5
$\begingroup$

Given a prime $p$, there is a well known homology theory $BP$, known as Brown-Peterson homology. Has several related theories, namely the Johnson-Wilson theories $E(n)$ and the truncated Brown-Peterson theories $BP\langle n\rangle$. Their homotopy groups are given by $$ E(n)_* = \mathbb{Z}_{(p)}[v_1, ..., v_{n-1}, v_n^{\pm 1}] $$ and $BP\langle n \rangle$ is a connective version of $E(n)$. In the case $n=1$, $p$-local $KU$ splits as a wedge of suspensions of $E(1)$, and $ku$ splits as a wedge of $BP\langle 1 \rangle$, and in this case the algebras $E(1)_*BP\langle 1\rangle$ and $E(1)_*E(1)$ are understood and the elements have interpretations as numerical polynomials.

Does anyone know of methods to compute these algebras for $n>1$, or even $n=2$? I think I can compute $E(1)_*E(2)$ and $E(1)_*BP\langle 2\rangle$, so I would also be interested to know if there is a method of producing $E(2)_*E(2)$ from $E(1)_*E(2)$ and $K(2)_*E(2)$.

$\endgroup$
5
  • $\begingroup$ Well the $E(n)$s are Landweber exact so $E(n)_*E(m)$ is just the scheme representing isomorphisms of their respective formal group laws. $\endgroup$ Commented Nov 14, 2015 at 17:40
  • $\begingroup$ In fact I think we can say more: $E(n)_*BP(m) = E(n)_*\otimes_{BP_*}BP_*BP(m) = E(n)_*\otimes_{BP_*}BP(m)_*BP$, so the only thing left to figure out is the map $BP_*\to BP(m)_*BP$, which should be related to the comultiplication on $BP_*$. $\endgroup$ Commented Nov 14, 2015 at 17:49
  • 1
    $\begingroup$ Your description of $BP\langle n \rangle$ as the connective cover of $E(n)$ is incorrect as they do not have the same $\pi_0$. For example, $\frac{v_1^{p^n-1}}{v_n}$ is in the homotopy of the connective cover of $E(n)$ but not in the homotopy of $BP\langle n \rangle$. I do not think this effects what anyone has said though. $\endgroup$ Commented Nov 16, 2015 at 11:48
  • $\begingroup$ @sean yes you're absolutely right! Thanks for pointing out my mistake :). I guess the right thing to say is that its the cofibre of the regular sequence $v_{n+1}, v_{n+2}, ...$. $\endgroup$
    – CWcx
    Commented Nov 16, 2015 at 12:13
  • $\begingroup$ @denis what you're describing is more or less how I computed $E(1)_*E(1)$, but I guess I am more curious in various relations that hold in $E(n)_*E(n)$. Based on some fiddling around I did, it seems hard to come up with explicit relations since the right unit is famously difficult to calculate. $\endgroup$
    – CWcx
    Commented Nov 16, 2015 at 12:20

1 Answer 1

5
$\begingroup$

Regarding $E(n)_*E(n)$, see "On the Structure of the Hopf Algebroid $E(n)_*E(n)$" by Keith Johnson. Johnson shows that $$E(n)_*E(n) \otimes \mathbb{Q} \simeq \mathbb{Q}[v_1,\cdots,v_{n-1},v_n^{\pm 1},\overline{v}_1,\cdots,\overline{v}_{n−1},\overline{v}_n^{\pm 1}],$$ where $v_i = \eta_L(v_i)$ and $\overline{v}_j = \eta_R(v_j)$.

Moreover, let $A$ be the ring of integers in an unramified degree $n$ extension of p-adic numbers, and let $\mathbb{S}_n$ be the $n$-th Morava stabilizer group. Johnson also shows that there is an embedding of $E(n)_*E(n)$ in $C(\mathbb{S}_n \times \mathbb{S}_n ,A)$.

$\endgroup$
1
  • $\begingroup$ That's so cool! thanks for the reference! $\endgroup$
    – CWcx
    Commented Nov 16, 2015 at 12:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .