"Phantom" non-equivalences of spectra?

Question

I would like an example of the following situation, or a proof that no such example exists.

$\textbf{Situation}$: Two connective (EDIT: I'm fine with dropping this condition) spectra $X$ and $Y$ such that the spaces $\Omega^\infty\Sigma^nX$ and $\Omega^\infty\Sigma^nY$ are equivalent (as spaces) for all $n$, but as spectra, $X$ and $Y$ are $\textit{not}$ equivalent. It was pointed out to me by Maxime that another way to phrase things is that $X$ and $Y$ are two (grouplike) $E_\infty$ spaces that are equivalent as $E_n$-spaces for all finite $n$ but not as $E_\infty$-spaces.

I think (as Charles pointed out in the comments, this is wrong) it would be sufficient to find an example of the following: An infinite loop space $A$ (with terms $A_0=\Omega A_1,A_1,...$ and structure maps $a_i:A_i\rightarrow \Omega A_{i+1}$ such that $a_0$ is the identity) and an equivalence $f:A_0\rightarrow A_0$ (of spaces) such that for any (infinite loop) equivalence $\phi:A\rightarrow A$ with 0th component $\phi_0:A_0\rightarrow A_0$, the composite $\phi_0\circ f$ is not homotopic to the component of any infinite loop map. Because then we can consider the infinite loop space $B$ with the same terms $B_i=A_i$ and the same structure maps $b_i=a_i$ for $i>0$ but with the $b_0$ changed to $f$. The condition on $f$ implies that there is no way to fill in the 0th component of any attempted equivalence between $A$ and $B$.

The problem is that due to my insanely low supply of brain cells the only infinite loop spaces I can really come to grips with are Eilenberg-Maclane, and among Eilenberg-Maclane spectra such an example can't exist since those determined by their homotopy groups...

Answer 1

For this we can use a swindle-type technique.

Let $B = \bigoplus_{n=2}^\infty H\Bbb Z/2$, and $A = \bigoplus_{n=2}^\infty \Sigma^{-n} H\Bbb Z/2$. We can construct maps $A \to B$ by specifying their effect on each summand of $A$.

• The first map, $f: A \to B$, is the sum of the composites $$\Sigma^{-n} H\Bbb Z/2 \xrightarrow{Sq^n} H\Bbb Z/2 \to \bigoplus_{n=2}^\infty H\Bbb Z/2,$$ where the second map includes as the $n$'th summand.
• The second map, $g: A \to B$, is the same, except the we map the $n$'th summand of $A$ to the $(n+1)$'st summand of $B$. (In particular, the summand with $n=2$ is completely missed by $g$.)

Define $X = cofib(f)$ and $Y = cofib(g)$. Since these sums start at $n=2$, $B$ is the connective cover of both $X$ and $Y$ by the long exact sequence in homotopy. I claim that $X$ and $Y$ have equivalent connective covers $\tau_{\geq m} X$ and $\tau_{\geq m} Y$ for all $m < 0$, and that they are inequivalent spectra. This suffices to give an example for your question, since taking the associated infinite loop spaces factors through some connective cover.

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First let's show that they have equivalent covers. Note that $\tau_{\geq m+1} X$ is the cofiber of a map $$ \bigoplus_{2 \leq n \leq -m} \Sigma^{-n} H\Bbb Z/2 \xrightarrow{\tau_{\geq m} f} \bigoplus_{n=2}^\infty H\Bbb Z/2 $$ and similarly for $\tau_{\geq m+1} Y$. However, the maps $\tau_{\geq m} f$ and $\tau_{\geq m} g$ are equivalent maps in the homotopy category: both are the the same finite list of Steenrod squares summed with countably many extra summands of $H\Bbb Z/2$ on the target. Thus, these connective covers are equivalent.

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Next, let's show that $X$ and $Y$ are not equivalent. To see this, first note that $Y$ has $H\Bbb Z/2$ as a summand: the map $g: A \to B$ completely misses the summand $H\Bbb Z/2$ and allows us to split it off. However, the cofiber sequence defining $X$ gives an exact sequence $$ [X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \to [A, H\Bbb Z/2] $$ which is more explicitly $$ [X,H\Bbb Z/2] \to \prod_{n=2}^\infty H^0(H\Bbb Z/2) \xrightarrow{\prod Sq^n} \prod_{n=2}^\infty H^n(H\Bbb Z/2). $$ This second map is injective, and so by exactness the restriction $$ [X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \cong Hom(\pi_0 X, \Bbb Z/2) $$ must be trivial. Therefore, $X$ can't have $H\Bbb Z/2$ as a summand and thus can't be equivalent to $Y$.

Answer 2

I am pretty sure that it will be rather delicate to find an example like you want. Here are some thoughts:

Let $X_n$ denote $\Omega^{\infty} \Sigma^n X$. One is assuming an equivalence $X_n \simeq Y_n$ for all $n$, and asking about an equvalence $X \simeq Y$.

Let $\mathcal S_n= \{f: X_n \simeq Y_n \}\subset [X_n,Y_n]$. Note that looping gives set maps $\mathcal S_{n+1} \rightarrow \mathcal S_n$. It is not hard to check that there is an equivalence $X \simeq Y$ if and only if the inverse limit of the $\mathcal S_n$ is nonempty. So one would have an equivalence if this were an inverse limit of finite sets, or profinite sets, etc., and with many spectra $X$ that come to mind, this would be the case.

Answer 3

Here's a connective example. It is also an example of Maxime's variant question in the comments (regarding $\tau_{\leq m}$ truncations). And thanks to Maxime for looking this argument over before I posted it (but any errors are mine, etc etc).

It's just a tweaked version of Tyler's example and the arguments are very similar but I repeated them for the sake of self-contained-ness. Define $B=\prod_{n=2}^\infty H\mathbb{Z}/2$ and $C=\prod_{n=2}^\infty\Sigma^nH\mathbb{Z}/2$. Consider the following two maps $h,j:B\rightarrow C$. The map $h$ is the infinite product of the maps $Sq^n:H\mathbb{Z}/2\rightarrow \Sigma^n H\mathbb{Z}/2$. The map $j$ is the precomposite of $h$ with the shift-down self-map of $B$. Said differently we can write $B=H\mathbb{Z}/2\oplus \prod_{n=3}^\infty H\mathbb{Z}/2$ and $C=0\oplus\prod_{n=2}^\infty\Sigma^nH\mathbb{Z}/2$ and $j$ sends the first summand of $B$ to zero, and on the second summand is the infinite product of $Sq^{n-1}$. Let $Z$ and $W$ be the fibers of $h$ and $j$. Note that $Z\simeq\prod_{n=2}^\infty\text{cofib}(Sq^n)$ and $W\simeq Z\oplus H\mathbb{Z}/2$.

The truncations $\tau_{\leq m}Z$ and $\tau_{\leq m}W$ are given by truncations of the fibers of the maps $$\prod_{n=2}^\infty H\mathbb{Z}/2\xrightarrow{\tau_{\leq m}h,\tau_{\leq m}j}\prod_{n=2}^m \Sigma^nH\mathbb{Z}/2.$$ But note that $\tau_{\leq m}h$ and $\tau_{\leq m}j$ are equivalent since they are the same finite list of $Sq^k$'s together with zero on infinitely many factors of the domain $B$. Swindled! Hence $\tau_{\leq m}Z\simeq\tau_{\leq m}W$ for all $m$.

Let's show that $Z$ and $W$ are not equivalent. Recall that $W$ is equivalent to $Z\oplus H\mathbb{Z}/2$. So it suffices to show that $Z$ has no $H\mathbb{Z}/2$ summand. For a contradiction, suppose $Z\simeq V\oplus H\mathbb{Z}/2$. That $H\mathbb{Z}/2$ summand has to map nontrivially to $B$, for if it mapped trivially then $C$ would have a $\Sigma H\mathbb{Z}/2$ summand, but $\pi_1C=0$. Now any nontrival map from $H\mathbb{Z}/2$ to $B$ maps via the nontrivial self map of $H\mathbb{Z}/2$ to a nonzero number of the defining $H\mathbb{Z}/2$-factors of $B$ (after all $B$ is a $\mathbb{Z}/2$-vector space). The composite of such a map with $h$ cannot be null since $h$ is non-null on each factor, and the factors do not interact in the target (they map to diferent factors). So there is no $H\mathbb{Z}/2$ summand in the fiber of $h$ (which is $Z$).

Lastly, let's show that for all $m$ we have $\Omega^\infty\Sigma^mZ\simeq\Omega^\infty\Sigma^mW$ as spaces. Now consider $B$ and $C$ in their standard $\Omega$-spectrum presentations by Eilenberg-Maclane spaces. We will use the fact that the fiber of a map of $\Omega$-spectra is computed levelwise and is again an $\Omega$-spectrum. So to compute that $k$-th space in an $\Omega$-spectrum presentation of $\Sigma^{-1}Z$ and $\Sigma^{-1}W$, we need to compute the fiber of the map $h_k, j_k:\prod_{n=2}^\infty K(\mathbb{Z}/2,k)\rightarrow\prod_{n=2}^\infty K(\mathbb{Z}/2,k+n)$ induces by $h$ and $j$. But for $n>k$, the map $Sq^n:K(\mathbb{Z}/2,k)\rightarrow K(\mathbb{Z}/2,k+n)$ is null, so again, $h_k$ and $j_k$ are the same finite lists of unstable $Sq^i$'s plus a bunch of nullhomotopic maps, so they are equivalent by another swindle.