Integral cohomology (stable) operations - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T14:41:51Z http://mathoverflow.net/feeds/question/50519 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/50519/integral-cohomology-stable-operations Integral cohomology (stable) operations Sean Tilson 2010-12-27T23:02:06Z 2013-06-05T11:34:20Z <p>There have been a couple questions on MO, and elsewhere, that have made me curious about integral or rational cohomology operations. I feel pretty familiar with the classical Steenrod algebra and its uses and constructions, and I am at a loss as to imagine some chain level construction of such an operation, other than by coupling mod p operations with bockstein and reduction maps. I am mostly just curious about thoughts in this direction, previous work, and possible applications. So my questions are essentially as follows:</p> <p>1) Are there any "interesting" rational cohomology operations? I feel like I should be able to compute $H\mathbb{Q}^*H\mathbb{Q}$ by noticing that $H\mathbb{Q}$ is just a rational sphere and so there are no nonzero groups in the limit. Is this right?</p> <p>2) Earlier someone posted a reference request about $H\mathbb{Z}^*H\mathbb{Z}$, and I am just curious about what is known, and what methods were used.</p> <p>3) Is there a reasonable approach, ie explainable in this forum, for constructing chain level operations? the approaches I have seen seem to require some finite characteristic assumptions, but maybe I am misremembering things.</p> <p>4) I am currently under the impression that a real hard part of the problem is integrating all the information from different primes, is this the main roadblock? or similar to what the main obstruction is?</p> <p>My apologies for the barrage of questions, if people think it would be better split up, I would be happy to do so.</p> <p>Thanks for your time.</p> http://mathoverflow.net/questions/50519/integral-cohomology-stable-operations/50536#50536 Answer by Tom Goodwillie for Integral cohomology (stable) operations Tom Goodwillie 2010-12-28T04:31:27Z 2010-12-28T13:54:45Z <p>$HZ^nHZ$ is trivial for $n&lt;0$. $HZ^0HZ$ is infinite cyclic generated by the identity operation. For $n>0$ the group is finite. So you know everything if you know what's going on locally at each prime. For $n>0$ the $p$-primary part is not just finite but killed by $p$, which means that you can extract it from the Steenrod algebra $H(Z/p)^{*}H(Z/p)$ and Bocksteins.</p> <p>EDIT Here's the easier part: The integral homology groups of the space $K(Z,n)$ vanish below dimension $n$, and by induction on $n$ they are all finitely generated. Also $H_{n+k}K(Z,n)$ is independent of $n$ for roughly $n>k$, so that in this stable range $H_{n+k}K(Z,n)$ is $HZ_kHZ$, which is therefore finitely generated. This plus the computation of rational (co)homology gives that $HZ_kHZ$ is finite for $k>0$. Here's the funny part: Of course one expects there to be some elements of order $p^m$ for $m>1$ in the (co)homology of $K(Z,n)$, and in fact there are; the surprise is that stably this is not the case.</p> http://mathoverflow.net/questions/50519/integral-cohomology-stable-operations/50537#50537 Answer by Charles Rezk for Integral cohomology (stable) operations Charles Rezk 2010-12-28T04:32:12Z 2010-12-28T04:32:12Z <ol> <li><p>That is right.</p></li> <li><p>The slightly easier calculation, I think, is $H\mathbb{Z}_*H\mathbb{Z}$, and this is easier to approach one prime at a time, i.e., by calculating <code>$H\mathbb{Z}_*H\mathbb{Z}_{(p)}$</code>, which is something you can do using the classical Adams spectral sequence. I don't have it handy to check, but I suspect that this calculation is carried out in Part III of Adams's "blue book" (<em>Stable homotopy and generalised homology</em>). The main thing to take away is that <code>$H\mathbb{Z}_nH\mathbb{Z}_{(p)}$</code> is $p$-torsion (i.e., in the kernel of multiplication by $p$) for all $n>0$. </p></li> <li><p>Steenrod's original definition was by a chain level construction, called the cup-i product. This is discussed in some other questions, such as <a href="http://mathoverflow.net/questions/19762/homotopy-commutativity-of-the-cup-product" rel="nofollow">here</a>. </p></li> </ol> <p>Note that I'm discussing the (stable) homology of the Eilenberg MacLane spectrum. The homology of the integral Eilenberg MacLane spaces $H_*K(\mathbb{Z},n)$ are quite a bit more complicated.</p> http://mathoverflow.net/questions/50519/integral-cohomology-stable-operations/105132#105132 Answer by Martin Frankland for Integral cohomology (stable) operations Martin Frankland 2012-08-20T23:27:11Z 2012-08-20T23:27:11Z <p>Here is another interesting reference on $H\mathbb{Z}_* H\mathbb{Z}$ and $H\mathbb{Z}^* H\mathbb{Z}$:</p> <p>Kochman, Stanley; Integral cohomology operations. Current trends in algebraic topology, Part 1 (London, Ont., 1981), pp. 437–478, CMS Conf. Proc., 2, Amer. Math. Soc., Providence, R.I., 1982.</p> <p>It contains in particular the theorem explained above by Tom, that the $p$-primary part is killed by $p$.</p> http://mathoverflow.net/questions/50519/integral-cohomology-stable-operations/132834#132834 Answer by John Rognes for Integral cohomology (stable) operations John Rognes 2013-06-05T11:34:20Z 2013-06-05T11:34:20Z <p>A possibly interesting analogue of the formula $HF_{2*} HF_2 = \otimes_{i\ge1} F_2[\xi_i]$ is $HZ_{(2)*} HZ_{(2)} = \otimes^L_{i\ge1} Z_{(2)*}[\xi_i^2]/(2\xi_i^2)$, where $\otimes^L$ means the derived tensor product. In other words, resolve $Z_{(2)*}[\xi_i^2]/(2\xi_i^2)$ by (flat or) free $Z_{(2)}$-modules, tensor the resolutions together, and pass to homology. If I recall correctly, the "first" interesting class $\xi_2^3 + \xi_1^2 \xi_3$ (in degree 9) arises as a torsion product of $\xi_1^2$ and $\xi_2^2$. I needed this for a Shukla homology calculation once. Presumably there is also an odd story.</p>