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Let $E $ be a (possibly nonconnective) spectrum. Suppose $E \wedge K = 0$ (where $K$ is complex $K$-theory). Does it follow that $E = 0$?

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Sure. Smashing a based space with a spectrum is equivalent to smashing its suspension spectrum with that spectrum. So it suffices to give a nontrivial space whose reduced $K$-homology is trivial. A classical example due to Luke Hodgkin is $Coker J$. See Hodgkin, Luke; Snaith, Victor. The K-theory of some more well-known spaces. Illinois J. Math. 22 (1978), no. 2, 270–278.

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    $\begingroup$ Peter, you don't mean "sure". You mean "surely not". $\endgroup$
    – Tom Goodwillie
    Commented Jun 10, 2012 at 1:56
  • $\begingroup$ Right you are Tom, I surely mean surely not. But Akhil understood. $\endgroup$
    – Peter May
    Commented Jun 10, 2012 at 2:07
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    $\begingroup$ Tom, actually the original Sure'' was in answer to the original question. And for that question the answer Sure'' is clearly right. $\endgroup$
    – Peter May
    Commented Jun 10, 2012 at 2:11
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Another type of example: Let $E$ be the mod $p$ homology spectrum. The integral homology groups of the periodic $K$-theory spectrum are rational vector spaces, i.e. the integral homology groups of the mod $p$ spectrum are zero.

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    $\begingroup$ Moreover, you can prove this without too much computation. Both $K$ and $H/p$ are complex oriented, so a quite conceptual argument shows that the additive formal group law (associated to $H/p$) and the multiplicative one (associated to $K$) become isomorphic over $\pi_*(K\wedge H/p)$. A simple algebraic argument now shows that $\pi_*(K\wedge H/p)=0$. $\endgroup$
    – Neil Strickland
    Commented Jun 10, 2012 at 11:36
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    $\begingroup$ Neil, that's nice. The proof that I know consists of examining the effect of the Bott map $S^2\wedge BU\to BU$ on integral homology. This comes down to calculating the Chern class(es) of the generator of $\tilde K(S^{2n})$, which can be done using the Chern character. $\endgroup$
    – Tom Goodwillie
    Commented Jun 10, 2012 at 15:01
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    $\begingroup$ @Neil: I know that argument works more generally. Would you mind posting it as a separate answer? $\endgroup$
    – Sean Tilson
    Commented Jun 10, 2012 at 15:32

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