Spectral sequences: opening the black box slowly with an example - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:17:26Z http://mathoverflow.net/feeds/question/45036 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example Spectral sequences: opening the black box slowly with an example Dylan Wilson 2010-11-06T06:33:59Z 2010-12-15T10:14:50Z <p>My friend and I are attempting to learn about spectral sequences at the moment, and we've noticed a common theme in books about spectral sequences: no one seems to like talking about differentials.</p> <p>While there are a few notable examples of this (for example, the transgression), it seems that by and large one is supposed to use the spectral sequence like one uses a long exact sequence of a pair- hope that you don't have to think too much about what that boundary map does.</p> <p>So, after looking at some of the classical applications of the Serre spectral sequence in cohomology, we decided to open up the black box, and work through the construction of the spectral sequence associated to a filtration. And now that we've done that, and seen the definition of the differential given there... we want some examples.</p> <p>To be more specific, we were looking for an example of a filtration of a complex that is both nontrivial (i.e. its spectral sequence doesn't collapse at the $E^2$ page or anything silly like that) but still computable (i.e. we can actually, with enough patience, write down what all the differentials are on all the pages).</p> <p>Notice that this is different than the question here: <a href="http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences" rel="nofollow">http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences</a>, though quite similar. We are looking for things that don't collapse, but specifically for the purpose of explicit computation (none of the answers there admit explicit computation of differentials except in trivial cases, I think). </p> <p>For the moment I'm going to leave this not community wikified, since I think the request for an answer is specific and non-subjective enough that a person who gives a good answer deserves higher reputation for it. If anyone with the power to thinks otherwise, then feel free to hit it with the hammer.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45038#45038 Answer by Eric Peterson for Spectral sequences: opening the black box slowly with an example Eric Peterson 2010-11-06T06:59:33Z 2010-11-06T07:25:48Z <p>This is more than you bargained for, but it's too good an opportunity to pass up plugging a couple cool and readable papers with impact beyond containing computable spectral sequences. In the 1980s Ravenel and Wilson famously used Hopf rings to compute some extraordinary homologies of a variety of families of infinite loopspaces. In the specific case where the loopspaces are Eilenberg-Mac Lane spaces, they used a bar spectral sequence (which arises as a filtration spectral sequence), together with Hopf ring information, to compute $E_* K(G, *)$ for various $E$ and $G$. One of the cool features of their work is that everything involved is explicit and identifiable.<sup>&dagger;</sup> You might try:</p> <p>Doug Ravenel and Steve Wilson, <em>The Morava $K$-theories of Eilenberg-Mac Lane spaces</em>, published in 1980 in the American Journal of Mathematics, vol. 102, no. 4, pages 691-748</p> <p>for a mildly complicated but very rewarding example. Or, the algebras $H_*(K(\mathbb{Z}/p, *); \mathbb{Z}/p)$ (along with a million other things!) are computed in Wilson's exceptionally nice book</p> <p>Steve Wilson, <em>Brown-Peterson Homology: An Introduction and Sampler</em>, published in 1980, no. 48 in the CBMS series of conference notes,</p> <p>which uses all the same machinery as the Morava $K$-theory paper but employs singular homology and tells you about some algebras which you already understand. The familiarity of these two things will probably ease digestion of the ideas.</p> <p><sup>&dagger;</sup> -- Well, this is kind of a lie, since they argue the existence or nonexistence of some of their differentials by knowing what the $E^\infty$ page must look like together with some kind of sparseness of the $E^2$ page. However, the way they build the spectral sequence does actually give you a formula for the differentials, and it's certainly possible, if difficult, for you to make the relevant calculations once you read their argument so you know whereabouts to look. At small primes (and small heights, in the Morava K-theory paper), this is probably even accessible.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45039#45039 Answer by Neil Strickland for Spectral sequences: opening the black box slowly with an example Neil Strickland 2010-11-06T08:02:20Z 2010-11-06T08:02:20Z <p>There are some examples at <a href="http://www.shef.ac.uk/nps/courses/bestiary/ss.pdf" rel="nofollow">http://www.shef.ac.uk/nps/courses/bestiary/ss.pdf</a> that you might or might not find useful.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45040#45040 Answer by Tilman for Spectral sequences: opening the black box slowly with an example Tilman 2010-11-06T08:22:46Z 2010-11-06T08:22:46Z <p>Computing differentials in spectral sequences can be fun and very useful, and I absolutely agree with OP that the literature does not cover this adequately -- surely not in a book, but even articles using this are few and far between (Ravenel-Wilson being an excellent one, as Eric mentioned). I think there are a number of reasons for this:</p> <ul> <li>Like diagram chases (but on a higher level), print form without the time coordinate doesn't lend itself to a good presentation of what's happening from page to page in a spectral sequence</li> <li>Spectral sequence charts are hard to draw (but, if you don't know it, take a look at the "sseq" TeX package)</li> <li>for many people, computing higher differentials in spectral sequences is the epitome of technicality, and they don't want to read it.</li> </ul> <p>So, it seems to be considered more of a craft than something that can be laid out in a textbook. Additionally, correct me if I'm wrong, this is only done in homotopy theory. I don't really understand why that is, there are plenty of spectral sequences in algebraic geometry, for example, but they tend to collapse. I notice you're based at Seattle, so my strong recommendation would be to go ask an expert to teach you -- you've got Ethan Devinatz and John Palmieri right there, right?</p> <p>I'll add a few random comments about things to learn. One of my favorite (simple) examples is to compute the homotopy groups of connective K-theory by 1) producing a free resolution of $F_2$ over $A(1)$, the subalgebra of the mod-2 Steenrod algebra generated by $Sq^1$ and $Sq^2$. Then notice that this is $H^*(bo;F_2)$ and run the Adams spectral sequence on this. You can also compute the homotopy groups of real K-theory as the fixed points of the conjugation action on complex K-theory using the homotopy fixed point spectral sequence. If you liked that, you can take this one chromatic step up (compute the homotopy groups of topological modular forms) and see lots of techniques in action -- you'll find that in some readily available notes by Charles Rezk or (excuse me for advertising again) my paper "Computation of the homotopy of the spectrum tmf".</p> <p>One thing that happens very frequently when there's more than just a few differentials is that there is a strong interplay between multiplicative extensions, Massey products/Toda brackets and differentials. Often, one needs to compute all three pieces of data at once to inductively derive longer differentials/longer extensions. This is what happens for tmf. By the way, there is something completely absent in any treatment of spectral sequences I know: multiplicative extensions in $E^r$ terms (with $r&lt;\infty$).</p> <p>I realize this is turning into some fairly unstructured rant, so I'll stop and reiterate: go ask the masters to teach you in person.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45041#45041 Answer by Neil Strickland for Spectral sequences: opening the black box slowly with an example Neil Strickland 2010-11-06T08:37:36Z 2010-11-06T08:37:36Z <p>Here's another answer that addresses the question more directly. Take $A^\ast=\mathbb{Z}[x]\otimes E[a]$ with $d(x^k)=kx^{k-1}a$ and $d(x^ka)=0$, and filter it by $F_kA^\ast=2^kA^\ast$. This gives a Bockstein spectral sequence with $E_1=(\mathbb{Z}/2)[x,a,t]/a^2$ (where $t$ represents $2$). It converges to the cohomology of the $2$-adic completion of $A^\ast$ rather than $A^\ast$, so it gives a natural example where the target is not the group you first thought of. Moreover, you can work out everything very explicitly from the definitions, the main point being that $x^{2^r(2k+1)}$ survives to the $E_r$ page and then $d_r(x^{2^r(2k+1)})=t^rx^{2^r(2k+1)-1}a$.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45046#45046 Answer by Allen Hatcher for Spectral sequences: opening the black box slowly with an example Allen Hatcher 2010-11-06T09:23:25Z 2010-11-06T09:23:25Z <p>Two simple examples with lots of interesting differentials are given by the Serre spectral sequences for integer homology (rather than cohomology) for the fibrations $$K({\mathbb Z}_2,1) \to K({\mathbb Z}_4,1)\to K({\mathbb Z}_2,1)$$ and $$K({\mathbb Z}_2,1) \to K({\mathbb Z},2) \to K({\mathbb Z},2)$$ where in the second case the map $K({\mathbb Z},2) \to K({\mathbb Z},2)$ induces multiplication by $2$ on $\pi_2$. In both cases one knows the homology of all three spaces and this allows one to work out what all the differentials must be. The differentials give a real shoot-out, with nontrivial differentials on more than one page, and in the second case there are nontrivial differentials on infinitely many pages. The best thing is to work everything out oneself, but if you want to check your answers these two examples are worked out as Examples 1.6 and 1.11 in Chapter 1 of my spectral sequence notes, available on my webpage.</p> <p>These examples may not really be the sort of thing you're looking for since they involve computing differentials purely formally, not by actually digging into the construction of the spectral sequence. But of course a lot of spectral sequence calculations have to be formal if one is to have any chance of succeeding.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45067#45067 Answer by Jeff Strom for Spectral sequences: opening the black box slowly with an example Jeff Strom 2010-11-06T12:35:44Z 2010-11-06T12:35:44Z <p>There's a paper of Fadell and Hurewicz (in the Annals, mid 1950's) identifying certain differentials with cap products. I can't recall the precise result.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45069#45069 Answer by Tyler Lawson for Spectral sequences: opening the black box slowly with an example Tyler Lawson 2010-11-06T12:57:08Z 2010-11-06T12:57:08Z <p>One fruitful example of differentials is the Lyndon-Hochschild-Serre spectral sequence associated to a short exact sequence of groups $1 \to N \to G \to G/N \to 1$, which is of the form <code>$$H^p(G/N, H^q(N, M)) \Rightarrow H^{p+q}(G, M)$$</code> and similarly for homology. Since you know specific interpretations for low-level groups in terms of crossed homomorphisms and extensions, you might try working out differentials (to put together the inflation-restriction sequence) more explicitly. You might also try full computations with mod-p cohomology for $G$ a small p-group, and $N$ something like its center or its commutator subgroup.</p> <p>But as most of the previous examples have concentrated on examples arising from topological considerations, let me advocate the idea that you should build examples yourself.</p> <p>When learning about chain complexes it's helpful to write down small, illustrative examples from algebra when learning to compute homology, and this is no different. One of the simplest ways to do this is to work with the spectral sequence associated to a <a href="http://en.wikipedia.org/wiki/Spectral_sequence#The_spectral_sequence_of_a_double_complex" rel="nofollow">double complex</a>. This has two different associated spectral sequences, one for filtering the total complex by rows and one for filtering by columns, and this gives you two ways to compute the result that you can play off each other to determine differentials (and extensions).</p> <p>For example, if you write down <em>any</em> double complex where the rows are exact, the "vertical differentials first" spectral sequence must ultimately result in the death of all classes.</p> <p>Other double complex examples are "limit/colimit" spectral sequences where you only have nonzero entries for two values of $p+q$, and where the nonzero horizontal differentials are all isomorphisms. These ones are simple enough that you can get some intuition for chasing higher differentials.</p> <p>Finally, another suggested exercise is to look carefully into the exact couple formalism. If you have an exact triangle $D \to D \to E \to D$ where you understand all the groups and maps, you can trace explicitly through the derived couples and see how they're building up the limiting object. (Unfortunately a lot of exact couples seem somewhat artificial in nature at first because they are formed by summing up a large number of separate long exact sequences. They're much more common than one might think.)</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45084#45084 Answer by Jeffrey Giansiracusa for Spectral sequences: opening the black box slowly with an example Jeffrey Giansiracusa 2010-11-06T18:12:58Z 2010-11-06T18:12:58Z <p>In Jim Stasheff's original papers on $A_\infty$-algebras he generalises the bar construction and its spectral sequence to $A_\infty$-algebras, and then he identifies all of the differentials explicitly in terms of things that look like (duals of) Massey products - he calls them Yessam products. </p> <p>I realise that this isn't exactly an explicit compution, but it is at least a general explicit description of the higher differentials that is not formal, and to get it requires a bit of the guts of the spectral sequence.</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45650#45650 Answer by Dev Sinha for Spectral sequences: opening the black box slowly with an example Dev Sinha 2010-11-11T05:56:15Z 2010-11-11T05:56:15Z <p>Let me elaborate on part of Tyler's answer, and "second" the importance of constructing examples yourself in a setting where that is feasible, namely that of bicomplexes. Even though as you say they seem more algebraic, they do come up in topology (as I note below) and you can "see everything" in first examples. Here's what I usually suggest to my students (often in or right after a first-year course).</p> <p>Exercise: Show that the bicomplex with entries of ${\bf k}$ (the ground ring) at (0,1), (1,1), (1,0) and (2,0) and all horizontal and veritcal differentials given by identity maps is acyclic, but the spectral sequence obtained by taking homology vertically first has a non-trivial $E^2$ page and a non-trivial $d_2$ differential which kills that $E^2$ page.</p> <p>Exercise: Construct bicomplexes with arbitrarily long differentials.</p> <p>Playing with these for a while you see what's going on, but they might seem too algebraic/ artificial. Here's something which you can understand at almost the same level of detail but which computes something interesting.</p> <p>Definition: Let $A \bigoplus A^i$ be a differential graded associative algebra (that is, a cochain complex with an associative multiplication for which the differential is governed by the Leibniz rule), defined in non-negative degrees with say $H^0(A) = {\bf k}$. Then Bar(A), the bar construction on $A$, is the bicomplex which in bidegree (-p, q) is the degree $q$ part of $\overline{A}^{\otimes p}$ - which we denote $a_1 | a_2 | \cdots | a_p$ (you can pretend at first that $\overline{A}$ is just $A$ - really, $\overline{A}$ agrees with $A$ in degrees two and greater, is 0 in degree zero, and in degree one we replace $A^1$ by $A^1/ im(d)$). The vertical differential is the standard one on $A^{\otimes p}$ (using the Leibniz rule to extend to the tensor product), and the horizontal differential is defined as a sum obtained by "removing bars and multiplying." Let ${\bf k}$ be $F_2$ if you don't want to worry about signs.</p> <p>Exercise: Show this is a bicomplex. If you're brave, do so with signs.</p> <p>Exercise: The spectral sequence for this bicomplex obtained by taking homology vertically first has $E^1$ given by $Bar(H_*(A))$, where the homology of $A$ is a differential graded algebra with zero differential. (This is an immediate application of one of the main theorems from a first algebraic topology course.)</p> <p>Theorem (Eilenberg-Moore, after Adams-Hilton): If $X$ is simply connected and $A$ is (quasi-isomorphic to) the cochains of $X$, then the homology of $Bar(A)$ is isomorphic to the cohomology of the loopspace of $X$.</p> <p>Exercises: first compute for $A$ with all products zero. Then compute when $A$ is a free associative algebra, and then free commutative algebra. Finally, try to make examples with higher differentials.</p> <p>(I might elaborate more later... and we can talk at the Cascade seminar this week-end).</p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/49508#49508 Answer by Matthias Künzer for Spectral sequences: opening the black box slowly with an example Matthias Künzer 2010-12-15T10:14:50Z 2010-12-15T10:14:50Z <p>Verdier and Deligne introduced more terms in a spectral sequence of a filtered complex (indexed by 4+1 indices instead of 2+1), thus in particular factoring the differentials into epis and monos. This enables one to splice certain short exact sequences to obtain Massey triangles or to obtain H(E_r) = E_{r+1}. Not sure whether this helps for calculations, but it does help with understanding what's going on with differentials. Cf. Deligne, Décompositions dans la catégorie dérivée, appendix, MR1265526; cf. also Verdier, Des catégories dérivées des catégories abéliennes, MR1453167.</p>