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Jul 25, 2014 at 15:17 answer added Urs Schreiber timeline score: 3
Oct 29, 2012 at 15:40 answer added Nick Kuhn timeline score: 7
Mar 13, 2012 at 14:33 vote accept Jonathan Beardsley
Mar 13, 2012 at 6:39 answer added Neil Strickland timeline score: 13
Mar 12, 2012 at 22:44 comment added Jonathan Beardsley So Tyler, doesn't that orthogonality mean that we glue together an E(n) local spectrum really easily from it's K(n) local pieces?
Mar 12, 2012 at 22:40 comment added Tyler Lawson On the other hand, you have that $K(n) \wedge K(m)$ is contractible for $n \neq m$, and the same identity holds for their Bousfield classes. The situation you're describing actually relies on something special - namely, that for $n > m$ anything $K(m)$-local is $K(n)$-acyclic.
Mar 12, 2012 at 22:26 comment added Jonathan Beardsley And yes... you're right about the composition, to build the $E(n)$ localizations. I guess... hmm, what am I saying. I guess it should be something like that. In that case, it should be like wedging right? Since that's how we build our $E(n)$'s?
Mar 12, 2012 at 22:04 comment added Tom Goodwillie You seem to have your arrows backwards. And it's possible that the well-known case you are thinking of involves a composition $L_E\circ L_F$ rather than $L_{E\wedge F}$ or $L_{E\vee F}$.
Mar 12, 2012 at 22:04 comment added Jonathan Beardsley As well as situations with the Morava $K$ and $E$ theories.
Mar 12, 2012 at 21:43 history edited Jonathan Beardsley CC BY-SA 3.0
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Mar 12, 2012 at 21:20 comment added Jonathan Beardsley I might add that there is the well known case where we do this with completion at primes and rationalization. I think...
Mar 12, 2012 at 21:16 history asked Jonathan Beardsley CC BY-SA 3.0