# Bockstein morphism of spectral sequences

Given an omega spectrum $E$, there is a type of chern character map given by its rationalization $$r:E\to E\wedge M\mathbb{R}\;,$$ where $M\mathbb{R}$ denotes a Moore spectrum. The cofiber of the map $$\mathbb{S}\to M\mathbb{R}$$ can be identified with an $M\mathbb{R}/\mathbb{Z}$ and smashing sequence on the left by $E$ yields a Bockstein sequence $$E\overset{r}{\to} E\wedge M\mathbb{R} \to E \wedge M\mathbb{R}/\mathbb{Z}\overset{\beta}{\to}\Sigma E\;.$$

Both $r$ and $\beta$ induce morphisms of corresponding Atiyah-Hirzebruch spectral sequences. Although $\beta$ is usually trivial in practice (its $0$ when the coefficients are concentrated in even degrees for example). If I assume that $\pi_*(E)$ is concentrated in even degrees and is (say free) in those degrees, then there is also a Bockstien morphism corresponding to the sequence $$0\to \pi_*(E)\to \pi_*(E)\otimes \mathbb{R} \to \pi_*(E)\otimes \mathbb{R}/\mathbb{Z}\to 0,$$ induces a map on the $E_2$ page of the corresponding theories. I was hoping to show that this map commutes with the differentials and that the induced map on higher pages commutes (essentially that it defines a morphism of spectral sequences). Does anyone know if this is true, or know of a reference where this is discussed?

• I'm confused. All morphisms of spectra induce a morphism of the corresponding AHSS, since you have a morphism of exact couples. Exactly where are you having problems? – Denis Nardin Apr 2 '16 at 13:23
• @DenisNardin I just realized I wasn't very clear. I edited my post – Daniel Grady Apr 2 '16 at 13:35
• I'm sorry if I'm being overly dense, but I don't understand. Are you looking for a map $H_*(X;(E\wedge M\mathbb{Z})_*)\to H_*(X;\Sigma E_*)$ that commutes with the differentials? If so I believe my argument (there's a map of exact couples) should provide that – Denis Nardin Apr 2 '16 at 13:38
• @DenisNardin Yes, $\beta$ induces a morphism of exact couples, but this is not the map I'm looking at. For example, take $K$ theory. Then $H^p(X;\pi_{-2q}(K\wedge M\mathbb{R}/\mathbb{Z}))\simeq H^p(X;\mathbb{R}/\mathbb{Z})$. $H^p(X;\pi_{-2q}(\Sigma K))\simeq H^p(X;0)\simeq 0$. So the map induced by $\beta$ is just $0$. What I really want is to look at is the Bockstien $\tilde{\beta}:H^p(X;\mathbb{R}/\mathbb{Z})\to H^{p+1}(X;\mathbb{Z})$. – Daniel Grady Apr 2 '16 at 14:23
• Then $\beta$ is not the Bockstein you're looking for. In fact it is still not clear what is the map you're looking for, even at the $E_2$-page level. Could you clarify what are the source and target of the map? – Denis Nardin Apr 2 '16 at 14:32