Does Waldhausen K-theory detect homotopy type?

Question

Recall that $A(X)$, the K-theory of a connected, pointed space X, is defined as the K-theory spectrum of the ring spectrum $\Sigma^\infty_+ \Omega X$ (or via a plethora of alternative definitions). Is it known if the homotopy type of $A(X)$ determines the homotopy type of $X$? If not, what is the best one can hope for?

Of course, since $X$ is connected the space $\Omega X$ with its loop space structure determines the homotopy type of $X$, but I am not sure if this is still true when we take $\Sigma^\infty_+$, I am worried we get might get $X$ only up to $\Sigma^n \Omega^n $. Then there is the question of if ring spectra of this type can have the same K-theory, perhaps we should assume $X$ simply connected to get a positive answer?

Answer 1

The answer to the question

Does the homotopy type of $𝐴(𝑋)$ determine the homotopy type of $𝑋$?

is No in general. As you say, $A(X)$ is determined by the homotopy type of $\Sigma^\infty \Omega X_+$ as an associative (or $A_\infty$) ring spectrum, and this ring spectrum does not uniquely determine $X$, even if $X$ is simply-connected.

For example, suppose $Z$ is a pointed space. Let $T(Z)$ be the free associative $S$-algebra generated by $Z$. I.e., $$T(Z)=\Sigma^\infty S^0\vee \Sigma^\infty Z \vee (\Sigma^\infty Z)^{\wedge 2} \vee \cdots .$$

If $Z$ is connected, there is an equivalence of associative ring spectra $$ T(Z) \simeq \Sigma^\infty \Omega\Sigma Z_+.$$ The equivalence is a version of the classical James splitting. It is induced by a map of spectra $\Sigma^\infty Z \to \Sigma^\infty \Omega\Sigma Z_+$, extended to a map of ring spectra $T(Z)\to \Sigma^\infty \Omega\Sigma Z_+$ using freeness.

It follows that if $X$ and $Y$ are connected spaces such that $\Sigma X$ and $\Sigma Y$ are not equivalent, but $\Sigma^\infty X$ and $\Sigma^\infty Y$ are equivalent, then there is an equivalence $A(\Sigma X)\simeq A(\Sigma Y)$ providing a counterexample.

A couple of comments:

• It is well-known that there exist non-isomorphic groups $G$ and $H$ such that the group rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are isomorphic. There are even examples with finite $G$ and $H$. One may wonder if for some of these examples the spherical group rings $\Sigma^\infty G_+$ and $\Sigma^\infty H_+$ are equivalent as associative ring spectra. If yes, then $BG$ and $BH$ would provide another counterexample.

• In general one can have non-equivalent ring spectra that have equivalent $K$-theories. For example, I believe that if $P$ and $Q$ are Morita equivalent in a suitable sense, then $K(P)\simeq K(Q)$. Can there be two spaces $X$ and $Y$ such that $\Sigma^\infty \Omega X_+$ and $\Sigma^\infty \Omega Y_+$ are not equivalent as ring spectra, but have equivalent categories of modules (in a strong enough sense to induce equivalence of $K$-theories)? It seems far fetched, but I don't know how to exclude this possibility. Added later: A paper by Roggenkamp and Zimmerman gives an example of two groups $G$ and $H$ for which the rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are not isomorphic, but Morita equivalent. It follows that the Quillen $K$-theory of these rings is isomorphic. One may ask whether the $K$-theory spectra of $\Sigma^\infty G_+$ and $\Sigma^\infty H_+$ are equivalent as well.

Answer 2

This doesn't get at $A$-theory specifically, but it seems to be already interesting to ask to what extent we can recover $X$ from knowing $\Sigma^\infty_+ \Omega X$. I believe that very often we can -- but take what I say with a grain of salt -- this is not my area of expertise:

EDIT: The following is wrong, as Gregory Arone's answer shows. Thanks to Maxime Ramzi for pointing out the error in the comments below. I will leave this up for the time being and possibly try to salvage some sort of positive statement later.

Claim 1: Let $X,Y$ be connected spaces. Suppose that $\Sigma^\infty_+ \Omega X \simeq \Sigma^\infty_+ \Omega Y$ as $A_\infty$ ring spectra. Then $X \simeq Y$.

Proof: Suppose that $\Sigma^\infty_+ \Omega X \simeq \Sigma^\infty_+ \Omega Y$ as $A_\infty$ ring spectra. Then we have an equivalence of module categories $Mod(\Sigma^\infty_+ \Omega X) \simeq Mod(\Sigma^\infty_+ \Omega Y)$ preserving the unit (the module category is not monoidal, but it does have a unit!). In other words, we have an equivalence of functor categories $Spectra^X \simeq Spectra^Y$ which preserves the representables. Preserving the representables amounts to saying that the equivalence commutes with the forgetful / evaluation functor to $Spectra$. This implies that the equivalence $Spectra^X \simeq Spectra^Y$ carries the comonad $\Sigma^\infty_X \Omega^\infty_X$ to the comonad $\Sigma^\infty_Y \Omega^\infty_Y$. The preservation of the unit also means that the equivalence preserves connectivities, so that we have an equivalence $(Spectra_{\geq 2}^X, \Sigma^\infty_X \Omega^\infty_X) \simeq (Spectra_{\geq 2}^Y, \Sigma^\infty_Y \Omega^\infty_Y)$, and so also an induced equivalence of categories of comonads. By a theorem of Blomquist and Harper, the $\Sigma^\infty \dashv \Omega^\infty$ adjunction is comonadic after restricting to the simply-connected case. We may apply this fact levelwise to conclude that we have an equivalence $Spaces_{\ast,\geq 2}^X \simeq Spaces_{\ast,\geq 2}^Y$ of categories of functors valued in pointed, simply-connected spaces, which preserves the representables. By Lemma 2 below, this implies that $\overline{X^+} \simeq \overline{Y^+}$, where $X^+$ is the $\infty$-category $X$ with a zero object added and $\overline C$ is the completion of $C$ under splitting of idempotents. But as $X$ is an $\infty$-groupoid, it is clear that there are no nontrivial idempotents in $X^+$, and similarly for $Y$. So $X^+ \simeq Y^+$. As the equivalence preserves the representables, we must have $X \simeq Y$.

Lemma 2: For any $\infty$-category $C$ enriched in $Spaces_{\ast,\geq 2}$, define $\overline{C^+}$ to be $C$ with a zero object added and idempotents split. Then if $C,D$ are two such $\infty$-categories, we have an equivalence $Spaces_{\ast,\geq 2}^C \simeq Spaces_{\ast,\geq 2}^D$ of $Spaces_{\ast,\geq 2}$-enriched categories of $Spaces_{\ast,\geq 2}$-enriched functor categories to $Spaces_{\ast,\geq 2}$ if and only if $\overline{C^+} \simeq \overline{D^+}$.

Proof: When $Spaces_{\ast,\geq 2}$ is replaced by $Spaces_\ast$, this is well-known, an application of the theory of Cauchy completion in enriched category theory. The same argument works in this case.