Timeline for "Natural" pairings between exterior powers of a vector space and its dual
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 27, 2021 at 13:50 | comment | added | LSpice | I did a bit of tidying while this was bumped to the front page. In particular, you referred to "the linked answer", but then spoke of something said by Aaron, which seemed to be a different answer, so I linked to that instead. | |
| Dec 27, 2021 at 13:50 | history | edited | LSpice | CC BY-SA 4.0 |
Dots, `\Lambda` -> `\bigwedge`, and names of links while this is on the front page
|
| Dec 19, 2021 at 23:37 | answer | added | Alexey Muranov | timeline score: 2 | |
| S Dec 3, 2021 at 7:59 | history | suggested | Alexey Muranov | CC BY-SA 4.0 |
latex: \to -> \mapsto
|
| Dec 2, 2021 at 20:36 | review | Suggested edits | |||
| S Dec 3, 2021 at 7:59 | |||||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
|
|
| Apr 2, 2015 at 18:17 | comment | added | Alexey Muranov | If i understand [differential forms and integrals] well, to integrate a differential form with the first pairing, one needs to cut the space into rectangles/parallelepipeds, and with the second pairing, one needs to cut the space into triangles/tetrahedra. :) | |
| Jun 17, 2011 at 12:51 | vote | accept | Qiaochu Yuan | ||
| Jun 17, 2011 at 10:08 | comment | added | Qiaochu Yuan | @Andrew: yes, that's what I thought too until I saw Torsten's answer. It seems the solution is to introduce more structure (namely the canonical comultiplication on the exterior algebra). I haven't checked this but the explicit formulas in Bourbaki agree with the first pairing when you do what Torsten suggests. | |
| Jun 17, 2011 at 9:59 | comment | added | Andrew Stacey | By asking for a pairing, you are wanting to use both definitions of "exterior product" the free functor and the alternating mapping one. The point of my answer there was that there will always be a conflict between these two and you can shuffle it around as much you like, but you will always get a 1/n! somewhere. | |
| Jun 17, 2011 at 9:41 | comment | added | Qiaochu Yuan | @Andrew: yes, one way to motivate this question is that I want all of my diagrams to commute by default. I am being very careful not to make any casual identifications, and my second pairing is precisely your filler map. | |
| Jun 17, 2011 at 8:16 | comment | added | Andrew Stacey | Have you read (my answer to) this question: mathoverflow.net/questions/37349/… There are a lot of things here that get casually identified which ought not to be. I haven't gone through your question to see if you are doing so, so this may not be relevant, but it feels like it might be. | |
| Jun 17, 2011 at 7:54 | comment | added | S. Carnahan♦ | (The above should not be taken as a general claim that choices are bad. They can be quite useful, but it is occasionally helpful to notice when one is making them.) | |
| Jun 17, 2011 at 7:51 | comment | added | S. Carnahan♦ | Random note: one explanation for the absence of a canonical Fourier-self-dual measure on $\mathbb{Z}/n\mathbb{Z}$ is that you are choosing a duality pairing that is non-canonical - one indicator of this non-naturality is that you find yourself choosing (among other things) a square root of minus one. The Pontryagin (or Cartier) dual of $\mathbb{Z}/n\mathbb{Z}$ is $\mu_n$, which is only isomorphic to $\mathbb{Z}/n\mathbb{Z}$ if you can choose a distinguished $n$th root of unity. If you leave the duals be, the Fourier transform works for arbitrary coefficients. | |
| Jun 17, 2011 at 7:37 | answer | added | Torsten Ekedahl | timeline score: 13 | |
| Jun 17, 2011 at 2:37 | answer | added | Theo Johnson-Freyd | timeline score: 1 | |
| Jun 16, 2011 at 22:52 | comment | added | Qiaochu Yuan | @james-parson: thanks! The Bourbaki approach is absolutely the one I wanted. Perhaps I should read the rest of Algebra after all... | |
| Jun 16, 2011 at 22:23 | comment | added | user2490 | Perhaps the treatment of these issues in Bourbaki Algebra Chapter III, Section 11 is a convenient reference for the approach that Ekedahl sketches. | |
| Jun 16, 2011 at 22:15 | comment | added | Torsten Ekedahl | I am not sure what you mean by "treat symmetrically" but one way is to get a graded algebra homomorphism $\Lambda^\ast V\to(\Lambda^\ast V^\ast)\ast$ by using the fact that $\Lambda^\ast V^\ast$ is a super-Hopf algebra such that the algebra structure on $(\Lambda^\ast V^\ast)\ast$ dual to the coalgebra structure is strictly graded commutative ("strictly" meaning that the square of an odd element is zero) and $\Lambda^\ast V$ is universal for strictly graded commutative algebras. This homomorphism is then verified to be an isomorphism. | |
| Jun 16, 2011 at 22:15 | comment | added | Qiaochu Yuan | @Torsten: that sounds like exactly what I'm looking for. Could you write that up in an answer? | |
| Jun 16, 2011 at 21:57 | answer | added | user15832 | timeline score: 1 | |
| Jun 16, 2011 at 21:57 | comment | added | Qiaochu Yuan | @KConrad: you can describe that isomorphism without specifying what happens on pure tensors, since $V \otimes W$ and $W \otimes V$ satisfy the same universal property. Again, there are situations in which that particular device is not available. | |
| Jun 16, 2011 at 21:48 | comment | added | KConrad | Why do you dislike specifying what happens on (all) pure tensors and then extending? After all, the simplest standard isomorphisms between tensor products of vector spaces, like the tensor product of $V$ and $W$ in both orders being isomorphic, involves saying what happens first on pure tensors and then extending. | |
| Jun 16, 2011 at 21:39 | comment | added | Qiaochu Yuan | I will dislike it until someone tells me how to construct it functorially. There are situations in which an analogue of exterior powers and duals exist but one can't work on the level of pure tensors (symmetric monoidal categories with left duals, say) and I want to know if this pairing is still defined in those situations. | |
| Jun 16, 2011 at 21:32 | comment | added | babubba | do you really dislike the first pairing so much? I would vote against the second pairing for (as you say) it's not defined in characteristic less than n. (not that I like positive characteristic, but it seems a good enough reason to discard it). anyway, the point of my comment is to advertise daniel murfet's notes therisingsea.org/notes/TensorExteriorSymmetric.pdf, I wish he'd produced more! | |
| Jun 16, 2011 at 21:18 | history | asked | Qiaochu Yuan | CC BY-SA 3.0 |