Let $X$ be a connected pointed topological space equipped with two different actions of $E_\infty$-operad. Each action provides a collection of deloopings $X_i$, where $X_0 = X$ and $\Omega X_i$ is homotopy equivalent to $X_{i-1}$, so there are two $\Omega$-spectra having $X$ as a zeroth space. Can these spectra have different cohomology or different cohomology operations? I would like an explicit example or just a better reason than "why not".

I don't have any applications for such spectra. This question is motivated by thinking about D. Rector's paper Loop Structures on the Homotopy Type of $S^3$, which produces uncountably many different deloopings of three-dimensional sphere. They can be distinguished using cohomology operations.

  • $\begingroup$ There are really easy counterexamples given by two nonisomorphic abelian groups with the same cardinality, so I'll assume you meant for $X$ to be connected. $\endgroup$ – Qiaochu Yuan Feb 10 '15 at 11:28
  • $\begingroup$ @QiaochuYuan, yes, $X$ should be connected. $\endgroup$ – Dmitry Pirozhkov Feb 10 '15 at 11:45

There are easier ways to distinguish connective spectra with the same underlying space, if that's all you want to do. Like spaces, spectra have a theory of Postnikov towers, and in the same way that the Postnikov tower of a (simply connected, say) space $X$ measures the extent to which it differs from the product

$$\prod_n B^n \pi_n(X)$$

of Eilenberg-MacLane spaces, the Postnikov tower of a connective spectrum $E$ measures the extent to which it differs from the product

$$\prod_n B^n \pi_n(E)$$

of Eilenberg-MacLane spectra.

Suppose $E$ is a spectrum with exactly two nontrivial homotopy groups $\pi_n(E), \pi_m(E), n < m$. Then $E$ naturally fits into a fiber sequence of spectra

$$B^m \pi_m(E) \to E \to B^n \pi_n(E)$$

and such fiber sequences are classified by a $k$-invariant given by a homotopy class of maps $B^n \pi_n(E) \to B^{m+1} \pi_m(E)$ of spectra, or equivalently by a stable cohomology operation

$$H^{\bullet + n}(-, \pi_n(E)) \to H^{\bullet+m+1}(-, \pi_m(E)).$$

If $n \ge 1$, then taking underlying spaces produces from the above fiber sequence of spectra a fiber sequence of connected spaces

$$B^m \pi_m(E) \to E \to B^n \pi_n(E)$$

which is a principal $B^m \pi_m(E)$-bundle, and in particular which is classified by a $k$-invariant given by a homotopy class of maps $B^n \pi_n(E) \to B^{m+1} \pi_m(E)$ of spaces, or equivalently by an unstable cohomology operation

$$H^n(-, \pi_n(E)) \to H^{m+1}(-, \pi_m(E)).$$

In particular, we can find nontrivial infinite loop space structures on the space $B^n \pi_n(E) \times B^m \pi_m(E)$ by finding stable cohomology operations whose corresponding unstable cohomology operation in some degree is trivial.

Example. The simplest nontrivial case where the underlying space is connected occurs when we pick the $k$-invariant to be the second Steenrod square

$$\text{Sq}^2 : B \mathbb{Z}_2 \to B^3 \mathbb{Z}_2$$

(taking $\text{Sq}^1$ just gives $\mathbb{Z}_4$ as a nontrivial extension of $\mathbb{Z}_2$ by $\mathbb{Z}_2$). This gives the unique nontrivial infinite loop space structure on the space $B \mathbb{Z}_2 \times B^2 \mathbb{Z}_2$, which arises naturally as the $2$-truncation of the connected component of the identity of the sphere spectrum; in particular, we know that the underlying space is $B \mathbb{Z}_2 \times B^2 \mathbb{Z}_2$ because $\text{Sq}^2$ as an unstable cohomology operation

$$H^1(-, \mathbb{Z}_2) \to H^3(-, \mathbb{Z}_2)$$

vanishes. Equivalently, $\text{Sq}^2$ is nontrivial as a map of spectra but trivial as a map of spaces.

The product spectrum structure on $B \mathbb{Z}_2 \times B^2 \mathbb{Z}_2$ gives the cohomology theory which is

$$H^1(X, \mathbb{Z}_2) \times H^2(X, \mathbb{Z}_2)$$

in degree $0$, with the product abelian group structure. By comparison, the nontrivial spectrum structure gives the cohomology theory which in degree $0$ has the same underlying set, but with the following modified abelian group structure, where $(w_1, w_2) \in H^1(X, \mathbb{Z}_2) \times H^2(X, \mathbb{Z}_2)$:

$$(w_1, w_2) + (w_1', w_2') = (w_1 + w_1', w_2 + w_2' + w_1 \cup w_1').$$

This group is in general a nontrivial extension of $H^1(X, \mathbb{Z}_2)$ by $H^2(X, \mathbb{Z}_2)$; it arises as a kind of super Brauer group of $X$. The multiplication can be thought of as the cup product on the invertible elements $1 + w_1 + w_2 \in H^{\bullet}(X, \mathbb{Z}_2) / H^{\bullet + 3}(X, \mathbb{Z}_2)$ of a truncated version of the cohomology ring over $\mathbb{Z}_2$.

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    $\begingroup$ Beat me to it!! $\endgroup$ – Chris Schommer-Pries Feb 10 '15 at 12:14
  • $\begingroup$ @Chris: did you have the same example in mind? $\endgroup$ – Qiaochu Yuan Feb 10 '15 at 12:20
  • $\begingroup$ Yes. Up to equivalence, there are exactly two spectra with $\pi_0 = pi_1 = \mathbb{Z}/2$ and all other homotopy groups zero. They have the same underlying topological space. $\endgroup$ – Chris Schommer-Pries Feb 10 '15 at 12:57
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    $\begingroup$ The point is that the second Steenrod square is a stable map $Sq^2: B^k \mathbb{Z}/2 \to B^{k+2}\mathbb{Z}/2$. It only exists when $k \geq 2$ (and is a mod 2 thing only in the stable range $k \geq 3$. When you loop the map enough, then it becomes null, which means that the underlying topological space splits as a product (as it must given the classification of 1-types). $\endgroup$ – Chris Schommer-Pries Feb 10 '15 at 13:02
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    $\begingroup$ Yeah, I noticed that I forgot to explain why we know that the underlying space is a product of Eilenberg-MacLane spaces in the above case; that's been edited in. $\endgroup$ – Qiaochu Yuan Feb 10 '15 at 13:04

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