# Commutativity in K-theory and cohomology

The Chern classes give a map f:BU \to \prod_n K(Z,2n), which is a rational equivalence. However, it is not an equivalence over Z because the cohomology of BU is just a polynomial algebra and has no Steenrod operations. In particular, the generators of the homotopy groups \pi_{2n}(BU)=Z will not map to generators of the homotopy groups \pi_{2n}(K(Z,2n))=Z. Another way to say this is that the duals of the Chern classes in homology are not in the image of Hurewicz; only certain multiples of them are. What multiples you have to take is determined by the order of the k-invariants of BU, which are certain Steenrod operations of the fundamental classes of K(Z,2n).

Steenrod operations can be understood as obstructions to the cup product on ordinary cohomology being strictly commutative. On the other hand, the fact that f is not an equivalence can also be understood as an obstruction to addition in K-theory being strictly commutative. Indeed, any space with a strictly commutative group structure is a product of K(\pi_n,n)s under the map given by a right inverse of the Hurewicz map.

So in some sense you could say that products in cohomology are only homotopy-commutative and sums in K-theory are only homotopy-commutative "for the same reason". Is there some deeper story behind this? I don't know exactly what I'm asking for, but I'd like to get a better understanding of what's going on in this picture.

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## 1 Answer

Perhaps the deeper story you want involves the notion of "E-infinity product". The cup product in cohomology, and the sum (and for that matter, the product) in K-theory are commutative (and associative, and unital) not merely up to homotopy, but "up to all possible higher homotopies". You can make this precise by saying that the appropriate binary operation on the representing space (the K(Z,n)'s, or Z x BU), is part of an E-infinity algebra structure on that space.

It seems that most "naturally occuring" sums or products in topology turn out to be E-infinity (addition for any generalized cohomology theory, multiplication for many nice ones such as ordinary cohomology, K-theories, bordism, elliptic cohomology). As you observe, having a strictly commutative operation is very special, and basically forces the representing spaces to be a product of K(A,n)'s, by the Dold-Thom theorem.

To me, the mystery here is that "stricly commutative" is so much more special than "E-infinity commutative". Associative products don't behave this way: "strictly associative" turns out to be no more special than "A-infinity associative", that is, any A-infinity product on a space can be "rigidified" to a strictly associative product on a weakly equivalent space.

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I guess what I was looking for more was an explanation for why the nonstrictness of commutativity of the product in ordinary cohomology and the sum in K-theory can be seen as coming from the same source. That is, assuming that a deep explanation exists. – Eric Wofsey Oct 19 '09 at 1:26
Oh. I can't think of an explanation like that. It seems to me that "non-strictness" is the generic phenomenon, and doesn't need a special explanation. – Charles Rezk Oct 19 '09 at 1:50