I am looking for a reference/proof that shows that the Morava $K$-theory spectra, $K(n)$ are not $E_{\infty}$ ring spectra. I suspect that this should be a calculation but I can't quite get it right.

Thank you as always.

  • 1
    $\begingroup$ At the prime 2 I don't even think they're homotopy commutative. In general they're probably not more than like... E_2. I don't know a reference... maybe Ravenel's orange book? $\endgroup$ – Dylan Wilson Aug 23 '14 at 17:01

The assertion is Lemma 5.6.4 in Rognes's "Galois extensions of structured ring spectra" available on the arXiv. In fact, the $K(n)$ spectra do not even admit $E_2$-algebra structures. The reason is that the free $E_2$-algebra with $p = 0$ is known to be (by a theorem of Hopkins-Mahowald) the Eilenberg-MacLane spectrum $H \mathbb{F}_p$. If $K(n)$ admitted an $E_2$-algebra structure, it would receive a map of $E_2$-rings $H \mathbb{F}_p \to K(n)$, which is not possible (for instance, as $H \mathbb{F}_p \wedge K(n) = 0$).

More generally, no $p$-power torsion $E_n$-local spectrum can admit an $E_\infty$-structure. This can be seen by playing with power operations and is worked out as the main result of joint work with J. Noel and N. Naumann. The closing remarks to that paper also outline a proof of the Hopkins-Mahowald theorem cited above (which was not formally documented in the literature for odd primes).

Analogous results for $E_2$-algebras (or $E_m$-algebras for $m \geq 2$) seem to be unknown. For example, consider the free $E_2$-algebra with $p^2 =0 $. Is it true that this is $K(n)$-acyclic for each $n$?

At the prime $2$, it is even true that any homotopy commutative ring spectrum with $2 =0$ is equivalent to a wedge of Eilenberg-MacLane spectra (and in particular is invisible to $L_n$-localization). This is a result of Wurgler.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.