Timeline for Corollaries of the Yoneda Lemma in Analysis?
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Aug 20, 2016 at 16:56 | vote | accept | Chill2Macht | ||
| Aug 15, 2016 at 22:18 | history | edited | Chill2Macht | CC BY-SA 3.0 |
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| S Jul 10, 2016 at 0:03 | history | bounty ended | Chill2Macht | ||
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| Jul 4, 2016 at 14:55 | answer | added | Tom LaGatta | timeline score: 6 | |
| Jul 3, 2016 at 0:10 | answer | added | godelian | timeline score: 17 | |
| S Jul 3, 2016 at 0:07 | history | bounty started | Chill2Macht | ||
| S Jul 3, 2016 at 0:07 | history | notice added | Chill2Macht | Canonical answer required | |
| Jul 2, 2016 at 20:56 | comment | added | Andrej Bauer | The Dedekind cuts are a little more complicated than that. You might be able to get some mileage out of the fact that presheavesa are the free cocompletion and use that to complete posets under suprema. But Dedekind cuts really are a bit more involved. | |
| Jul 2, 2016 at 19:47 | history | edited | Chill2Macht | CC BY-SA 3.0 |
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| Jul 1, 2016 at 14:01 | comment | added | David Roberts♦ | Riesz doesn't seem to me to be an example of Yoneda, though certainly it looks like there would be an abstract characterisation of it at the category level. I could be wrong though, I haven't thought about it deeply. | |
| Jul 1, 2016 at 3:20 | comment | added | Chill2Macht | @DavidRoberts You are right-- vector spaces and their bases do seem to be a special case of the principle of equivalence. What about the Riesz Representation Lemma? In that case the dual objects are literally defined by how they act on all of the regular vectors. Also any other example from analysis would be greatly appreciated/very interesting. | |
| Jul 1, 2016 at 1:22 | comment | added | David Roberts♦ | In particular, you don't then care whether your vector space is $\mathbb{R}^n$ or any random $n$-dimensional real vector space $V$ (or if you want the Hilbert space structure, any $n$-dimensional real Hilbert space). You only get coördinates when you choose an isomorphism $V \simeq \mathbb{R}^n$, equivalently a basis. | |
| Jul 1, 2016 at 1:20 | comment | added | David Roberts♦ | Regarding your examples, specifically considering the automorphism group of a vector space, you are probably more thinking of the principle of equivalence rather than the Yoneda lemma: you can replace the one-object category with object the specific vector space $V$ of interest, with the category with objects all possible vector spaces that are isomorphic to $V$. A representation of a group $G$ on $V$ is isomorphic to any functor from the one-object category with arrows $G$ to this larger category of vector spaces: you lose nothing. | |
| Jul 1, 2016 at 1:16 | comment | added | David Roberts♦ | I added a link to the relevant section of the YT video of LaGatta's talk. | |
| Jul 1, 2016 at 1:16 | history | edited | David Roberts♦ | CC BY-SA 3.0 |
Added direct link to timestamp in LaGatta's talk
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| Jun 30, 2016 at 23:13 | history | asked | Chill2Macht | CC BY-SA 3.0 |