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I have a countable sequence of finite suspension spectra $X_i$, whose $BP$-homology is a $BP_*(BP)$-comodule. Let's assume $BP_*(X_i) = \Sigma^{d_i} BP_* / (v_0^{k_0}, \dots v_i^{k_i}),$ for some $d_n$ big enough that makes them being suspension spectra.

Then, if we let $X = \bigvee_i X_i$,

$$BP_*( X)= \bigoplus_i BP_* / (v_0^{k_0}, \dots v_i^{k_i}).$$

Is that right? If yes/no, why?

Now, let's assume i have a spectrum $Y = \prod_i X_i$, which means

$$BP_* (Y) = \prod_i BP_* / (v_0^{k_0}, \dots v_i^{k_i})$$

and the obvious map $$i: X \to Y$$

How would the cofiber $ C_i$ of this map look like? More in detail, i would like $C_i$ to have non torsion elements. Is this the case?

Thank you!

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So far as I can tell, it depends.

The map $\oplus BP_*(X_i) \to \prod BP_*(X_i)$ is injective and the cokernel $M$ is $BP_*$ of the cofiber. In particular, for any $N$ we have an isomorphism $$ M \cong \prod_{i \geq 0} BP_*(X_i) \Big/ \bigoplus_{i \geq 0} BP_*(X_i) \cong \prod_{i \geq N} BP_*(X_i) \Big/ \bigoplus_{i \geq N} BP_*(X_i) $$ As a result, properties of the modules $BP_*(X_i)$ that hold for all but finitely many exceptions tend to be inherited by the module $M$.

The first case is if the natural numbers $d_n$ go to $\infty$ as $n$ goes to $\infty$. In the identities $BP_*(X) = \oplus BP_* (X_i)$ and $BP_*(Y) = \prod BP_*(X_i)$, the direct sum and product are taken in the category of graded abelian groups. If the $d_n$ go to infinity, then for any fixed $k$ there are only finitely many $X_i$ with $BP_k(X_i) \neq 0$. As a result the group $M$ is zero in grading $k$. (The same is true for homotopy groups, and so the cofiber of the map $X \to Y$ is contractible). If you're asking for the $X_i$ to be suspension spectra, then I suspect that you're forced to have $d_n \to \infty$ (but I don't have a proof).

The second case is when the integers $d_n$ do not go to $\infty$. In this case, multiplication by $v_j^{k_j}$ acts by zero on all but finitely many $BP_*(X_i)$, and so the same is true for $M$. Thus means that $M$ consists entirely of things that are torsion.


What you might have had in mind was an "ultrafilter" type argument that did the exact opposite. If, instead, you have $$ BP_*(X_i) \cong \Sigma^{d_i} BP_* / (v_0^{k_{0,i}}, v_1^{k_{1,i}}, \dots, v_i^{k_{i,i}}) $$ such that, for any fixed $n$, $k_{n,i} \to \infty$, then you have the opposite case: for all but finitely many $i$, $BP_m(X_i)$ has no $v_i^k$-torsion. As a result, the module $M$ has no torsion.


(You probably already know this, but I feel obligated to mention that the identity $BP_*(\prod X_i) \cong \prod BP_*(X_i)$ isn't always true, and relies on the $X_i$ being connective and $BP$ having ($p$-local) finite type.)

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  • $\begingroup$ Thank you very much for the clear answer! Looks that this ultrafilter argument may be what i'm looking for, could you please suggest me a text-paper-book where to find something more about it? For example, the spectrum you mentioned with the $k_{n,i}$ going to infinity, is still finite and suspension? Thanks!! $\endgroup$ – Alfred Dec 7 '17 at 17:20
  • $\begingroup$ I've written my question in the wrong way. What i meant was: Is it true that for every $i \geq 0$ there exist a finite (and then, with the proper $d_n,$ suspension) spectrum $X_i$ with the properties you wrote in the "ultrafilter" case, so with the $k_{n,i}$ going to infinity? $\endgroup$ – Alfred Dec 7 '17 at 19:08
  • $\begingroup$ @Alfred The existence of finite spectra $X_i$ like you described is guaranteed by results of Devinatz-Hopkins-Smith, and you can read about this in Ravenel's Nipotence and periodicity in stable homotopy theory (the "orange book"). But to get non-torsion you need the $d_n$ to not go to $\infty$ while maintaining the property of being suspension spectra and I cannot guarantee that: in fact, I suspect that it might be false. $\endgroup$ – Tyler Lawson Dec 7 '17 at 19:40
  • $\begingroup$ Thank you again! Last question, promise: Do you think that is there a way to define these spectra such that this cofiber has at least one non torsion element? What i can think is a way to make $BP_k(X_n) \neq 0$ for almost every $n$ by having a double-indexed sum $\bigvee X_{n,i}$ such that $BP_*(X_{n,i})= \Sigma^{d_n} BP_* / (V_0^{k_{0,n}}, \dots, v_n^{k_{n,n}})$, where every $k_j > i$ $\endgroup$ – Alfred Dec 7 '17 at 20:08
  • $\begingroup$ @Alfred Again, I'm still not sure that you can ensure that the $d_n$ aren't forced to go to $\infty$ in that case. For example, when $n=2$ you are looking for Moore spaces $M(p^k,d)$ with $v_1^l$-self-maps -- with $k$ independent of $d$ -- and even if you can accomplish that, I'm not sure if you can get $v_2^m$-self-maps uniformly on the results, because the powers $m$ you must choose grow as $k$ and $l$ do. However, I don't know much unstable chromatic theory. $\endgroup$ – Tyler Lawson Dec 8 '17 at 16:39

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