# Products and the skeletal filtration in K-theory

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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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.
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