Timeline for Rigidity of the category of schemes
Current License: CC BY-SA 3.0
27 events
| when toggle format | what | by | license | comment | |
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| Dec 23, 2019 at 23:15 | vote | accept | Martin Brandenburg | ||
| Dec 23, 2019 at 21:10 | answer | added | R. van Dobben de Bruyn | timeline score: 16 | |
| Dec 23, 2019 at 18:27 | vote | accept | Martin Brandenburg | ||
| Dec 23, 2019 at 23:15 | |||||
| Nov 25, 2018 at 16:47 | answer | added | user131755 | timeline score: 8 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Sep 9, 2013 at 19:26 | comment | added | Hans-Peter Stricker | I come in from over there: math.stackexchange.com/questions/487272/…. You write "[...] that a category is rigid if every object can be defined in a categorical way". What did you have in mind with "defined in a categorical way"? | |
| Jul 18, 2012 at 11:03 | comment | added | Martin Brandenburg | Under finiteness conditions everything works out fine: See for example the answer by Laurent Moret-Bailly, my questions about regular monomorphisms etc., and the paper Categorical Representation of Locally Noetherian Log Schemes by Shinichi Mochizuki. | |
| Feb 19, 2012 at 17:21 | comment | added | Martin Brandenburg | @Tom: Corrected. @Charles: Actually one gets Aut(k) as the automorphism class group | |
| Feb 19, 2012 at 6:55 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Feb 18, 2012 at 23:28 | comment | added | Tom Leinster | Martin, sorry to keep correcting, but I think the category of groups is rigid (at least, I can't think of any counterexample). The endofunctor sending a group to its opposite is isomorphic to the identity, by taking inverses. | |
| Feb 18, 2012 at 15:22 | answer | added | Laurent Moret-Bailly | timeline score: 13 | |
| Feb 18, 2012 at 9:05 | comment | added | Martin Brandenburg | @Tom: Yes, of course. Thanks for this correction. @Charles: Probably. But in order to prove this, we can't just copy the previous proofs (the rigidity of the category of rings used quotient fields). Instead, the following should work: $\mathbb{A}^1$ has only one ring object structure, thus $F(\mathbb{A}^1) \cong \mathbb{A}^1$ even as ring objects. Then as explained above, $F$ restricts to an equivalence of the category of irreducible finite type $k$-algebras which preserves $k[t]$. Then we may recover an arbitrary $A$ as $\mathrm{Hom}(k[t],A)$, and conclude $F(A) \cong A$. | |
| Feb 18, 2012 at 8:08 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Feb 17, 2012 at 19:40 | comment | added | Charles Staats | Checking for understanding: Does your February 2012 edit imply that the category of varieties (i.e., irreducible varieties) over an algebraically closed field is rigid? If so, this seems like it is already a very interesting result. | |
| Feb 17, 2012 at 16:59 | comment | added | Tom Leinster | Martin, re your third paragraph, the category of semigroups isn't rigid, is it? It has an automorphism sending each semigroup to its opposite. But this automorphism is not isomorphic to the identity. Ditto (not necessarily commutative) rings. | |
| Feb 17, 2012 at 13:52 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| May 11, 2011 at 23:13 | comment | added | Todd Trimble | Theo, I've heard people say "rigid monoidal category", but not "rigid category" to mean the same thing. (If a Wikipedian wrote that, then I think that's a mistake!) | |
| May 11, 2011 at 9:51 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Mar 1, 2011 at 9:21 | answer | added | S. Carnahan♦ | timeline score: 11 | |
| Mar 1, 2011 at 3:54 | comment | added | David Roberts♦ | Interestingly, the category of categories is almost rigid, in that it only has one autoequivalence (up to iso, I suppose) which isn't isomorphic to the identity, namely the functor $C \mapsto C^{op}$. | |
| Feb 28, 2011 at 15:46 | comment | added | Theo Johnson-Freyd | Oh, +1 by the way. | |
| Feb 28, 2011 at 15:46 | comment | added | Theo Johnson-Freyd | This is a better use of the word "rigid" than is normally used. The old word "rigid category" refers to a monoidal category with well-behaved duals [en.wikipedia.org/wiki/Rigid_category]. It's not a very good word. In general, particularly bad words in category theory come from Australia, and some of the good ones are Russian, but in this case the name is from two Germans (Dold and Puppe). Another reasonable meaning for "rigid" is "no nontrivial infinitesimal deformations", whereas you mean "no nontrivial automorphisms". | |
| Feb 28, 2011 at 14:50 | answer | added | John Iskra | timeline score: 3 | |
| Feb 28, 2011 at 14:19 | answer | added | Harry Gindi | timeline score: 13 | |
| Feb 28, 2011 at 13:22 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Feb 28, 2011 at 11:30 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Feb 28, 2011 at 11:23 | history | asked | Martin Brandenburg | CC BY-SA 2.5 |