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Given a finite CW complex X, there is a filtration of the topological K-theory of X given by setting $K_n(X) = \ker \left(K(X) \to K(X^{(n-1)})\right)$, where $X^{(n-1)}$ is the (n-1)-skeleton of X. (The choice of indexing here is from Atiyah-Hirzebruch.)

My question is:

How does this filtration interact with the external product $K(X)\times K(Y)\to K(X \times Y)$? I believe the answer should be that $K_n (X) \cdot K_m (Y) \subset K_{n+m} (X\times Y)$.

Just to be clear, and to set notation, this external product is the one induced by sending a pair of vector bundles $V\to X$ and $W\to Y$ to the external tensor product, which I'll write $V\widetilde{\otimes} W = \pi_1^* V \otimes \pi_2^* W \to X\times Y$.

Of course, if $V\in K_n (X)$ and $W \in K_m (Y)$, then $V\widetilde{\otimes} W$ restricts to zero in both $K(X^{(n-1)} \times Y)$ and $K(X \times Y^{(m-1)})$, and $(X\times Y)^{(n+m-1)}$ is contained in the union of these two subsets. Is there some way to deduce from this information that the class $V\widetilde{\otimes} W$ is actually trivial in $K((X\times Y)^{(n+m-1)})$?

Here's the reason I'm asking (which is really a second question, I guess). In Characters and Cohomology Theories, Atiyah states (without comment) that for the internal product $K(X)\times K(X)\to K(X)$, one has $K_n (X) \cdot K_m (X) \subset K_{n+m} (X)$. In Atiyah-Hirzebruch, they state this formula and say that it "admits a straighforward proof."

I thought I remembered that the straighforward proof was the following:

  1. Show that the external product satisfies $K_n (X) \cdot K_m (Y) \subset K_{n+m} (X\times Y)$

  2. Observe that if $f:X\to X\times X$ is a cellular approximation to the diagonal $X\to X\times X$, then $f(X^{(n+m-1)}) \subset (X\times X)^{(n+m-1)}$. So for any $V, W\in K(X)$, we have $V\otimes W = f^*(V\widetilde{\otimes} W)$, and if $V\in K_n (X)$ and $W\in K_m (X)$, it then follows from 1. that $V\otimes W\in K_{n+m} (X)$.

Am I barking up the wrong tree here?

Presumably these questions will turn out to have an easy answer, but I've been thinking about them for a while now and haven't gotten any further. Any suggestions or references would be great! I haven't found any sources other than the two mentioned above that talk about the relation between skeleta and products, and neither of these sources mentions case of external products.

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up vote 5 down vote accepted

Hi Dan, welcome to Math Overflow.

The group you denote $K_m(X)$ is the image of the relative K-group $K(X,X^{(m-1)})$, which for nice spaces (e.g. finite CW-complexes) consists of equivalence classes of formal differences $V - W$ of vector bundles equipped with an isomorphism $V|_{X^{(m-1)}} \cong W|_{X^{(m-1)}}$. The product on K-groups lifts to an exterior pairing $$ K(X,A) \otimes K(Y,B) \to K(X \times Y,A \times Y \cup X \times B). $$ In particular, if $X$ and $Y$ are CW then using the standard CW structure on the product we have $$(X \times Y)^{(n+m-1)} \subset (X^{(n-1)} \times Y) \cup (X \times Y^{(m-1)}).$$ This gives us an exterior pairing $$ K(X,X^{(n-1)}) \otimes K(Y,Y^{(m-1)}) \to K(X \times Y,(X \times Y)^{(n+m-1)}) $$ that lifts the ordinary K-theory product, and implies the result you want about the image of the group $K_n(X) \times K_m(Y)$. This answers your part (1), and (2) follows just as you said.

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Thanks, Tyler! I'll just point out for anyone interested that this relative exterior pairing becomes quite easy to think about if you use reduced K-theory (and then the non-reduced case follows from the reduced case). The key point is that `$(X\times Y)/(A\times Y \cup X \times B) \simeq X/A \wedge Y/B$'. As with most problems in (algebraic topological) life, this could have been solved by a more careful inspection of Hatcher's notes (in this case, p. 55 of his Vector Bundles book). – Dan Ramras Feb 28 '10 at 18:31

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