User greg friedman - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T14:11:24Z http://mathoverflow.net/feeds/user/6646 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product References for Eilenberg-Zilber shuffle product Greg Friedman 2012-10-08T16:51:28Z 2012-10-17T10:21:51Z <p>Most of the treatments I can find in the literature for the Eilenberg-Zilber shuffle product approach it from the point of view of simplicial sets (including the original Eilenberg-MacLane paper). I was wondering if anyone knows of a reference written directly in the language of singular or PL chains. By this I mean that one should show that the (p,q)-shuffles each produce a distinct embedded p+q simplex in $\Delta^p\times \Delta^q$ and that, collectively, this produces a triangulation of $\Delta^p\times \Delta^q$. Furthermore, this should be a chain map, in the sense that the boundary of the triangulation is compatible with the same triangulation process applied to the boundaries (or in other words, the map $S_p(X)\otimes S_q(X)\to S_{p+q}(X\times Y)$ obtained by applying this process is a chain map). Dold constructs this map explicitly in his book, but it's in an exercise and he doesn't provide any proofs of properties (he does helpfully indicate with an asterisk that it is a challenging exercise to do so!)</p> <p>The reason I ask is that I'm working on some lecture notes (for publication) for which I need to have an explicit construction of the homology cross product at the chain level, but the intended audience is not expected to know about simplicial sets. </p> <p>If what I'm looking for does not exist, can anyone recommend the simplicial set treatment that provides the best description of the connection to the explicit geometric triangulation side of things?</p> http://mathoverflow.net/questions/104379/are-open-convex-pl-subsets-of-rn-pl-homeomorphic-to-rn Are open convex PL subsets of R^n PL homeomorphic to R^n? Greg Friedman 2012-08-09T22:58:34Z 2012-08-09T22:58:34Z <p>This is a basic issue of PL topology that I assume must be true, but I can't find a written reference: is a convex open PL subset of $\mathbb R^n$ PL homeomorphic to $\mathbb R^n$? I've scanned through Rourke-Sanderson, Hudson, Zeeman, and Stallings, but can't quite find it in there. I'd appreciate a quotable reference if possible. </p> http://mathoverflow.net/questions/102323/integer-cohomology-of-the-grassman-manifold-of-n-planes-in-r-infty/102329#102329 Answer by Greg Friedman for Integer cohomology of the Grassman manifold of n planes in $R^\infty$ Greg Friedman 2012-07-16T03:37:15Z 2012-07-16T03:37:15Z <p>I don't know if these have everything that you want, but see the following:</p> <p>Brown, Edgar H., Jr. The cohomology of BSOn and BOn with integer coefficients. Proc. Amer. Math. Soc. 85 (1982), no. 2, 283–288.</p> <p>Feshbach, Mark The integral cohomology rings of the classifying spaces of O(n) and SO(n). Indiana Univ. Math. J. 32 (1983), no. 4, 511–516.</p> http://mathoverflow.net/questions/98535/tensor-product-of-acyclic-complexes-of-free-abelian-groups/98542#98542 Answer by Greg Friedman for Tensor product of acyclic complexes of free abelian groups Greg Friedman 2012-06-01T05:18:29Z 2012-06-01T05:46:43Z <p>The homotopy approach works if one of your chain complexes, say $K$, is bounded below (assuming homological indexing). In this case $K$ is chain homotopy equivalent to the $0$ complex (see Munkres, Elements of Algebraic Topology, Theorem 46.2) In fact, let $f: 0\to K$ and $g:K\to 0$ be the unique maps, which must then be the homotopy inverses. In particular then, there must be a degree 1 $D:K\to K$ such that $\text{id}_K=\text{id}_K-gf=\partial D+D\partial$ (so $D$ is the chain contraction). Now consider $\bar D=D\otimes \text{id}_L:K\otimes L\to K\otimes L$. Then for any $x\otimes y\in K\otimes L$, we have \begin{align*} (\partial\bar D+\bar D \partial)(x\otimes y) &amp; =\partial (Dx\otimes y)+\bar D(\partial x\otimes y+(-1)^{|x|}x\otimes \partial y)\newline &amp; =\partial Dx\otimes y+(-1)^{|x|+1}Dx\otimes\partial y+D\partial x\otimes y+(-1)^{|x|}Dx\otimes \partial y\newline &amp; =\partial Dx\otimes y+D\partial x\otimes y\newline &amp; =((\partial D+D\partial)(x))\otimes y\newline &amp; =\text{id}_K(x)\otimes y=x\otimes y \end{align*} So in other words $\bar D$ is a chain contraction of $K\otimes L$</p> <p>Sorry - I don't know how to do aligned equations on here. I'm sure someone will come spruce that up (please?). </p> http://mathoverflow.net/questions/90660/where-can-i-see-the-proof-that-the-homology-groups-of-the-moore-complex-of-a-simp/90663#90663 Answer by Greg Friedman for Where can I see the proof that the homology groups of the Moore Complex of a simplicial group coincide with the homotopy groups of its geometric realization? Greg Friedman 2012-03-09T06:34:07Z 2012-03-09T06:34:07Z <p>There seems to be a proof in Moore's notes on Algebraic Homotopy Theory. There's a copy up on my web site as the first set of links here: <a href="http://faculty.tcu.edu/gfriedman/notes/" rel="nofollow">http://faculty.tcu.edu/gfriedman/notes/</a> The material you want seems to be the beginning of Chapter 2.</p> http://mathoverflow.net/questions/89981/geometric-meaning-of-l-genus/90001#90001 Answer by Greg Friedman for Geometric meaning of L-genus Greg Friedman 2012-03-02T00:31:03Z 2012-03-02T00:31:03Z <p>Hirzebruch himself has a very nice paper explaining (if I remember correctly) how he came up with the signature theorem and why the formulas arise in a fairly reasonable way. Here's the reference: MR0368023 (51 #4265) Hirzebruch, F. The signature theorem: reminiscences and recreation. Prospects in mathematics (Proc. Sympos., Princeton Univ., Princeton, N.J., 1970), pp. 3–31. Ann. of Math. Studies, No. 70, Princeton Univ. Press, Princeton, N.J. 1971.</p> http://mathoverflow.net/questions/89774/chain-homotopy-classes-as-n-homology-of-a-double-complex/89809#89809 Answer by Greg Friedman for Chain Homotopy classes as n-homology of a double complex Greg Friedman 2012-02-28T23:15:50Z 2012-02-28T23:15:50Z <p>Part of the problem is that you're probably only used to seeing chain homotopies of degree zero chain maps. In this case $n=1$ (because the homotopy itself raises degree by 1 and so is an element of $Hom^1$) and the formula reduces to the standard formula for a chain homotopy that you'd find, for example, in Hatcher or Munkres or any other introduction to algebraic topology. In this more general setting, elements of $Hom^n(C^\ast, D^\ast)$ (I'm generalizing the setting to pure algebra - though note that algebraically $C^\ast=C_{-\ast}$ by convention) raise degree by $n$ and certain signs come in when you work with homomorphisms of chain complexes that shift degrees like this. </p> <p>Probably the best explanation for why you want the signs in the general case is the explanation I first found <a href="http://www.math.umn.edu/~tlawson/signs.pdf" rel="nofollow">here</a> in Section 12 of Lawson's paper about signs (though no doubt many people have known this for a long time). The point is that given the sign conventions for tensor products (which are also reasonable from the theory of symmetric monoidal categories) this is the only choice of sign convention that will make the evaluation map $Hom(C^\ast, D^\ast)\otimes C^\ast\to D^\ast$ a chain map, which is a pretty reasonable demand. Try to show yourself that this evaluation is a chain map and you'll see where the mysterious sign comes into it. </p> http://mathoverflow.net/questions/88842/when-is-a-finite-cw-complex-a-compact-topological-manifold/88856#88856 Answer by Greg Friedman for When is a finite cw-complex a compact topological manifold? Greg Friedman 2012-02-18T20:09:57Z 2012-02-18T20:09:57Z <p>In a certain sense, this is one of the starting questions for surgery theory, which aims to classify all manifolds, to the extent possible. You can get some idea of the complexity of this problem (including your problem) by reading some surveys about surgery theory. A good place to start might be Surveys on Surgery Theory, edited by Cappell, Ranicki, and Rosenberg, or even have a look at the Wikipedia article on surgery theory. In general, even when you know your CW complex is homotopy equivalent to a manifold it requires non-trivial work to show that it's homeomorphic to a manifold. For example, every finite CW complex can be embedded in Euclidean space and stably this yields a unique spherical fibration over the space, called the Spivak spherical fibration. If your space is a manifold, this should be the stable normal bundle (or its appropriate PL or topological analogues). So things like this start coming into play and it gets complicated pretty quickly. </p> http://mathoverflow.net/questions/81740/modern-source-for-spectra-including-ring-spectra/81756#81756 Answer by Greg Friedman for Modern Source for Spectra (including Ring Spectra) Greg Friedman 2011-11-23T21:09:49Z 2011-11-23T21:09:49Z <p>Probably neither of these will be exactly what you're looking for, but here are two references that come to mind and might have some of what you want:</p> <p>Algebraic Topology by Robert M. Switzer is a good classical source. It doesn't have the newer things you're looking for, but it's less hand-wavey than Adams tends to be. </p> <p>On Thom Spectra, Orientability, and Cobordism by Yuli Rudyak. I don't remember exactly what's in there (probably not symmetric spectra), but I've found it to be a useful source in the past.</p> <p>Certainly both of these handle ring spectra and module spectra. </p> http://mathoverflow.net/questions/76940/how-to-disjoint-two-cycles-with-zero-intersection/76991#76991 Answer by Greg Friedman for How to disjoint two cycles with zero intersection? Greg Friedman 2011-10-02T16:14:16Z 2011-10-02T16:14:16Z <p>Let $M$ be the union of two oriented circles. Let $\Gamma$ be the orientation class of $M$. Let $Z$ be the $0$-dimensional homology class represented by a positively oriented point in one of the circles and a negatively oriented point in the other circle (so $Z$ representes the class $&lt;1,-1>\in H_0(M)\cong \mathbb{Z}\oplus \mathbb{Z}$). Then the intersection number of $\Gamma$ with $Z$ should be $0$, but $\Gamma$ and $Z$ can't be made disjoint. In fact, the basic property of a fundamental class of a compact manifold is that it's supported at every point.</p> http://mathoverflow.net/questions/74097/alternative-approaches-to-the-universal-coefficient-theorem Alternative approaches to the universal coefficient theorem Greg Friedman 2011-08-30T19:51:11Z 2011-08-30T19:51:11Z <p>Let $A$ be a chain complex of free $R$-modules over a PID $R$, and let's assume $A$ has finite cohomological type, by which I mean $H^\ast(A)$ is finitely generated in each dimension and $0$ for large enough $|\ast|$ (though this may be stronger than necessary for the question. </p> <p>This is enough for the universal coefficient theorem to tell us that $$H^i(Hom^\ast(A,R))\cong Hom(H^{-i}(A),R)\oplus Ext(H^{-i+1}(A),R).$$ Some more basic homological algebra tells us that $$Ext(H^{-i+1}(A),R)\cong Hom(T^{-i+1}(A),Q(R)/R),$$ where $T^\ast$ is the torsion subgroup of $H^\ast$ and $Q(R)$ is the field of fractions of $R$. So, $$H^i(Hom^\ast(A,R))\cong Hom(H^{-i}(A),R)\oplus Hom(T^{-i+1}(A),Q(R)/R).$$ Let's call this formula $(\ast)$.</p> <p>Now let $I$ be the complex with $Q(R)$ in degree $0$, $Q(R)/R$ in degree 1, and the projection as the only non-trivial map. Then the obvious coaugmentation $R\to I$ (thinking of $R$ as a complex concentrated in degree $0$) is a quasi-isomorphism, and since $A$ is free, $Hom^\ast(A,R)$ should be quasi-isomorphic to $Hom^\ast(A,I)$. Here, of course, the $i$th cohomology groups of this latter complex are chain homotopy equivalence classes of degree $i$ chain maps from $A$ to $I$. </p> <p>So my question is whether there might be a more direct homological algebra argument to get formula $(\ast)$ from this starting point. It's not hard to get a map $H^\ast(Hom^\ast(A,I))\to Hom(H^{-\ast}(A),R)$, so the hard parts are seeing that this is onto and working in the torsion pairing somehow (I haven't yet stumbled upon the correct map $H^\ast(Hom^\ast(A,I)) \to Hom(T^{-\ast+1}(A),Q(R)/R)$). </p> <p>Does anyone have any ideas or know of some references where this approach to the universal coefficient theorem has been taken before?</p> <p>Ultimately my interest is in topology and how linking pairings on manifolds arise algebraically from intersection pairings. Of course the linking pairing is well-established in the literature, but I'm interested in getting at it from this point of view with the goal of some ultimate applications to sheaf-theoretic versions of duality theorems. </p> <p>Thanks.</p> http://mathoverflow.net/questions/72495/where-to-start-with-research-regarding-maslov-index-class/72506#72506 Answer by Greg Friedman for Where to start with research regarding maslov index/class Greg Friedman 2011-08-09T19:12:08Z 2011-08-09T19:12:08Z <p>This paper outlines several different approaches to the Maslov index: Cappell, Sylvain E.; Lee, Ronnie; Miller, Edward Y. On the Maslov index. Comm. Pure Appl. Math. 47 (1994), no. 2, 121–186. Link <a href="http://onlinelibrary.wiley.com/doi/10.1002/cpa.3160470202/abstract" rel="nofollow">here</a></p> <p>If you're looking for general background on cohomology and coming from a physics background, a good starting point might be Differential Forms in Algebraic Topology by Bott and Tu.</p> http://mathoverflow.net/questions/70965/singular-analog-of-cellular-homology/70967#70967 Answer by Greg Friedman for Singular analog of cellular homology Greg Friedman 2011-07-22T07:18:29Z 2011-07-22T07:18:29Z <p>Just thinking on my feet: From what I can tell, the trouble with setting up a "singular cell" version of homology is that simplices, whether part of a simplicial complex or on their own, have boundaries that are themselves simplices. So the boundary of a singular simplex is a singular complex simply by restricting. On the other hand, the boundary formula for a cell in a cell complex is much more complicated, and taking the boundary of a (mapping of an) abstract cell doesn't continue to carry the same sort of information that the boundary of a (singular) simplex does. In other words, the same thing that let's us glue together CW complexes in ways that are more complex than what we do for simplicial complexes is a hindrance here because we don't have enough extra structure to say that the boundary of a cell is a cellular object in a nice way. I suppose one could try to impose extra conditions, like prescribing that we describe the boundary of a cell in a certain way and then declaring that the singular cellular boundary be made of the kinds of pieces that we get, but now we're well on the road to just recreating singular simplicial homology, though perhaps with some shapes that are a bit different from the standard generalized tetrahedra. This, I believe, can be done: there's such a thing as cubical homology. I don't know much about it, but the basic idea is one uses n-cubes instead of n-tetrahedra. If I remember correctly, as one might imagine, it works out about the same as the tetrahedral version and in its generalized form leads to a notion of cubical sets instead of simplicial sets. My understanding is that each has its technical advantanges, depending what you're trying to do. For example, arguments about products (including homotopies) are going to work more nicely, but then once you've swept that difficulty under the rug, some other lump pops up somewhere else to make something else more difficult (perhaps the simplicial set people will chime in and tell us what). </p> <p>Sorry that this is more philosophizing than serious answer, but perhaps it will provide some ideas about why we don't generally see the kind of thing you're asking about.</p> http://mathoverflow.net/questions/69582/signs-and-functoriality-of-tensor-products Signs and functoriality of tensor products Greg Friedman 2011-07-05T22:57:00Z 2011-07-05T23:24:29Z <p>Let $C,C',D,D'$ be chain complexes of $R$-modules (let's say with upper indexing, so perhaps I should call them cochain complexes, though they're not duals of anything). Let $f\in Hom^\ast(C,C')$ and $g\in Hom^*(D,D')$. Then the standard convention is that $$(f\otimes g)(x\otimes y)=(-1)^{|g||x|}f(x)\otimes g(y),$$ where $|g|$ is the degree of $g$ and $|x|$ is the degree of $x$. As observed on page 171 of Dold, this is consistent with having a degree 0 chain map $$Hom^\ast(C,C')\otimes Hom^\ast(D,D')\to Hom^\ast(C\otimes C',D\otimes D').$$ </p> <p>What bothers me, though, is that this formula forces $$(h\otimes k)\circ(f\otimes g)=(-1)^{|k||f|}hf\otimes kg,$$ which seems to violate the definition of a bifunctor as given, for example, on page 17 of Kashiwara and Schapira's "Categories and Sheaves", which would seem to require (adapting the notation) $$(1_{C'}\otimes g)(f\otimes 1_D)=(f\otimes 1_{D'})(1_C\otimes g).$$ (Here I suppose we assume that the relevant categories are the category of chain complexes of $R$ modules with $Mor(X,Y)=Hom(X,Y)$ (certainly such things can be composed functorially and the identity behaves properly) and the products of this category with itself). If I'm reading it correctly, this requirement in Kashiwara-Schapira seems to be the same as what Mac Lane is asking for in Proposition II.3.1 of "Categories for the Working Mathematician".</p> <p>So are we to believe $\otimes$ is not a functor or is there a way to reformulate all of this to be consistent (or am I just getting something wrong)?</p> <p>Thanks in advance!</p> http://mathoverflow.net/questions/68689/more-questions-about-verdier-duality-and-related-math/68695#68695 Answer by Greg Friedman for More questions about Verdier duality (and related math) Greg Friedman 2011-06-24T05:12:02Z 2011-06-24T05:12:02Z <p>I'm not sure if this will answer your question or not, but let $\mathbb{D}$ be the Verdier dualizing sheaf on the locally compact space $X$. If $M$ is a manifold, then $\mathbb{D}[-n]$, where $n$ is the dimension of $M$, is isomorphic in the derived category to the orientation sheaf $\omega$ on $M$ (see, for example, Borel's book on interesection cohomology, I think around section V.7). The Borel-Moore homology of $X$, which is equivalent to the ordinary homology of $X$ if $X$ is compact (otherwise it's the homology theory built from locally-finite chains) is defined to be $H^{BM}_{k}(X)=H^{-k}(X; \mathbb{D})$. If you unwind all that indexing business, it says that, for a compact manifold, $H_k(M)=H^{n-k}(M; \omega)$, which is probably the duality statement you're looking for. </p> <p>Of course showing that this has anything to do with classical Poincare duality (say via the cap product) is pretty far from obvious. This is part of the content of a paper I currently have in preparation with Jim McClure. </p> http://mathoverflow.net/questions/66809/stratified-pseudomanifold/66856#66856 Answer by Greg Friedman for Stratified Pseudomanifold Greg Friedman 2011-06-03T22:36:44Z 2011-06-03T22:36:44Z <p>It's not!</p> <p>Okay, that's a little too glib. You're right that it's often required that $X^{n-1}=X^{n-2}$, but it depends somewhat on your purpose. In most of my recent papers, including those I'm current working on with Jim McClure, and in the papers of Saralegi (which look at intersection homology from an analytic point of view), there's freedom to include codimension one strata, but you need to be more careful about your definition of intersection (co)homology.</p> <p>In even more detail: there are a variety of connected reasons why one would classically not work with codimension one strata. One reason has to do with the properties of classical intersection homology theory. Many of the results of intersection homology, including the ones that are approached sheaf theoretically, depend, at heart, on having a nice formula for the intersection homology on cones. If there are no codimension one strata around and you're working with perversities as defined by Goresky and MacPherson, then this cone formula works out, the sheaf of intersection chains is quasi-isomorphic to the Deligne sheaf, and away you go. However, with codimension one strata, and Goresky-MacPherson perversities, the cone formula doesn't quite work out. I have a detailed exposition about this in the following paper: </p> <p>Greg Friedman, <em>An introduction to intersection homology with general perversity functions</em>, in <strong>Topology of Stratified Spaces</strong>; Greg Friedman, Eugénie Hunsicker, Anatoly Libgober, Laurentiu Maxim (editors) Mathematical Sciences Research Institute Publications 58, Cambridge University Press 2011, 177-222</p> <p>Here's the link to the copy on my web site:</p> <p><a href="http://faculty.tcu.edu/gfriedman/papers/MSRI-revised-2.pdf" rel="nofollow">http://faculty.tcu.edu/gfriedman/papers/MSRI-revised-2.pdf</a></p> <p>One argument that I make there looks carefully at what the perversity should be on such a stratum: if the perversity is 0 (as one might expect), then the perversity is greater than the top perversity allowed by Goresky and MacPherson (which would have to be -1 for a codimension one stratum), and that leads to the cone formula issues. On the other hand, if the perversity assigned is $\leq -1$, now you're below the bottom perversity $\bar 0$ allowed by Goresky and MacPherson, and that causes other issues. </p> <p>Another way to think about codimension one strata is that they could arise if you took a manifold with boundary and let the boundary be a codimension one stratum. Then, as algori notes, you have a problem getting your fundamental class as a geometrically defined cycle (unless you sufficiently modify your definition of intersection homology). And that causes trouble with Poincare duality. </p> <p>So, after all that, in summary: a pseudomanifold can have codimension one strata, but much of intersection homology breaks down. Unless you modify your intersection homology a bit, and then everything does work! See my papers or feel free to e-mail if you have further questions. </p> http://mathoverflow.net/questions/66401/singular-homology-cohomology-as-a-derived-functor/66410#66410 Answer by Greg Friedman for Singular Homology/Cohomology as a derived functor? Greg Friedman 2011-05-30T01:51:57Z 2011-05-30T01:51:57Z <p>Elaborating a bit on Qiaochu Yuan's comment, if X is "nice enough" and $\mathcal{R}$ is the constant sheaf with stalks in $R$, then the singular cohomology agrees with the derived functor of the global section functor: $H^\ast(X;\mathcal{R})\cong H^\ast(X;R)$. This result is scattered throughout Bredon's book on sheaf theory, though I grant that it's not super-intuitive there. Alternatively, it's not hard to show the sheafification of the presheaf complex of singular cochains $U\to C^\ast(U;R)$ is a resolution of $\mathcal{R}$ and it can be shown to be homotopically fine, by a dual argument to the proof that the sheaf complex of singular chains is homotopically fine. Furthermore the presheaf of cochains is conjunctive and, while it's not a mono-presheaf, the cohomology with zero supports of the cochain presheaf is trival. Putting all those things together, the singular cohomology is isomorphic to the hypercohomology of the cochain sheaf complex, which is a derived functor of the global section functor, though in the "hyper" sense. Similarly things can be done with homology, the sheaf of germs of singular chains being homotopically fine. Swan's book on sheaf theory is a good reference for that. </p> http://mathoverflow.net/questions/59677/cobordism-categories-that-dont-involve-manifolds/59896#59896 Answer by Greg Friedman for Cobordism categories that don't involve manifolds Greg Friedman 2011-03-28T21:08:53Z 2011-03-28T21:08:53Z <p>One famous (in my field!) example is Witt space bordism. Witt spaces are not manifolds but rather pseudomanifolds (which aren't so far off from manifolds, but they can have singularities). A pseudomanifold is a Witt space if certain local rational intersection homology groups vanish. The bordism group of Witt spaces is important because it turns out to be a geometric model for ko-homology after inverting 2. The original reference is Paul Siegel's thesis: <a href="http://www.jstor.org/stable/2374334" rel="nofollow">http://www.jstor.org/stable/2374334</a> There's a slightly fancier version due to Pardon using "IP spaces" which satisfy integral Poincare duality: <a href="http://www.springerlink.com/content/6m5j386lr5hx2444/" rel="nofollow">http://www.springerlink.com/content/6m5j386lr5hx2444/</a></p> http://mathoverflow.net/questions/56932/what-characteristic-class-information-comes-from-the-2-torsion-of-hbsonz What characteristic class information comes from the 2-torsion of $H^*(BSO(n);Z)$? Greg Friedman 2011-02-28T22:42:30Z 2011-03-01T00:12:16Z <p>This is just a general curiosity question:</p> <p>In the standard textbook treatments of characteristic classes, and in particular the treatment of universal Pontrjagin classes, it's standard to consider $H^\ast(BSO,Z[1/2])$ (or $H^\ast(BSO(n),Z[1/2])$) in order to kill the 2-torsion. But I'm curious about that 2-torsion, since it should still give us some extra characteristic class-type information about real oriented bundles. If nothing else, it would give a class that's characteristic in the sense that it behaves the right way with respect to pullbacks (though I imagine it might be too much for these to be stable or have any kind of product formulas). So I suppose my questions are:</p> <ol> <li><p>What's known about such classes? </p></li> <li><p>Are they useful for anything? </p></li> <li><p>Do these interact in any interesting way with the Stiefel-Whitney classes when everything is reduced mod 2? </p></li> <li><p>Why are they usually ignored (or obliterated by coefficient change)?</p></li> </ol> http://mathoverflow.net/questions/56809/what-is-the-best-way-to-study-rational-homotopy-theory/56844#56844 Answer by Greg Friedman for What is the best way to study Rational Homotopy Theory Greg Friedman 2011-02-27T18:24:43Z 2011-02-27T18:24:43Z <p>While not comprehensive, the following book has a nice introduction to the subject in its first chapter or so: J. Oprea, A. Tralle, Symplectic manifolds with no Kähler structure, Lecture Notes in Math. 1661,. Springer–Verlag, 1997.</p> http://mathoverflow.net/questions/53419/are-there-oriented-4k2-manifolds-such-that-imh-2k1m-z-2-to-h-2k1m Are there oriented $4k+2$ manifolds such that $im(H_{2k+1}(M; Z/2)\to H_{2k+1}(M, \partial M; Z/2))$ has odd dimension? Greg Friedman 2011-01-26T23:04:22Z 2011-01-29T04:38:30Z <p>The following fairly specific question comes up in a bordism computation I'm trying to do:</p> <p>Are there compact $\mathbb Z$-oriented $4k+2$ dimensional manifolds with boundary $M$ such that $im(H_{2k+1}(M; \mathbb Z/2)\to H_{2k+1}(M, \partial M; \mathbb Z/2))$ has odd dimension as a $\mathbb Z/2$ vector space?</p> <p>Clearly the answer is no if $k=0$. Also, I can show that this can't happen if $\partial M=\emptyset$ using a combination of Poincare duality and the universal coefficient theorem. But I haven't been able to rule out the possibility if the boundary is non-empty, or to construct examples.</p> <p>Thanks. </p> http://mathoverflow.net/questions/53315/references-for-sign-conventions-in-homological-algebra/53327#53327 Answer by Greg Friedman for References for sign conventions in homological algebra Greg Friedman 2011-01-26T06:34:52Z 2011-01-26T06:41:14Z <p>This is much more of a comment than an answer, but I ran out of room in the comment box:</p> <p>Tyler, my new standard reference for sign issues is YOUR paper, which I post here for the benefit of others <a href="http://www.math.umn.edu/~tlawson/signs.pdf" rel="nofollow">http://www.math.umn.edu/~tlawson/signs.pdf</a></p> <p>Jim McClure and I have done a lot of wrestling with signs lately, and it seems that your intuition in your paper is the right one: start with the axiom that evaluation should be a degree 0 chain map $C^\ast\otimes C_*\to R$, and most of the rest of the standard conventions will follow. By the way, Jim and I have also come to the conclusion that the Poincare duality map should not be a cap product but a signed cap product $\alpha\to (-1)^{|\alpha|\dim(M)}\cap [M]$ in order to make it a degree $-\dim(M)$ chain map (with cohomological indexing) when it acts on the left (like all good maps should). This is assuming the cap product as defined in Dold, which Jim and I have concluded is a good definition (after playing with some other possible definitions and having them backfire for one reason or another).</p> <p>Other than this issue, while he doesn't really provide much in the way of explanation, Dold's classic textbook tends to be at least accurate and consistent most of the way through. The exception is in his definition of the transfer (though he notes about three pages later that he should really have a sign there as well - this sign issue seems to correspond to the issue of making the duality map a properly signed chain map). This slightly messes with the signs in the remainder of the book, for example in the definition of the intersection product, although this issue doesn't occur until near the end so there's not much damage.</p> <p>Sorry to ramble without providing much of an answer, but this is an issue close to my heart. At one point I decided that I want $(-1)^?$ chiseled into my tombstone.</p> http://mathoverflow.net/questions/51801/when-do-we-need-the-axiom-of-compact-support-for-a-homology-theory-to-be-uniquely When do we need the axiom of compact support for a homology theory to be uniquely defined? Greg Friedman 2011-01-12T00:34:02Z 2011-01-12T01:13:35Z <p>Looking over the treatment of the Eilenberg-Steenrod axioms in a few of my favorite introductory algebraic topology texts, I see that some include an "axiom of compact support", while others do not. Whether or not one needs such an axiom for the homology theory to be uniquely defined (assuming it satisfies the dimension axiom) seems to depend on the category of pairs for which one is defining the homology theory. For example, if one is only looking at the homology of compact pairs, the axiom of compact supports is certainly unnecessary. Spanier invokes the axiom to handle arbitrary polyhedral pairs, and this also seems to be the situation in Munkres. But then Hatcher (if I'm reading it correctly) proves uniqueness (assuming the dimension axiom) on CW pairs, without any mention of the compact support axiom, which I find somewhat surprising without limiting to compact pairs. </p> <p>So for which categories containing possibly non-compact spaces does one need the compact support axioms (along with the dimension axiom) in order to know that the homology theory is unique? In particular, what is the status for polyhedral pairs, CW pairs, and arbitrary topological pairs? And if I'm reading Hatcher correctly, is there an intuitive reason why the compact support axiom isn't necessary for CW pairs but is for polyhedral pairs? </p> <p>It's amazing the things one is forced to think about when teaching a graduate algebraic topology class! </p> http://mathoverflow.net/questions/49315/what-group-is-a-b-a2b2 What group is <a,b | a^2=b^2>? Greg Friedman 2010-12-13T20:57:18Z 2010-12-14T12:58:07Z <p>In teaching my algebraic topology class, this group showed up as part of an easy fundamental group computation: $\langle a,b\mid a^2=b^2\rangle$. My first instinct was that this must be $\mathbb{Z}*\mathbb{Z}/2$ because clearly every element can be written as a product of $b$s (only to the power 1) and powers of $a$. But this turns out to be far from clear (and likely wrong). I assume this must be a well-known group to group theorists, so I'm curious if it's isomorphic to something that can be described by other means (or what's known about it in general). </p> <p>Thanks!</p> http://mathoverflow.net/questions/48328/ranicki-symmetric-l-groups-of-finite-fields Ranicki symmetric L-groups of finite fields? Greg Friedman 2010-12-04T23:55:03Z 2010-12-05T12:06:56Z <p>Can anyone tell me what the Ranicki symmetric L-groups $L^*(F)$ are when $F$ is a finite field? (and maybe provide a reference?) Thanks!</p> http://mathoverflow.net/questions/48103/is-intersection-homology-the-usual-homology-of-something-else/48121#48121 Answer by Greg Friedman for Is intersection homology the usual homology of something else? Greg Friedman 2010-12-03T01:45:50Z 2010-12-03T01:45:50Z <p>The short answer is "no". But you might be interested in some work Markus Banagl is doing along these lines. He has a functor that assigns to a space X an "intersection space" $I^{\bar p}X$. Then rather than studying $I^{\bar p}H_\ast(X)$, he looks at $H_\ast(I^{\bar p}X)$. However, this is not generally the same thing as $I^{\bar p}H_\ast(X)$. </p> http://mathoverflow.net/questions/36708/conventions-for-definitions-of-the-cap-product Conventions for definitions of the cap product Greg Friedman 2010-08-26T00:44:20Z 2010-08-26T04:29:33Z <p>In singular (co)homology, if <code>$\alpha\in C^*(X)$</code> and $x\in C_*(X)$, then the cap product $\alpha \cap x$ is generally defined by the following process:</p> <ol> <li><p>Apply to $x$ the diagonal map <code>$C_*(X)\to C_*(X\times X)$</code> followed by some choice of Alexander-Whitney chain equivalence <code>$C_*(X\times X)\to C_*(X)\otimes C_*(X)$</code> to obtain an element $\sum y_i\otimes z_i$. </p></li> <li><p>Apply $\alpha$ to $\sum y_i\otimes z_i$ by the slant product, or, in other words and roughly speaking, apply $\alpha$ to half of the factors''. </p></li> <li><p>Depending on your favorite conventions (and what you're trying to accomplish), there may also be a sign (which might also be part of your definition of the slant product - but let's ignore this).</p></li> </ol> <p>The reason I'm being so cagey with wording in part 2 is directly related to my question: In almost all major textbook sources I have consulted, step 2 is performed by forming $\sum y_i \alpha(z_i)$, which strikes me as somewhat unnatural, forcing the $\alpha$, which starts off on the left to jump all the way over the $y_i$ terms to get to the $z_i$ terms on the right. Is there a good mathematical reason for this convention? Why not define the cap product to be $\sum \alpha(y_i) z_i$?</p> <p>The one major exception to this convention seems to be Hatcher. He does form $\sum \alpha(y_i) z_i$, but he also writes cap products as $x\cap \alpha$, so his cochain also has to jump, but it jumps over the $z$s instead!</p> <p>(For the record, I'm not asking this question out of idle pickiness. Jim McClure and I have been doing a lot of work with cap products recently, and we're trying to be consistent amongst various conventions for various issues, but preferably with good reasons thrown in!) </p> http://mathoverflow.net/questions/27429/how-does-the-lefschetz-poincare-dual-torsion-linking-pairing-on-manifolds-with-bo How does the Lefschetz-Poincare dual torsion linking pairing on manifolds with boundary interact with the maps of the long exact sequence of the manifold-boundary pair? Greg Friedman 2010-06-08T04:36:48Z 2010-08-13T00:42:19Z <p>I'm wondering if anyone can point me to a reference on how the various Lefschetz-Poincare dual torsion pairings of a manifold with boundary fit together.</p> <p>To explain in more detail, consider a connected compact oriented n-manifold $M$ with boundary. Then we have the various dualities with rational coefficients $H_i(M;Q)=H_{n-i}(M, \partial M; Q)$ and $H_i(\partial M;Q)=H_{n-i-1}(\partial M; Q)$, but we also obtain, from some easy tinkering with the long exact sequence of the pair and other basic observations that, e.g. $im(H_i(M)\to H_i(M, \partial M))$ is dual to $im(H_{n-i}(M)\to H_{n-i}(M, \partial M))$ (in fact this is how we define signatures on manifolds with boundary) and similar results hold for the other "shared" terms in the long exact sequence.</p> <p>I'd like to know more about the generalization of this over Z and, in particular, about what torsion pairings to Q/Z exist on the various images, cokernels, etc. of the long exact sequence and which are nonsingular. Does anyone know of any place in the literature where this is worked out?</p> <p>Thanks!</p> <p>Updated: Thanks, Tom, for your response. I've been thinking about it, and while I agree that the situation is much murkier over Z, I think there are still some things that can be said.</p> <p>For example, it's true that the map $H_n(M)\to H_n(M, bd M)$ (let's assume $n$ is the middle dimension to simplify the discussion), only yields a nondegenerate pairing on the image (which you call $A$) mod torsion and not a perfect (nonsingular) pairing. But now since $H_n(M)\to A/\text{torsion}$ has a free group as its image, we have a (non-unique) splitting that lets us consider $A/\text{torsion}$ as a direct summand of $H_n(M)$. Let's fix this summand for now (the choice turns out not to matter). Since $H_n(M)/\text{torsion}$ and $H_n(M, bd M)/\text{torsion}$ are (perfectly) Z-dual, $A/\text{torsion}$ must be dual to something in $H_n(M, bd M)/\text{torsion}$, and I claim that the thing it's dual to is isomorphic to the kernel of $H_n(M,bd M)\to H_{n-1}(bd M)/\text{torsion}$ (mod torsion). Notice that any non-torsion element of this kernel has a multiple represented by a cycle in $M$, and so it must have non-zero intersection number with any cycle in the boundary, and this shows that as far as intersection pairings are concerned, the choice of $A/\text{torsion}$ as a summand of $H_n(M)$ doesn't matter. I don't really want to get into the details of the rest of my claim here, but the basic idea should be that these are the things that should be dual once we tensor with Q but over Z we need to make sure we have all the appropriate Z-primitives, which I'm pretty sure this does (since if a multiple of $x\in H_n(M,bd M)$ is in $A$, then $x$ becomes torsion in $H_{n-1}(M)$).</p> <p>So what's the point of all this? Well, even over Z we can say that there is a nonsingular pairing between a certain cokernel and a certain kernel, as long as we mod out the torsion in the right places. I suspect that there's a more subtle version of this for the torsion linking pairings where instead of quotienting out all torsion we just kill certain torsion subgroups. I think I've seen things somewhat of this nature in papers relating Witt groups of Z pairings to Witt groups of Q/Z pairings. </p> http://mathoverflow.net/questions/33843/reference-for-intersection-and-linking-in-algebraic-topology/33863#33863 Answer by Greg Friedman for Reference for intersection and linking in algebraic topology Greg Friedman 2010-07-29T22:30:24Z 2010-07-29T22:30:24Z <p>I'm not quite sure if this is what you're thinking of, but intersection homology has a good theory of intersection products for simplicial pseudomanifolds. See Goresky-MacPherson, "Intersection Homology Theory"</p> http://mathoverflow.net/questions/33833/what-is-a-principal-refinement-of-a-postnikov-system What is a principal refinement of a Postnikov system? Greg Friedman 2010-07-29T19:29:58Z 2010-07-29T20:50:43Z <p>I've been reading the book of Hilton, Mislin, and Roitberg on Localization of Nilpotent Groups and Spaces. In Section II.2 they define a principal refinement at stage $n$ of a Postnikov system $$\cdots \to X_n\overset{p_n}{\to} X_{n-1}\to \cdots$$ to be a factorization of $p_n$ into a finite sequence of fibrations $$X_n=Y_c \overset{q_c}{\to}\cdots\to Y_1\overset{q_1}{\to}Y_0=X_{n-1}$$ whose fibers are Eilenberg-MacLane spaces $K(G_i,n)$.</p> <p>But isn't the point of a Postnikov system that $p_n$ is already a fibration whose fibers are Eilenberg-MacLane spaces? So I don't understand why the condition of the definition isn't satisfied trivially at every stage. Perhaps there's some subtlety involving the condition also given that each $q_i$ be induced by a map $g_i: Y_{i-1}\to K(G_i, n+1)$, but aren't all fibrations with fiber $K(G_i, n)$ induced this way since $K(G_i, n+1)$ is the base of a path-space fibration with fiber $K(G_i, n)$?</p> http://mathoverflow.net/questions/124278/mayer-vietoris-sequence-in-homology-with-local-coefficients Comment by Greg Friedman Greg Friedman 2013-03-15T18:27:30Z 2013-03-15T18:27:30Z I realize this has already been answered, but Hatcher's Algebraic Topology also has a nice section on local coefficients, relating the covering space and bundle-of-groups points of view http://mathoverflow.net/questions/122630/topology-of-mathbb-rn/122640#122640 Comment by Greg Friedman Greg Friedman 2013-02-22T21:22:20Z 2013-02-22T21:22:20Z If one of $M$ or $N$ isn't simply connected, the same argument will work with $\pi_1$; this result isn't as general but might be more accessible for undergraduates, as it's probably easier for them to quickly grasp the idea of fundamental groups than of homology (if they haven't seen either before). Then you can explain to them that there are &quot;similar but a little more complicated&quot; things one can use if both spaces are simply connected. http://mathoverflow.net/questions/119796/any-map-of-a-contractible-complex-to-itself-has-a-fixed-point/119800#119800 Comment by Greg Friedman Greg Friedman 2013-02-08T08:11:24Z 2013-02-08T08:11:24Z @Tom, that's a nice construction! http://mathoverflow.net/questions/114286/visualising-locally-flat-embeddings-of-surfaces-in-r4 Comment by Greg Friedman Greg Friedman 2012-12-05T06:12:22Z 2012-12-05T06:12:22Z Coming at this with a lot of ignorance, this raises the following related question: are there locally-flat embeddings of surfaces in $\mathbb R^4$ that are not smoothable embeddings but such that the intersection with every hyperplane parallel to a given one are all sufficiently &quot;nice&quot; (either smoothly embedded curves, smoothly immersed curves, or finite sets of points (or unions of such things))? http://mathoverflow.net/questions/111231/a-dual-universal-coefficient-theorem/111236#111236 Comment by Greg Friedman Greg Friedman 2012-11-03T02:36:36Z 2012-11-03T02:36:36Z R is a principal ideal domain and G is an R-module. You also need $H_\ast(X;R)$ to be of finite type, meaning each $H_i(X;R)$ is a finitely generated $R$-module. http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109665#109665 Comment by Greg Friedman Greg Friedman 2012-10-17T20:50:32Z 2012-10-17T20:50:32Z Thanks! This is a very interesting approach. Is it hard to verify that the type B Coxeter arrangement is a simplicial hyperplane arrangement. Can you recommend a reference for that? http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109891#109891 Comment by Greg Friedman Greg Friedman 2012-10-17T20:45:03Z 2012-10-17T20:45:03Z I just had a look and it seems that Bredon also uses acyclic models - he does not provide an explicit triangulation of the product $\Delta^p\times \Delta^q$. http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109163#109163 Comment by Greg Friedman Greg Friedman 2012-10-15T20:56:09Z 2012-10-15T20:56:09Z Thanks, Ronnie. I had a look, but Tonks's paper also seems to be mostly about simplicial sets. Perhaps I can reverse engineer what I want from a proof that, for simplicial sets, $|K|\times |L|=|K\times L|$. I suppose what I really want is a proof of this for semi-simplicial sets (Delta sets). http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product Comment by Greg Friedman Greg Friedman 2012-10-15T20:54:07Z 2012-10-15T20:54:07Z @Justin. To clarify, I'm not looking for the definition of the shuffle product (I've found that in, e.g., Dold). I'm looking for a proof that the construction provides a triangulation of $\Delta^p\times \Delta^q$ that's compatible with boundaries (in the sense that the cross product induced from this point of view is a chain map). http://mathoverflow.net/questions/109159/references-for-eilenberg-zilber-shuffle-product/109162#109162 Comment by Greg Friedman Greg Friedman 2012-10-09T18:43:52Z 2012-10-09T18:43:52Z Thanks. I had a look at Hilton and Wylie, and unfortunately, they handle the Eilenberg-Zilber theorem in much the same way other texts do - abstractly by the method of acyclic models. http://mathoverflow.net/questions/107692/closed-dual-of-vector-fields/107702#107702 Comment by Greg Friedman Greg Friedman 2012-09-20T17:44:12Z 2012-09-20T17:44:12Z Assuming you can construct a Riemmanian metric! (though I grant that one can in most cases of interest) http://mathoverflow.net/questions/106779/beginning-reference-for-configuration-spaces Comment by Greg Friedman Greg Friedman 2012-09-10T16:12:02Z 2012-09-10T16:12:02Z I second the suggestion of Dev Sinha's paper. It's very readable and makes very nice use of geometric intuition to get at some very nice homology and cohomology computations (even the product structures!). http://mathoverflow.net/questions/105958/directed-system-as-local-coefficient-system Comment by Greg Friedman Greg Friedman 2012-09-04T22:44:51Z 2012-09-04T22:44:51Z Can you explain in what sense you're thinking of $\hat F$ as a local coefficient system? http://mathoverflow.net/questions/105527/about-relative-homotopy-group/105546#105546 Comment by Greg Friedman Greg Friedman 2012-08-27T21:11:57Z 2012-08-27T21:11:57Z This seems like overkill to me. Why not just use the long exact sequence of the pair and that $\pi_0(RP^2)$ and $pi_1(S(RP^2))$ are trivial? http://mathoverflow.net/questions/102161/proving-the-existence-of-good-covers/102168#102168 Comment by Greg Friedman Greg Friedman 2012-07-28T06:35:13Z 2012-07-28T06:35:13Z Nothing in the original question says that the manifold is smooth or triangulable.