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I know that the individual cohomology groups are representable in the homotopy category of spaces by the Eilenberg-MacLane spaces. Is it also true that the entire cohomology ring is representable? If so, is there a geometric interpretation of the cup product as an operation on the representing space?

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The total cohomology of spaces should be thought of as a graded ring, or even more precisely as a graded E* algebra where E* is the cohomology of a point. It is representable in the sense that there is a graded E* algebra object in hTop representing it.

Let's unpack that a little.

First, you have to understand about group objects and so forth. The full story is in Lawvere theories, but the basic idea is that a, say, group object in a category 𝒞 is an object, say C, of 𝒞 together with all the structure needed to ensure that the contravariant hom-functor 𝒞(-,C) lifts from a functor to Set to a functor to Group. Providing 𝒞 has enough products of C, it's simple to write down a few morphisms that C must have, together with some diagrams that must commute. Lawvere theories are the correct way of thinking about these things in general, but in a specific instance it can be instructive to just write everything down.

Now, that works fine for groups, abelian groups, modules, rings, and so forth, but cohomology isn't any of those things. Cohomology is a graded ring. So we need graded objects in our category. To talk about graded objects we need a grading set, say Z. This doesn't have to have any structure whatsoever. A Z-graded object in 𝒞 is just a functor Z → 𝒞 where Z is viewed as a discrete category. So it's a family of objects in 𝒞, indexed by the elements (objects) of Z. A Z-graded object in 𝒞 represents a functor from 𝒞 into SetZ, the category of Z-graded sets. While we can send a Z-graded set to a set either by its coproduct or its product, we shouldn't do so. We should keep the labelling.

That's because we now want to mix these two things and talk of graded rings, or more generally graded theories, also called many-sorted theories. In a single-sorted theory, such as groups or rings, the standard set-based (or more generally 𝒞 based) groups or rings consist of a set (object) together with certain functions (morphisms), called operations, from certain n-fold products of that set (object) to itself. In a graded theory, the operations go from products of components of the graded set (object) to other components. Thus in a graded abelian group, say A, we have operations A(z) × A(z) → A(z) but not A(z) × A(z') → A(z''). That is, we can only add terms in the same component.

So back to cohomology. Cohomology is a representable functor into the category of graded E* algebras, where the grading set is ℤ. So for each integer n we have a space En in hTop and for each operation of a graded E* algebra we have an operation En × Em → El. In particular, multiplication corresponds to a map (technically, a homotopy class of maps as we're in hTop) En × Em → En+m. For ordinary cohomology, these spaces are the Eilenberg-Maclane spaces.

Several remarks are in order here.

Sometimes, the operations in a graded object come in families. Then it can be severely tempting to put the graded pieces together into one single thing, either by taking the coproduct or the product. However, this destroys information. Just because the primary operations come in families doesn't mean that all of the operations come in families. For example, the primary structure of a graded E* algebra come in suitable families and it can be tempting to put the pieces of the cohomology together. (One has to choose what method to use: product or coproduct. Product generally works better because in a product one doesn't have to worry about things remaining finite, which is good, because you have no control over this.) But cohomology theories don't just have their primary structure operations, they also have a whole raft of other operations. The main division of these two is into stable and unstable operations. By putting your theory together into a single object, you are effectively saying that you will only consider stable operations and will ignore unstable ones. While this is basically okay for cohomology, for K-theory it is a disaster because we don't know what the stable ones are!.

Secondly, to correct something Chris said, the spaces En do keep track of the suspension isomorphisms. The fact that there are suspension isomorphisms in cohomology theories is encoded in the fact that that there is an equivalence ΩEn ≃ En-1. Also, the Mayer-Vietoris maps are part of the fact that cohomology is representable (you get MV from the long exact sequence for pairs, and that's needed for representability).

Of course, knowing that ΩEn ≃ En-1 allows you to construct a spectrum from the spaces En, even an Ω-spectrum, and those operations that come in families, namely the stable ones, become morphisms of spectra. Once you have that, you can define the associated homology theory and lots of other wonderful things. But the point is that there's a lot that you can do before going to spectra and, up to a point, the language of spectra is merely a way of keeping track of the grading.

Finally, this probably isn't geometric enough for the cup product in ordinary cohomology, but there you're asking a difficult question. To give a truly geometric interpretation, you would first need to find good geometric models of all the K(π,n)s. While this can be done for a few, I don't know of a good family for all of them ("good" in the sense of "easy to think about" rather than anything technical). One can start out fairly well, say with π = ℤ, with ℤ, S1, ℂℙ, but then it gets a bit sticky. Another model for K(ℤ,2) is the projective unitary group on a Hilbert space and this acts on the general linear group for the space of Hilbert-Schmidt operators on that same Hilbert space so you could make K(ℤ,3) as the quotient of GL(H⊗H)/ℙ(H) and this is essentially the start of gerbe theory, but, as I'm sure you'll agree, it's hard to look at that and say "Aha! Now I understand K(ℤ,3)." in quite the same way as one can look at ℂℙ and understand the connection between H2(X;ℤ) and line bundles.

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Could you explain what you have in mind when saying that putting a cohomology theory into a single object loses unstable operations? Which information precisely is lost, and how come? –  Sam Derbyshire Oct 20 '09 at 10:42
    
You lose the action of those operations that are unstable but not stable. For example, in K-theory you lose all but two of the Adams operations. The problem is that when you put your representing objects into a whole then your morphisms have to be globally defined (otherwise you aren't really putting your objects into one object). There are lots of interesting operations (such as Adams operations) which cannot be globally defined. –  Loop Space Oct 20 '09 at 14:05
    
So the cup product H^1 x H^1 -> H^2, should come from a homotopy class of maps S^1 x S^1 -> CP^inf ? Is it clear what this map should be? –  Dinakar Muthiah Oct 20 '09 at 15:21
    
Dinakar: The cup product is represented by the tensor product of the two universal classes in H^*(S^1 \times S^1)=H^*(S^1) \otimes H^*(S^1). This is just a generator of H^2. Now the 2-skeleton of CP^\infty is just CP^1=S^2, and the inclusion of the 2-skeleton gives an isomorphism on H^2. Thus the universal class on CP^\infty pulls back to the fundamental class on S^2, which will then pull back to the desired class on the torus via any degree 1 map. So you can get your map by taking S^1 \times S^1 \to S^2=CP^1 \to CP^\infty, where the first map has degree 1. –  Eric Wofsey Oct 20 '09 at 18:22

The cohomology ring is representable by the product of all the Eilenberg-Mac Lane spaces K(Z, n) as n varies. Note that this gives the product of the abelian groups HnX, not their direct sum. For instance when X = CP we get the power series ring Z[[t]], not the polynomial ring Z[t]. This is in some ways more natural (it's properly dual to taking the direct sum of homology groups) and is often what you want anyways (as in the relationship between cohomology theories and formal group laws).

There is a homotopy class of maps which represents the cup product, but I don't know of a geometric interpretation.

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Yes, but it is much better to look at the representing spectrum. Cohomology in degree n is represented by the (pointed) space K(Z, n), as you pointed out. Then the product R of all the K(Z, n) where n ranges over all non-negative integers is the representing object for the whole cohomology ring. By the Yoneda lemma (applied to the homotopy category of spaces) the cup product is represented by maps:

K(Z, n) x K(Z, m) --> K(Z, n + m)

These assemble to give a single map R x R ---> R. This works for any (connective) multiplicative generalized cohomomology theory.

What you DON'T get from this single space R are any of the suspension isomorphisms, or equivalently the Mayer-Vietoris maps. A (naive) ring spectrum also keeps track of these maps.

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