleague. He was checking that some map existed (and was unique) by checking element by element, but actually all he really had to do was to check that some other map was bilinear. In this case it would have been much easier to use the universal property of the tensor product. On the other hand I do agree that as an object itself it's easier to understand the tensor product as linear combinations modulo some equivalence relation. </p>

http://mathoverflow.net/questions/21171/when-is-a-homology-class-represented-by-a-submanifold/21185#21185 Don Stanley 2010-04-13T06:36:46Z 2010-04-13T06:36:46Z

<p>If you really want a submanifold then I guess you can't always do it. For a closed manifold $M$ consider two times the fundamental class $2[M]$. It's easy to see you can't represent this class as a submanifold when $M=S^1$. Perhaps if you take any class in $a\in H_*(M)$ with nonzero self intersection, then $2a$ can't be represented as a submanifold? </p>

http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3 Don Stanley 2010-03-22T10:28:33Z 2010-03-22T12:35:01Z

<p>Using Alexander duality, you can show that the Klein bottle does not embed in $\mathbb{R}^3$. (See for example <a href="http://www.math.cornell.edu/~hatcher/AT/ATch3.pdf" rel="nofollow">Hatcher's book Chapter 3</a> page 256.) Is there a more elementary proof, that say could be understood by an undergraduate who doesn't know homology yet?</p>

http://mathoverflow.net/questions/3237/japanese-chinese-for-mathematicians/17071#17071 Don Stanley 2010-03-04T09:59:09Z 2010-03-04T09:59:09Z

<p>I use <a href="http://usa.mdbg.net/chindict/chindict.php" rel="nofollow">http://usa.mdbg.net/chindict/chindict.php</a> most for translating Chinese. Although I don't think it has mathematical terms, you can cut and paste whole paragraphs into its "translate" page. </p>

http://mathoverflow.net/questions/16750/resolution-of-a-free-lie-algebra-as-a-module-over-its-universal-enveloping-algebr Don Stanley 2010-03-01T10:31:01Z 2010-03-03T09:41:13Z

<p>Let $L=L(V)$ be a free Lie algebra on a vector space $V$ and $A=T(V)$ the tensor algebra. Make $L$ into a module over $A$ consistent with the formula $a\cdot \alpha=[a,\alpha]$ for $a\in V$ and $\alpha\in L$. </p> <p>What is a canonical resolution of $L$ by free $A$ modules? I'm really most interested in the case where there is a grading and a differential. </p> <p>Edited:</p> <p>After thinking I realize there is the bar construction $B(A,A,L)$. Is there anything smaller in this special case?</p>

http://mathoverflow.net/questions/16779/chas-sullivan-string-topology/16782#16782 Don Stanley 2010-03-01T20:49:18Z 2010-03-01T20:49:18Z

<p>I think this was one of the main motivations for the following paper of McClure.</p> <p>math/0410450 On the chain-level intersection pairing for PL manifolds. J. E. McClure. Geom. Topol. 10 (2006) 1391-1424 and Geom. Topol. 13 (2009) 1775-1777. math.QA (math.GT).</p>

http://mathoverflow.net/questions/16094/reference-for-a-result-on-thick-subcategories-and-t-structures/16139#16139 Don Stanley 2010-02-23T08:13:55Z 2010-02-23T08:13:55Z

<p>First of all I'm assuming that in your first $Ext^j(A,B)$ that you mean $Ext$ in $C$ and not in the heart. </p> <p>Then there's something I don't understand. Isn't $B$ in the thick subcategory? So then in $C/N$, $B=0$. So then in your equation aren't both sides just $0$?</p> <p>I like this concept of being $i$-irrelevant. Can you use it to characterize the codimension of a subvariety?</p>

http://mathoverflow.net/questions/15087/computing-fundamental-groups-and-singular-cohomology-of-projective-varieties/15099#15099 Don Stanley 2010-02-12T12:45:43Z 2010-02-12T12:45:43Z

<p>I believe you can have a elliptic curve and a singular elliptic curve both described by equations of degree three. Some people who know more should be able to answer this. If you change the coefficients I guess the homotopy type can change, just imagine that some subvariety degenerates into something singular and then expands back out to something else on the other side. </p> <p>Even at the algebraic level I think you can have (a minimal set of) polynomials of different degrees defining the same ideal. </p> <p>Another thought (or variant of your question) ... if you start with some subvariety Y and take a hyperplane section X, then X is cut out by the same things that cut out Y together with another linear polynomial. All the cohomology of X is determined except in the middle dimension by the Lefschetz hyperplane theorem. How do the coefficients describing the hyperplane determine this middle cohomology, including the cup products? (I guess the answer is well known?) </p>

http://mathoverflow.net/questions/15083/what-can-be-said-about-the-homotopy-groups-of-a-cw-complex-in-terms-of-its-coho/15096#15096 Don Stanley 2010-02-12T12:08:33Z 2010-02-12T12:08:33Z

<p>If you don't want to make any assumptions about $\pi_1$, then I think the question is hard. Maybe Hurewicz is most of what you can say. If assume something like simply connected, you can say a lot more. There are many things you can say rationally, some of which were pointed bout my Kevin, but even integrally or mod p there are a lot of techniques. Can you be more precise about the setup you are interested in?</p> <p>You can always replace a connected CW complex with a homotopy equivalent one that has a single 0-cell, so this is not really an issue. </p>

http://mathoverflow.net/questions/14629/what-is-the-relationship-between-t-structure-and-torsion-pair/14939#14939 Don Stanley 2010-02-10T19:51:22Z 2010-02-10T20:04:23Z

<p>As I understand it, all torsion classes correspond to t-structures in the way described by Greg, but there is almost always more t-structures than torsion classes (even taking into account the shifts). I think t-structures are closer to some kind of filtration on the abelian category. There's a paper on stability conditions (Gorodentsev, A. L.; Kuleshov, S. A.; Rudakov, A. N. $t$-stabilities and $t$-structures on triangulated categories. (Russian) Izv. Ross. Akad. Nauk Ser. Mat. 68 (2004), no. 4, 117--150; translation in Izv. Math. 68 (2004), no. 4, 749--781) that talks about this a bit. </p>

http://mathoverflow.net/questions/652/homological-algebra-and-calculus-as-in-newton/13633#13633 Don Stanley 2010-02-01T08:02:18Z 2010-02-01T08:02:18Z

<p>There is a whole theory of calculus of functors started by Goodwillie. There are Taylor approximations of functors and so on. Here's the wikipedia page which contains over references:</p> <p><a href="http://en.wikipedia.org/wiki/Calculus_of_functors" rel="nofollow">http://en.wikipedia.org/wiki/Calculus_of_functors</a></p>

http://mathoverflow.net/questions/9146/noncommutative-rational-homotopy-type/13246#13246 Don Stanley 2010-01-28T10:36:16Z 2010-01-28T10:36:16Z

<p>As far as I know for positively graded connected and simply connected complexes with differential of degree $+1$ even in characteristic $0$ (corresponding to the rational homotopy theory case), this is an open problem.</p> <p>So it would be nice to have a published paper that settles it. </p>

http://mathoverflow.net/questions/10771/killing-the-torsion-in-homotopy/13118#13118 Don Stanley 2010-01-27T10:40:50Z 2010-01-27T10:40:50Z

<p>I believe it follows from Theorem 4.4 of "Neeman, Amnon Stable homotopy as a triangulated functor, Invent. Math. 109 (1992), no. 1, 17--40" that this procedure of killing the torsion will work if you invert the prime 2. </p>

http://mathoverflow.net/questions/13005/what-is-formal/13045#13045 Don Stanley 2010-01-26T14:57:56Z 2010-01-26T15:30:32Z

<p>Formal can mean slightly different things in different contexts.</p> <p>A commutative differential graded algebra (CDGA) is formal if it is quasi-isomorphic to it's homology. This is stronger than having all the higher Massey products equal to 0 (I think there are such examples in the Halperin-Stasheff paper). </p> <p>To a space you can associate a CDGA (via Sullivan's $A_{pl}$ functor) which is basically the deRham complex when the space is a manifold. In nice cases this functor induces an equivalence from the rational homotopy category to the homotopy category of CDGA. Quasi-isormorphic CDGA correspond to (rationally) homotopy equivalent spaces. You can also tensor with the reals to get real CDGA. </p> <p>If A is a CDGA which is quasi-isomorphic to $A_{pl}(X)$ for a space $X$ then A is often called a model of X. A space is formal if $A_{pl}$ of it is formal. So a formal space is modeled by its cohomology. In that sense its rational homotopy type is a formal consequence of its cohomology.</p> <p>I think you have to be slightly careful with using $C^*$. This functor lands in differential graded algebra which are not commutative, so possibly the notion of formality could be different. In particular if you consider two CDGA there may be more strings of quasi-isomorphisms between them as DGAs then as CDGAs. I believe it is unknown if two CDGA that are quasi-isomorphic as DGA have to be quasi-isomorphic as CDGA. </p>

http://mathoverflow.net/questions/53595/software-for-rational-homotopy-theory Don Stanley 2011-06-26T18:12:11Z 2011-06-26T18:12:11Z

Did you ask Yves Felix or Pascal? I have heard rumors of a probably like this written in C. Don

http://mathoverflow.net/questions/27144/why-didnt-vladimir-arnold-get-the-fields-medal-in-1974/28215#28215 Don Stanley 2010-06-15T06:50:09Z 2010-06-15T06:50:09Z

It seems like many of the comments about being passed by are from Arnold himself.

http://mathoverflow.net/questions/23857/what-does-the-classifying-space-of-a-category-classify Don Stanley 2010-05-07T21:43:21Z 2010-05-07T21:43:21Z

I don't know the paper of Michael Weiss, however when I was writing the title of this question I had the feeling I have read this title before. So probably I at least read the title of that paper.

http://mathoverflow.net/questions/23857/what-does-the-classifying-space-of-a-category-classify/23863#23863 Don Stanley 2010-05-07T15:23:29Z 2010-05-07T15:23:29Z

I will look at this too.

http://mathoverflow.net/questions/23857/what-does-the-classifying-space-of-a-category-classify/23865#23865 Don Stanley 2010-05-07T15:23:20Z 2010-05-07T15:23:20Z

Thank you. I will look at this.

http://mathoverflow.net/questions/21888/when-is-every-solid-perfect-complex-faithful Don Stanley 2010-04-22T07:05:33Z 2010-04-22T07:05:33Z

Doesn't (4) give an equivalent condition? ie $E$ is solid if and only if for every $f$, $f\otimes E=0$ implies that for some $n$, $f^{\otimes n}=0$. (The converse direction follows since $E\otimes k(p)\not=0$.)

http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils/21339#21339 Don Stanley 2010-04-19T09:56:41Z 2010-04-19T09:56:41Z

I guess there are some dimension restrictions that make my last suggestion not possible in general.

http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils/21339#21339 Don Stanley 2010-04-19T09:32:53Z 2010-04-19T09:32:53Z

Is it possible to do (3) fiberwise over the image of the moment map? So basically constructing the Lefschetz pencil on each symplectic reduction in a way that only changes a &quot;little bit&quot; as you change the values of the moment map. (I wasn't sure if this comment is appropriate for MO or if I should have written you personally.)

http://mathoverflow.net/questions/21331/hamiltonian-circle-actions-and-lefschetz-pencils/21629#21629 Don Stanley 2010-04-19T08:36:19Z 2010-04-19T08:36:19Z

Thanks for your response. I guess it's possible to come up with examples using toric manifolds and various projections of polytopes onto intervals, but this is not something I really know about.

http://mathoverflow.net/questions/21171/when-is-a-homology-class-represented-by-a-submanifold/21185#21185 Don Stanley 2010-04-14T12:31:26Z 2010-04-14T12:31:26Z

Thanks Paul and Tim for letting me know.

http://mathoverflow.net/questions/21171/when-is-a-homology-class-represented-by-a-submanifold Don Stanley 2010-04-13T06:59:11Z 2010-04-13T06:59:11Z

Well it's not exactly the same question. The other question asks what is generated by images of fundamental classes of submanifolds. This question asks what is the image of the fundamental classes of submanifolds.

http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3/19005#19005 Don Stanley 2010-03-22T15:53:15Z 2010-03-22T15:53:15Z

Yup this is really elementary. Thanks!

http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3/18997#18997 Don Stanley 2010-03-22T13:00:52Z 2010-03-22T13:00:52Z

This means that the linking number of the meridian of $M_1$ and $\partial M_2$ is twice the linking number with $M_2$. However we know that this linking number is odd by the link appearing theorem. Maybe you had something more elementary in mind?

http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3/18997#18997 Don Stanley 2010-03-22T13:00:26Z 2010-03-22T13:00:26Z

This is very nice, and I read the proofs on Maehara and the Conway and Gordon paper it refers too. So the proof that $P^2$ doesn't embed in $\mathbb{R}^3$ is elementary. So thanks for the reference. However I don't understand your comment &quot;therefore the two-cells must intersect&quot;. You can isotope the boundary link to something very close to the meridian of $M_2$, but which wraps around it twice.

http://mathoverflow.net/questions/18987/why-cant-the-klein-bottle-embed-in-mathbbr3/18991#18991 Don Stanley 2010-03-22T11:18:08Z 2010-03-22T11:18:08Z

I mean without using some characteristic class theory ...