# Integral cohomology (stable) operations

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:

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?

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.

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.

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?

My apologies for the barrage of questions, if people think it would be better split up, I would be happy to do so.

• All stable rational cohomology operations are trivial, i.e. $H\mathbb Q^{\ast}H\mathbb Q$ is $\mathbb Q$ concentrated in degree $0$. – Fernando Muro Jun 5 '13 at 22:14

$HZ^nHZ$ is trivial for $n<0$. $HZ^0HZ$ 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.

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.

• I do not understand how those groups can be finite for all n larger than 0. It seems that there should be an operation $\beta_p$ for each prime, a composition of the reduction mod p and then the integral bockstein, in $H\mathbb{Z}^1 H\mathbb{Z}$. – Sean Tilson Dec 28 '10 at 6:59
• The composite of reduction mod p and the integral Bockstein is zero by construction as the integral Bockstein is the boundary map coming from the sequence $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/p\to0$. – Torsten Ekedahl Dec 28 '10 at 9:08
• I also believe that it is "well-known" that stable operations of degree $1$ are just given by Ext-groups of the two coefficient groups between which you want to calculate operations. Of course this implies that $H\mathbb{Z}^1H\mathbb{Z} = 0$. – Markus Land Mar 17 '13 at 10:31

A possibly interesting analogue of the formula $$H\mathbb{F}_{2*} H\mathbb{F}_2 = \otimes_{i\ge1} \mathbb{F}_2[\xi_i]$$ is $$H\mathbb{Z}_{(2)*} H\mathbb{Z}_{(2)} = \bigotimes^\mathbb{L}_{i\ge1} \mathbb{Z}_{(2)*}[\xi_i^2]/(2\xi_i^2)$$, where $$\otimes^{\mathbb{L}}$$ means the derived tensor product. In other words, resolve $$\mathbb{Z}_{(2)*}[\xi_i^2]/(2\xi_i^2)$$ by (flat or) free $$\mathbb{Z}_{(2)}$$-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.

• This is a great answer. It is a fair bit suggestive of some sort of decomposition of $H\mathbb{Z}_{(2)*}H\mathbb{Z}_{(2)}$. What I mean is that the right hand side looks like it would really like to be a K\"{u}nneth spectral seqeunce. – Sean Tilson Jun 5 '13 at 15:18
• Sorry for the bump on this old answer, but just to point out that this does indeed come from a K\"unneth thingy: we have $\mathbb{Z} \wedge \mathbb{Z}= (\mathbb{Z}\wedge \mathrm{BP})\otimes_{\mathrm{BP}} \mathbb{Z}$. Then $\mathbb{Z}$ is the tensor product in BP-modules of all the BP/v_i, and the image of v_i in homology is p times $\overline{\xi}_i$ (at odd primes) and 2 times $\overline{\zeta}_i^2$ at 2, whence the decomposition above. (everything p-local above) – Dylan Wilson Nov 24 at 21:24

Here is another interesting reference on $H\mathbb{Z}_* H\mathbb{Z}$ and $H\mathbb{Z}^* H\mathbb{Z}$:

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.

It contains in particular the theorem explained above by Tom, that the $p$-primary part is killed by $p$.

1. That is right.

2. The slightly easier calculation, I think, is $H\mathbb{Z}_\ast H\mathbb{Z}$, and this is easier to approach one prime at a time, i.e., by calculating $H\mathbb{Z}_\ast H\mathbb{Z}_{(p)}$, 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" (Stable homotopy and generalised homology). The main thing to take away is that $H\mathbb{Z}_nH\mathbb{Z}_{(p)}$ is $p$-torsion (i.e., in the kernel of multiplication by $p$) for all $n>0$.

3. Steenrod's original definition was by a chain level construction, called the cup-i product. This is discussed in some other questions, such as here.

Note that I'm discussing the (stable) homology of the Eilenberg MacLane spectrum. The homology of the integral Eilenberg MacLane spaces $H_\ast K(\mathbb{Z},n)$ are quite a bit more complicated.

• So the cup-i construction works integrally? Also, I was under the impression that this data was hard to assemble together into one object. I get the impression from your answer, as well as Toms, that this is not the case. Is that a fair assessment? – Sean Tilson Dec 28 '10 at 7:01
• Oh, integrally. The exact cup-i's may or may not be defined integrally ... you'd have to check. But they descend from a chain level construction which exists integrally. – Charles Rezk Dec 28 '10 at 14:35
• A reference for 1. is Kraines, David Rational cohomology operations and Massey products. Proc. Amer. Math. Soc. 22 1969 238–241. – Mark Grant Dec 28 '10 at 15:57