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Nov 16, 2015 at 12:20 comment added CWcx @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.
Nov 16, 2015 at 12:13 comment added CWcx @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 $v_{n+1}, v_{n+2}, ...$.
Nov 16, 2015 at 11:48 comment added Sean Tilson 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.
Nov 15, 2015 at 15:06 answer added Drew Heard timeline score: 5
Nov 14, 2015 at 17:49 comment added Denis Nardin 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 $BP_*$.
Nov 14, 2015 at 17:40 comment added Denis Nardin 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.
Nov 14, 2015 at 17:37 review First posts
Nov 14, 2015 at 17:55
Nov 14, 2015 at 17:36 history asked CWcx CC BY-SA 3.0