Example of a space whose complex K-theory is not easily computable from singular cohomology

Question

I am looking for a counterexample to the formula $$ K^n(X) \cong \prod_{i\equiv n \mod 2} H^i(X) $$ where $K^*$ denotes complex topological $K$-theory, $H^*$ singular cohomology and $X$ a compact space. I would hope for an example where $X$ is a finite $CW$-complex.

There should be such an example, as I was told that the $K$-theoretic Atiyah-Hirzebruch sprectral sequence does not collapse at $E^2$ in general.

Answer 1

Good example is $X=\mathbb RP^n$ (see e.g. $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$).

Answer 2

Since you mention the Atiyah-Hirzebruch spectral sequence, I want to point out how you might construct a (minimal) example if you somehow didn't know any examples of topological spaces (but do know quite a bit of algebraic topology).

As we're working with the $K$-theory of spaces, we would like to understand the structure of the connective complex $K$-theory spectrum $ku$ better and find examples of $X$ for which the AHSS does not degenerate at $E_2$. It's well-known that the first differential in the AHSS is given by the bottom nontrivial $k$-invariant in the Postnikov tower of $ku$.

Suppose you somehow knew that this $k$-invariant is given by $$Q_1: \Sigma^{-1} H\mathbb{Z} \to \Sigma^2 H\mathbb{Z}.$$

Then, counterexamples to your question can be constructed if we can find spaces whose cohomology has a nontrivial $Q_1$ operation. Let's work mod $2$; we have $$Q_1 = \mathrm{Sq}^1 \mathrm{Sq}^2 + \mathrm{Sq}^2 \mathrm{Sq}^1$$ in mod $2$ cohomology. Therefore, a minimal cell complex with a nontrivial $Q_1$ has at least three cells connected by $\mathrm{Sq}^1$ and $\mathrm{Sq}^2$ in either order (i.e., a complex that looks like a question mark, or inverted by Spanier-Whitehead duality). We can build such complexes cell-by-cell by taking iterated cofibers of maps between spheres. In particular, since $\mathrm{Sq}^1$ detects $2$ and $\mathrm{Sq}^2$ detects $\eta$ in the homotopy groups of spheres, we can build the required three cell complex in two steps by taking cofibers of maps representing $2$ and $\eta$. That this construction is possible is guaranteed by the relation $2\eta = 0$ in $\pi^s_*$.

So we now have a finite complex with a nontrivial $Q_1$ operation, and hence a counterexample to your formula in the question. How does this relate to the example $\mathbb{R}P^n$ given in the other answer? Well, if you look at the cell structure of $\mathbb{R}P^n$, you would see exactly this kind of question mark complex as a frequently occuring motif.

Of course, this answer isn't very helpful since by the time you come up with it you would have other simpler ways to see this result. On the other hand, maybe it's a good showcase of how the tools developed in homotopy theory can answer such questions "intrinsically".