Simple examples for the use of spectral sequences - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T20:14:29Z http://mathoverflow.net/feeds/question/23297 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences Simple examples for the use of spectral sequences Hanno Becker 2010-05-02T23:39:00Z 2012-01-29T19:29:48Z <p>I'm looking for basic examples that show the usefulness of spectral sequences even in the simplest case of spectral sequence of a filtered complex. </p> <p>All I know are certain "extreme cases", where the spectral sequences collapses very early to yield the acyclicity of the given complex or some quasi-isomorphism to another easier complex (balancing tor, for example).</p> <p>Is there an example of a useful filtration where one really computes something nontrivial also in the higher sheets?</p> <p>The examples I have in mind come from topology. For example, the calculation of <code>$H_{\ast}(\Omega{\mathbb S}^n;{\mathbb Z})$</code> is simply beautiful using the Serre spectral sequence, and one needs to pass to the $n$-th sheet until something happens. Another more difficult example would be the computation of the rational cohomology of <code>$K({\mathbb Z},n)$</code> by induction on $n$ (depending on the parity of $n$, we get a polynomial algebra or an exterior algebra, if I remember correctly). </p> <p>Are there similar, but purely algebraic examples which could show the usefulness of spectral sequences to those seeing them the first time?</p> http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences/23316#23316 Answer by Greg Muller for Simple examples for the use of spectral sequences Greg Muller 2010-05-03T04:19:59Z 2010-05-03T04:19:59Z <p>This isn't exactly what you asked, but its a very simple example that (to me) demonstrates some of the necessity of the complexities of spectral sequences. Consider the ring $R=\mathbb{C}[x,y]$, and consider the module $M=(Rx+Ry)\oplus R/x$. Then the double dual spectral sequence converges to the original module: $$Ext^{-i}_R(Ext^j_R(M,R),R) \Rightarrow M$$ The second page of this spectral sequence has </p> <ul> <li>$R$ in degree $(0,0)$</li> <li>$R/x$ in degree $(-1,1)$</li> <li>$R/(Rx+Ry)$ in degree $(-1,2)$</li> <li>A non-trivial knights-move map (differential on the second page) from $(0,0)$ to $(-1,2)$ which is the natural quotient map.</li> </ul> <p>The spectral sequence collapses on the third page, with $(Rx+Ry)$ in degree $(0,0)$ and $R/x$ in degree $(-1,1)$. One shortcoming of this example is that you get the same module back, split apart into different components; rather than the associated graded of some interesting filtration. If memory serves, there was a way to tinker with this example to give it that property too, but it escapes me at the moment.</p> http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences/23324#23324 Answer by Daniel Moskovich for Simple examples for the use of spectral sequences Daniel Moskovich 2010-05-03T06:39:17Z 2010-05-03T06:39:17Z <p>The best example I can think of is the <a href="http://en.wikipedia.org/wiki/Lyndon%E2%80%93Hochschild%E2%80%93Serre_spectral_sequence" rel="nofollow">Lyndon-Hochschild-Serre spectral sequence</a> in group cohomology. See for instance Chapter VII Section 6 of Brown's <em>Cohomology of Groups</em>.<br> The spectral sequence, for a group extension $1\to H\to G\to Q \to 1$ (I'm using Brown's notation) and for a G-module M, is of the form $$E^2_{pq}=H_p(Q,H_q(H,M))\Rightarrow H_{p+q}(G,M).$$ Let's use this to calculate the third integral homology of the dihedral group D<sub>2n</sub>, for $n$ an odd integer. The group extension is $$1\to C_n\to D_{2n}\to C_2 \to 1,$$ and the corresponding Lyndon-Hochschild-Serre spectral sequence is $$E^2_{pq}=H_p(C_2,H_q(C_n,\mathbb{Z}))\Rightarrow H_{3}(D_{2n},\mathbb{Z}),$$ where p and q add up to 3. The integral homology of a cyclic group C<sub>m</sub> is $\mathbb{Z}$ for $q=0$, vanishes for $q\in\mathbb{N}^\ast$ even, and is $\mathbb{Z}/m\mathbb{Z}$ for $q$ odd. Plug this information into the Lyndon-Hochschild-Serre spectral sequence, and you find $$H_{3}(D_{2n},\mathbb{Z})=H_0(C_2,\mathbb{Z}/n\mathbb{Z})\oplus H_3(C_2,\mathbb{Z})\simeq \mathbb{Z}/2n\mathbb{Z}.$$ This is easy and elegant calculation in my opinion, and occurs in practice in knot theory ($H_{3}(D_{2n},\mathbb{Z})$ turns out to be isomorphic to the relative bordism group of Fox n-coloured knots).</p> http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences/23329#23329 Answer by Kevin Buzzard for Simple examples for the use of spectral sequences Kevin Buzzard 2010-05-03T08:22:03Z 2010-05-03T08:22:03Z <p>This is not the most profound answer but it's something that came up last week when I was reading a paper. The author wanted to prove that a certain obstruction to a problem was zero, and the obstruction lived in an $H^2$ that looked a bit scary: it was $H^2(W,V)$ with $W$ a local Weil group and $V$ a finite-dimensional vector space over a field of characteristic zero (I'll tell you all you need to know about this Weil group in a sec, in case you don't know what one is; it's a topological group coming up in number theory). But then I realised the obstruction was really in the image of $H^2(W/C,V)$ with $C$ a compact open subgroup of $W$ (a finite index subgroup of inertia). </p> <p>But now I'm done because $W/C$ has a two-step filtration with a finite sub and a quotient isomorphic to $\mathbf{Z}$, and I know finite groups have no cohomology in char 0 in degrees 1 or more, and $\mathbf{Z}$ is the fundamental group of a 1-dimensional thing so it has no cohomology in degree 2 or more, and so by Hochschild-Serre, a calculation I can even do in my head in this example, there are no non-zero terms to build $H^2(W/C,V)$ from in $E_2$ and hence in $E_\infty$ so this group vanishes. I can do this calculation without even pulling out a piece of paper.</p> <p>I'm sure if I were more "group-theoretic" I would have a much clearer picture about what was going on, but Hochschild-Serre just explains to me in a very concrete way how the cohomology of a group is built from the cohomology of its subs and quotients, and is definitely something I carry around in my "useful tools" bag.</p> http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences/86972#86972 Answer by Timo Keller for Simple examples for the use of spectral sequences Timo Keller 2012-01-29T18:51:10Z 2012-01-29T18:51:10Z <p>Non-trivial spectral sequences occur when calculating the homotopy groups of tmf at 2 or 3: <a href="http://www.math.uiuc.edu/~rezk/512-spr2001-notes.pdf" rel="nofollow">http://www.math.uiuc.edu/~rezk/512-spr2001-notes.pdf</a> §16 ff., see also Tilman Bauer's article <a href="http://arxiv.org/abs/math/0311328" rel="nofollow">http://arxiv.org/abs/math/0311328</a></p> http://mathoverflow.net/questions/23297/simple-examples-for-the-use-of-spectral-sequences/86976#86976 Answer by Lalit Jain for Simple examples for the use of spectral sequences Lalit Jain 2012-01-29T19:29:48Z 2012-01-29T19:29:48Z <p>I can't recommend the following document by Tom Weston enough. It introduces spectral sequences rapidly and at a comfortable level of generality. It then applies the Hochschild-Serre sequence to group cohomology.</p> <p>www.math.mcgill.ca/goren/SeminarOnCohomology/infres.pdf</p>