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If $F \rightarrow E \rightarrow B$ is a Serre fibration, and I know the complex $K$-theory of two of these spaces, what can I learn about the $K$-theory of the third? I wish there was a spectral sequence $$ K^i(B,K^j(F)) \implies K^{i+j}(E) $$ but it doesn't look like it. What can I do instead?

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    $\begingroup$ You could try the Atiyah Hirzebruch spectral sequence, which is like $H^i(B, K^j(F)) \Rightarrow K^{i+j}(E)$. $\endgroup$ Dec 16, 2012 at 3:32
  • $\begingroup$ Thanks Dylan. There are cases where the cohomology of B is harder to compute that the K-theory, like the classifying space of a finite group. Atiyah-Hirzebruch has me stumped in that case. $\endgroup$
    – ya-tayr
    Dec 16, 2012 at 3:48
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    $\begingroup$ There might be a sort of Eilenberg-Moore spectral sequence that looks like $Tor_{K^*B}(K^*E, K^*) \Rightarrow K^*E$, since $K$-theory has a nice Kunneth formula... but I'm not completely sure about that $\endgroup$ Dec 16, 2012 at 4:09
  • $\begingroup$ I'm also not sure about how to deal with convergence... $\endgroup$ Dec 16, 2012 at 4:09

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Here are a couple possible answers:

(1). You could follow Dylan's suggestion about the Atiyah-Hirzebruch spectral sequence for

$$H^{\ast}(B, K_{\ast}) \implies K^{\ast}(B)$$

and use the fact that you know the K-theory of $B$ to conclude something about the differentials in this spectral sequence. Then plug this information in to the Atiyah-Hirzebruch-Serre spectral sequence for the fibration.

(2). You can't (easily) use the Eilenberg-Moore spectral sequence

$$Tor_{K^{\ast} (B)} (K^{\ast} (E), K_{\ast}) \implies K^{\ast}(F),$$

since it won't generally converge (because K-theory is not connective). To my understanding, this paper by Tilman Bauer is the state of the art on these sorts of convergence questions.

(3). This is pretty distant from your desired answer, but it's too tempting not to share. In the case that $F =G$ is a group and $E \to B$ is a principal $G$-bundle, there is a bar spectral sequence

$$Tor_{K_{\ast}(G)} (K_{\ast} (E), K_{\ast}) \implies K_{\ast}(B),$$

at least in the case that $K_{\ast}(G)$ is flat over $K_{\ast}$, since $B$ is the Borel construction for the action of $G$ on $E$ (When the flatness assumption doesn't hold, you can either make do with a simplicial construction that's trying to be the bar complex; i.e., $K_{\ast}(G^{\times n} \times E)$, or reduce mod $p$ and do it one prime at a time). This of course requires that you know $K_{\ast}(G)$, and how it acts on $K_{\ast}(E)$, but can be fantastically successful when you do (see, e.g., Ravenel-Wilson's paper on the Morava K-theories of Eilenberg-MacLane spaces).

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    $\begingroup$ An addendum about the last suggestion: if it's not $K_{\ast}(B)$ that you're after, but rather $K_{\ast}(E)$, you can back up the fibre sequence by one to $\Omega B \to F \to E$, and run the indicated SS on that. $\endgroup$ Dec 16, 2012 at 9:56
  • $\begingroup$ This may be particularly useful in the case that $B = BG$ is the classifying space of a group whose group ring is easily presented. $\endgroup$ Dec 16, 2012 at 11:43

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