# Morphisms from $bstring$ to $X\otimes \mathbb{Q}$ and sequences $s_n\in\pi_n(X)\otimes \mathbb{Q}$

I'm currently studying Ando-Hopkins-Rezk's work Multiplicative orientations of KO-theory and of the spectrum of TMFs. At a point a presumably obvious isomorphism is mentioned, which I'm however not able to immediately see (I guess I'm missing something obvious here, forgive me if the question is too naive).

I'm referring to the identification $$s: [bstring, X\otimes \mathbb{Q}] \cong \mathbf{D}(X):=\left\{s\in\prod_{k\geq 4} \pi_{2k}X\otimes \mathbb{Q}\,\biggr\vert\, s_k=0 \text{ if k is odd}\right\}$$ (equation (5.10) in the article), where $$X$$ is a spectrum and $$bstring$$ is the spectrum realizing the standard infinite loop space structure on $$BString=BO\langle 8\rangle$$.

While I clearly see there's a natural morphism $$s: [bstring, X\otimes \mathbb{Q}] \to \mathbf{D}(X)$$, that fully encodes the datum of the sequence of group homomorphisms $$\phi_{\ast,n}: \pi_n(bstring) \to \pi_n(X)\otimes \mathbb{Q}$$ associated with (the homotopy class of) a morphism $$\phi\colon bstring \to X\otimes \mathbb{Q}$$, I don't clearly see why any such a sequence should be induced by a unique (homotopy class of a) morphism from $$bstring$$ to $$X_\mathbb{Q}$$, i.e., I don't immediately see why $$s$$ should be a bijection. I guess the answer is in some property of $$bstring_\mathbb{Q}$$ so well known not to be explicitly mentioned in the article, or in some even more fundamental property of rational homotopy theory of infinite loop spaces (or even general spectra) I'm missing at the moment, but I have not been able to work out this answer myself.

Thanks a lot for any possible hint to help me solve my puzzlement

• Isn't this just using that every rational spectrum splits as a sum of Eilenberg-MacLane spectra, so that $bstring_{\mathbb Q}\cong \bigoplus_{n\ge 0} H(\pi_n(bstring)\otimes \mathbb Q)[-n]$? – Bertram Arnold Sep 25 '19 at 15:40
• This is precisely the kind of "fundamental property of rational homotopy theory of spectra" I had in mind (and actually by Yoneda this would have been the only possibility), but all my attempts at finding this result in the sources I looked into or to prove it myself had failed, so I ended up asking the Oracle (MO). Could you point me to a reference? Thanks! – domenico fiorenza Sep 25 '19 at 15:45

This is an assemblage of known results, I'll try to put a reference for all of them.

• By a classical theorem of Serre all stable homotopy groups are finite in positive degree. In particular we have $$\mathbb{S}_\mathbb{Q}=H\mathbb{Q}$$.

• Rationalization is a smashing localization, so that $$\pi_*(\mathbb{S}_\mathbb{Q}\otimes E)\cong \pi_*E \otimes_{\mathbb{Z}}\mathbb{Q}$$. This follows immediately from the fact that we can write $$E_\mathbb{Q}=\mathrm{colim}\left(E\xrightarrow{2}E\xrightarrow{3}E\xrightarrow{4}E\xrightarrow{5}\cdots\right)$$ and that tensor (smash) products and homotopy groups commute with filtered colimits of spectra.

• By Schwede-Shipley Morita theory, the category of modules over $$H\mathbb{Q}$$ is equivalent to the derived category of $$\mathbb{Q}$$. (Theorem 5.1.6 in Stable model categories are categories of modules) Moreover under this equivalence homotopy groups correspond to homology groups (because $$H\mathbb{Q}$$ is sent to $$\mathbb{Q}$$ in degree 0).

• In the derived category $$D(\mathbb{Q})$$ for every object $$M$$ there is an equivalence $$M\cong \bigoplus_{n\in\mathbb{Z}} (H_nM)[n]$$. This is an easy exercise, using the fact that in $$\mathbb{Q}$$-vector spaces every short exact sequence splits.

In particular for every spectrum $$E$$, the spectrum $$E\otimes H\mathbb{Q}$$ has homotopy groups $$\pi_*(E\otimes H\mathbb{Q})\cong \pi_*E\otimes_{\mathbb{Z}}\mathbb{Q}$$ and so there is an equivalence of $$H\mathbb{Q}$$-modules (in particular of spectra)

$$E\otimes H\mathbb{Q}\cong \bigoplus_{n\in\mathbb{Z}} \Sigma^nH(\pi_nE\otimes_{\mathbb{Z}}\mathbb{Q})\cong\prod_{n\in\mathbb{Z}} \Sigma^nH(\pi_nE\otimes_{\mathbb{Z}}\mathbb{Q})$$ (it's not necessary, but it is very convenient to observe that in this case for degree reasons the coproduct and the product coincide)

So, to conclude, we have that $$[bstring, X\otimes H\mathbb{Q}]\cong \left[ bstring, \prod_{n\in\mathbb{Z}}\Sigma^nH(\pi_nX\otimes_{\mathbb{Z}}\mathbb{Q})\right]\cong \prod_{n\in\mathbb{Z}} H^n(bstring;\pi_nX\otimes_{\mathbb{Z}}\mathbb{Q})\cong \prod_{n\in\mathbb{Z}} \pi_nX\otimes_{\mathbb{Z}} H^n(bstring;\mathbb{Q})\,.$$