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Jul 28, 2014 at 3:03 comment added Peter May Right, Dustin: the space F = GL_1(S) is both additive (F) and multiplicative (GL_1(S)), and it is split by an exponential equivalence from J with its additive structure to J with its multiplicative structure.
Mar 29, 2014 at 22:33 comment added Dustin Clausen Hi Elden, I mean that in your map of cohomology theories, if you take the direct sum on the left (not tensor product!) and the fiberwise smash product on the right, then the map preserves these structures. It's the smash product on the right that I was calling "multiplicative", because it's related to the product in the E-infinity ring S^0.
Mar 20, 2014 at 21:05 comment added Elden Elmanto Hi @DustinClausen, I'm sorry but I don't quite get why having "direct sum of vector bundles getting sent to smash product of spheres" tells us that only the multiplicative structure is relevant. Could you elaborate on that? My (possibly wrong) understanding would be: to see that the J-map is a multiplicative map would be to check that the map on cohomology theories: $K(X) \rightarrow Sph(X)$ is a map on multiplicative but not additive structures. This is tensor product of bundles on the left but I am not quite sure what it is on the right. It would be great if you could clarify this!
Feb 6, 2014 at 17:52 history edited Dustin Clausen CC BY-SA 3.0
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Feb 1, 2014 at 16:44 comment added Dustin Clausen Well, understandable would be better than impressive! I hope someone else can come along and answer your question as intended.
Feb 1, 2014 at 15:56 comment added Johannes Ebert +1, even if I have to admit that I understand even less than in Adams' work. But it sounds impressive.
Feb 1, 2014 at 14:06 history edited Dustin Clausen CC BY-SA 3.0
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Feb 1, 2014 at 13:55 history edited Dustin Clausen CC BY-SA 3.0
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Feb 1, 2014 at 13:52 comment added Dustin Clausen Oh, by the way, I shouldn't give the wrong impression here. I breezily talk about all these "computations". But I'm terrible at computations, and besides, I did these computations a while ago, and I'm partly relying on memory. So they could be wrong!
Feb 1, 2014 at 13:41 history answered Dustin Clausen CC BY-SA 3.0