Is a filtered colimit of rational spaces again rational?

Question

Let me first explain the statement of the question and then give some indication why the answer might be 'yes'. By a space I mean, say, a simplicial set and by rational I mean rational in the sense of Bousfield, i.e. local with respect to the homology theory $H\mathbb{Q}$. If we denote the category of spaces by $\mathcal{S}$ then the category of rational spaces $\mathcal{S}_\mathbb{Q} \subseteq \mathcal{S}$ is a full subcategory and the question is whether it is closed under filtered colimits. This would in particular imply that the $\infty$-category of rational spaces is compactly generated. Note that it is known that this $\infty$-category is generated by $\kappa$-compact objects for $\kappa$ the successor cardinal of $\omega$ (by results of Bousfield). But whether it is compactly generated is unclear to me and this is really what I want to know, even if the inclusion $\mathcal{S}_\mathbb{Q} \subseteq \mathcal{S}$ is not necessarily $\omega$-continuous.

Now let me give some indication why the answer could be 'yes': first a simply connected (or more generally nilpotent) space is rational if and only if its homotopy groups are uniquely divisible. This property is obviously closed under filtered colimits. Thus a filtered colimit of simply connected, rational spaces is again rational. Bousfield characterizes rationality for arbitrary spaces as a property of homotopy groups, but this property is also homotopy theoretic in nature so that I am unable to decide if it is preserved by filtered colimits.

The second indication is that one can study the stable analogue of my question: is a filtered colimit of rational spectra again rational. The answer is yes. This is in fact equivalent to the fact that rationalization is a smashing localization. This observation also shows that the class of $\mathbb{S}/p$-local spectra for a prime $p$ cannot be closed under filtered colimits since $p$-completion is not smashing. A consequence is that $\mathbb{F}_p$-local spaces (a.k.a. $\mathbb{S}/p = M(\mathbb{Z}/p)$-local spaces) are not closed under filtered colimits.

Answer 1

The answer to this, as a result of some discussions with Thomas, turns out to be no.

Consider the abelian groups $A_n = \Bbb Q[x]/(x^n)$. Multiplication by $x$ embeds $A_n$ into $A_{n+1}$, and the direct limit of the sequence $$ A_0 \to A_1 \to \dots $$ is the group $A_{\infty} \cong \Bbb Q[x^{\pm 1}] / \Bbb Q[x]$.

The group $\Bbb Q$ acts compatibly on the $A_n$ and $A_\infty$ if we define $$ q \ast m = (1+x)^q m $$ where $(1 + x)^q = \sum \binom{q}{n} x^n$ is defined using the binomial theorem.

The action of $\Bbb Q$ on each $A_n$ is nilpotent: there is a finite filtration whose layers are acted on trivially. In fact, $\Bbb Q$ acts trivially on the quotient groups $A_n/A_{n-1} \cong \Bbb Q$ because $x$ acts trivially on this group.

By contrast, the action of $\Bbb Q$ on $A_\infty$ is perfect: for any element $m \in A_\infty$, we find $$ m = x (x^{-1} m) = 1 \ast (x^{-1} m) - (x^{-1} m) $$ and so the set of elements of the form $q \ast y - y$ generate all of $A_\infty$. Therefore, $A_\infty / \Bbb Q = 0$.

There is an associated sequence of Eilenberg--Mac Lane spaces $K(A_n,2)$ and $K(A_\infty,2)$ with compatible actions of $\Bbb Q$, for example by using a functorial construction of E-M spaces. We can form the Borel constructions $X_\alpha = K(A_\alpha,2) \times_{\Bbb Q} E\Bbb Q$. These spaces have the property that $$ \pi_n X_\alpha = \begin{cases} \Bbb Q &\text{if }*=1,\\ A_\alpha &\text{if }*=2,\\ 0 & \text{otherwise}. \end{cases} $$ In particular, $X_\infty$ is the homotopy colimit of the spaces $X_n$.

Each $X_n$ is a nilpotent space: it has abelian (hence nilpotent) fundamental group, the fundamental group acts nilpotently on the higher homotopy groups, and the homotopy groups are rational vector spaces. This means that $X_n$ has a Postnikov-type construction formed by iteratively taking fibers of maps to $K(\Bbb Q,m)$, and so $X_n$ is local with respect to rational homology.

By contrast, $X_\infty$ is not local with respect to rational homology. The fact that $\pi_1(X_\infty)$ acts perfectly on $\pi_2(X_\infty)$ allows us to use an analogue of Quillen's plus-construction to find a map $X_\infty \to Y$, by attaching $3$-dimensional and $4$-dimensional cells, which is a rational homology isomorphism but which induces the zero map $\pi_2(X_\infty) \to \pi_2(Y)$. In particular, the map $X_\infty \to (X_\infty)_{\Bbb Q}$ to its rationalization factors through $Y$, and so it must send all of $\pi_2$ to zero.