Timeline for Ostensibly different products on Ext-groups
Current License: CC BY-SA 2.5
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2020 at 21:07 | comment | added | Robert Bruner | Lovely argument that $E \otimes E'$ is equivalent to $E \circ E'$, reminiscent of mathoverflow.net/questions/35868/… | |
| Mar 18, 2011 at 14:17 | comment | added | Wilberd van der Kallen | @Ralph Thanks! I am not at all sure that flatness is enough. | |
| Mar 18, 2011 at 12:25 | comment | added | Ralph | @Wilberd: In the 2nd edition of Benson's book the beginning of 3.2 requires: "Suppose $\Lambda$ is a Hopf algebra over $R$ which is projective as an $R$-module." Are you sure that flat is sufficient ? | |
| Mar 18, 2011 at 10:27 | comment | added | Wilberd van der Kallen | My favorite reference is D.J.Benson, Representations and cohomology, section 3.2. It requires $A$ to be flat over $k$, which makes sense to me. | |
| Mar 18, 2011 at 8:28 | comment | added | Ralph | Another reference for the material described by Mariano is section VIII.4 in MacLane's book "Homology". A second application of the argument above (besides commutativity of cup product as remarked by John) is that cup product only depends on the product of $A$, not on the coproduct (though used in the definition!), i.e. if there are two k-Hopf algebras with the same underlying k-module $A$ those products agree then their cup products agree, too. I find this property quite remarkable. | |
| Mar 18, 2011 at 6:58 | comment | added | Theo Buehler | Thanks for this answer! I didn't know that paper and it's something that should be really useful for me. Why do you call it Hilton-Eckmann and not the other way around? I can't resist to add that the name Eckmann-Hilton has a long tradition in Zurich. The first location of the Forschungsinstitut (the research institute created by E. for hosting foreign scientists) was often called like that for what should be obvious reasons. Eckmann was great: He used to come to the institute on a daily basis until about three years ago when he was 91 already. He always took his time for answering questions. | |
| Mar 18, 2011 at 6:52 | comment | added | Ben Williams | I suppose I should really try to understand the exterior product construction as a construction using the derived tensor product in the derived category of $A$-modules. | |
| Mar 18, 2011 at 6:47 | comment | added | Ben Williams | Thanks especially for the preprint, which I think meets my needs. I have qualms about the concrete construction though, because neither the functor $\otimes_A$ or $\otimes_k$ is exact (I assume $\otimes_k$ is what is meant by $\otimes$ in the explicit calculation). | |
| Mar 18, 2011 at 6:37 | vote | accept | Ben Williams | ||
| Mar 18, 2011 at 4:38 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Mar 18, 2011 at 4:15 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Mar 18, 2011 at 4:09 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Mar 18, 2011 at 4:08 | comment | added | John Palmieri | The argument in the first paragraph also shows that the product is (graded) commutative, right? It's just like the proof that the fundamental group of a topological group is abelian. | |
| Mar 18, 2011 at 3:53 | comment | added | Ben Williams | Thanks for the answer, which is both helpful and promising. I am afraid I don't understand it, being inexperienced with this sort of thing. What is $E \otimes E'$? I think if I understood this I would understand everything. Thanks again. | |
| Mar 18, 2011 at 3:28 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Mar 18, 2011 at 2:58 | comment | added | Sean Tilson | that is always the way isn't it. | |
| Mar 18, 2011 at 2:57 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
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| Mar 18, 2011 at 2:49 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |