QUESTION EDITED: There was a mistake, the spectrum i had written before didn't even exist, so a big thanks to the people who made me notice that in the comments.

Let $X_n$ be the spectrum such that $BP_*(X_n) = \Sigma^{d_n}BP_*/(v_0^{p^{i_0}},v_1^{p^{i_1}},\dots,v_{n-1}^{p^{i_{n-i}}})$. Recall that it's always possible to have a sequence $i_0, i_1, i_2 \dots$ such that a spectrum with the given BP-homology exists (this was my mistake before editing the question). $d_n$ is a natural number big enough which makes $X_i$ be a suspension spectrum.

Let also $X = \bigvee_i X_i$.

Now consider the following fibration:

$ \bigvee_i X_i \to L_n \bigvee X_i \to \Sigma C_n \bigvee X_i $

which gives me the following inverse system of short exact sequences:

$$ 0 \to \bigoplus_{i \leq n} BP_*(X_i) \longrightarrow \ \bigoplus_{i \leq n} BP_*L_n(X_i) \ \longrightarrow \bigoplus_{i \leq n} BP_{*-1}(\Sigma C_iX_i) \to 0$$

$$ 0 \to \bigoplus_{i \leq n-1} BP_*(X_i) \to \bigoplus_{i \leq n-1} BP_*L_{n-1}(X_i) \to \bigoplus_{i \leq n-1} BP_{*-1}(\Sigma C_{n-1}X_i) \to 0$$

with vertical maps $$ \bigoplus_{i \leq n} BP_{*-1}(\Sigma C_iX_i) \to \bigoplus_{i \leq n-1} BP_{*-1}(\Sigma C_{i-1}X_i)$$

So here is my second question: How can i study the inverse limit of the right vertical tower $$lim_n \bigoplus_{i \leq n} BP_{*-1}(\Sigma C_iX_i)?$$

Thank you

  • 2
    $\begingroup$ First of all, do they exist? $\endgroup$ – user43326 Nov 24 '17 at 18:12
  • $\begingroup$ Are you talking about the spectra $X_i$ or about the limit? $\endgroup$ – Alfred Nov 25 '17 at 12:46
  • $\begingroup$ I was talking about $X_i$'s, but Neil showed that they didn't exist, so this comment is now pointless. $\endgroup$ – user43326 Nov 26 '17 at 13:40

There are no spectra with the indicated $BP$-homology. The $BP$-homology of a spectrum is always a $BP_*BP$-comodule, and $BP_*/(v_0^i,v_1^j)$ only admits a comodule structure if $\eta_R(v_1)^j=v_1^j\pmod{v_0^i}$. Here $\eta_R(v_1)$ can be calculated from the relation $$ \sum^F_{i,j}t_i\eta_R(v_j)^{p^i}x^{p^{i+j}} = \sum^F_{i,j}v_it_j^{p^i}x^{p^{i+j}} $$ as explained in Section 4.3 of Ravenel's "Complex Cobordism and Stable Homotopy Groups of Spheres", for example: we get $$ \eta_R(v_1) = v_1 + (p-p^p) t_1. $$ (This is with Araki generators; the numbers are slightly different with Hazewinkel generators, but the overall picture is the same. Of course $v_0=p$ here.)

From this we find that $\eta_R(v_1)^2\neq 0\pmod{v_0^2}$, so $BP_*/(v_0^2,v_1^2)$ is not a comodule. However, $\eta_1(v_1)^p=0\pmod{v_0^2}$, so $BP_*/(v_0^2,v_1^p)$ is a comodule. But it is still quite delicate to decide whether there is a spectrum with $BP_*(X)=BP_*/(v_0^2,v_1^p)$, and it is not known how to settle such questions for ideals involving $v_0,\dotsc,v_n$ when $n$ is large. The best that you can do is to say that there exists a sequence $i_0,i_1,i_2,\dotsc$ and spectra $X_n$ for all $n$ such that $$ BP_*(X_n) = BP_*/(v_0^{p^{i_0}},\dotsc,v_{n-1}^{p^{i_{n-1}}}). $$ This follows from the work of Hopkins and Smith on existence of $v_n$-self maps. As $X_n$ is a finite spectrum, you can also choose $d_n$ such that $\Sigma^{d_n}X_n$ is a suspension spectrum. I guess that the minimum possible $d_n$ grows quite rapidly with $n$, much faster than $n$ itself. I don't think that there are any results in the literature that would give a useful upper bound on $d_n$.

| cite | improve this answer | |
  • $\begingroup$ This was a great answer, I'm going to edit the question with all the info you gave here, using the spectrum $BP_*(X_n) = BP_*/(v_0^{p^{i_0}},v_1^{p^{i_1}},\dots,v_{n-1}^{p^{i_{n-i}}})$ Thank you very much! $\endgroup$ – Alfred Nov 25 '17 at 13:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.