7
$\begingroup$

The wedge sum $\bigvee_{k \in 2 \mathbb{Z}} S^{k}$ is an $A_\infty$-ring spectrum: the connective cover is the free $A_\infty$-ring on the sphere $S^2$, if I'm not mistaken, and then one inverts the element in $\pi_2$. It is also homotopy commutative. Can it be made into an $E_\infty$-ring spectrum? More generally, given an $E_\infty$-ring spectrum $R$, when can $\bigvee_{k \in 2 \mathbb{Z}} \Sigma^k E$ be made into an $E_\infty$-ring?

$\endgroup$
2
  • 6
    $\begingroup$ I'm pretty sure it is not, why would it be? Sorry not to have time to say more. Essentially, even with the $2$, the commutativity is only homotopical, not as strict as would be required. $\endgroup$
    – Peter May
    Nov 2, 2012 at 3:11
  • 2
    $\begingroup$ Here's a point of terminology that will help you find references: an $H_\infty$ structure on $\bigvee_{k\in d\mathbb{Z}}\Sigma^kE$ is essentially what is called an $H^d_\infty$ structure in the $H_\infty$ book (Springer LNM 1176). Of course $H_\infty$ is weaker than $E_\infty$ but this is a start. $\endgroup$ Nov 2, 2012 at 7:07

1 Answer 1

16
$\begingroup$

Ah, tracked it down. Here is an argument. I should mention that Peter once pointed me towards an original source due to McClure, or page 238, Prop. 6.1, of SLN 1176 (the $H_\infty$ book).

Suppose you had such a ring object $R$. We examine its mod-2 homology. This has several features:

  • It is the ring $\mathbb Z/2[t^{\pm 1}]$.

  • It has trivial action of the Steenrod operations $P_r$ (this is dual to the cohomology action), because it's a wedge of spheres.

  • From the $H_\infty$ book, it has Dyer-Lashof operations $Q^s$. These satisfy $Q^{|x|} x = x^2$, and the Nishida relations $$ P_r Q^s = \sum \binom{s - r}{r - 2i} Q^{s-r+i}P_i. $$

In particular, these together would say $$ 0 = P_2 Q^4 t = \binom{2}{2} Q^{2}P_0 t + \binom{2}{0} Q^{4}P_2 t = Q^2 t = t^2. $$

If $E$ is a commutative $MU$-algebra, then you can use the map from $MU$ to its periodic version $MUP$ to produce $R \wedge_{MU}MUP$, which is 2-periodified. Unfortunately, we know very few genuine $MU$-algebras. Barry Walker proved that complex K-theory is one of them, based on Matthew Ando's work on studying $H_\infty$ structures on Lubin-Tate cohomology theories. My understanding is that the problem is still open for almost all of the Lubin-Tate cohomology theories.

$\endgroup$
2
  • 2
    $\begingroup$ Thank you! LNM 1176 is on my reading list, but I haven't gotten very far in it. These types of methods seem like quite a useful tool for answering such questions. $\endgroup$ Nov 2, 2012 at 11:32
  • 6
    $\begingroup$ Perhaps worth pointing out: the ring spectrum in question can be made E_2, since it can be described as the Thom spectrum of a 2-fold loop map Omega^2(BU(1)) -> Omega^2(BU) = Z x BU. $\endgroup$ Nov 4, 2012 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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