Timeline for Exponentiation of vector spaces?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
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| Jun 25, 2023 at 1:00 | comment | added | Daniel Asimov | I would start by defining exp(V). A reasonable definition of exp(V) is the direct sum over n ≥ 0 of the symmetric tensor products Sym^n(V) = the nth tensor power of V quotiented out by the symmetric group in the obvious way. | |
| Jun 24, 2023 at 22:19 | answer | added | LSpice | timeline score: 1 | |
| Feb 10, 2018 at 5:43 | vote | accept | მამუკა ჯიბლაძე | ||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 30, 2015 at 23:40 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Sep 30, 2015 at 23:34 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
inequality
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| Sep 30, 2015 at 6:27 | comment | added | მამუკა ჯიბლაძე | This is even more reminiscent of plethories (see my link in a comment above) but I really don't know | |
| Sep 30, 2015 at 6:26 | comment | added | user76479 | Even $\Bbb K_1=\Bbb K_2$ is an interesting case. | |
| Sep 30, 2015 at 6:26 | comment | added | მამუკა ჯიბლაძე | Over different fields?? | |
| Sep 30, 2015 at 6:25 | comment | added | user76479 | Polynomial algebras over fields. Let us look at specific cases before generalizing. | |
| Sep 30, 2015 at 6:25 | comment | added | მამუკა ჯიბლაძე | You mean rings of polynomials? | |
| Sep 30, 2015 at 6:22 | comment | added | user76479 | I dont even see how you would define $\Bbb K_1[x]^{\otimes\Bbb K_2[x]}$. | |
| Sep 30, 2015 at 6:22 | comment | added | მამუკა ჯიბლაძე | @Arul Sorry I don't see how algebra structures might enter the picture (except I have some vague association with the formalism of plethories) | |
| Sep 30, 2015 at 6:14 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Sep 30, 2015 at 6:11 | comment | added | user76479 | Even if you consider $V,W$ to be algebras (not just vector spaces) can you define $V^{\otimes W}$? | |
| Sep 30, 2015 at 6:10 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
added action on operators
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| Sep 30, 2015 at 5:16 | answer | added | Qiaochu Yuan | timeline score: 16 | |
| Sep 30, 2015 at 5:01 | comment | added | Qiaochu Yuan | My answer was incorrect; the correct observation (that $GL(W)$ can't act faithfully) is in the comment above. Anyway, also note that if you want functoriality in $W$ with respect to non-invertible morphisms then there are generally no nontrivial $GL(V)$-equivariant maps $V \to V^{\otimes 2}$ or $V^{\otimes 2} \to V$, etc. In that exterior algebra observation the "base" of the exponential is not a vector space but a graded vector space, so my comment above doesn't apply. | |
| Sep 30, 2015 at 4:59 | comment | added | მამუკა ჯიბლაძე | I liked your answer very much and almost accepted it, but since you deleted it let me add one thing here :D It occurred to me that $\Lambda^*(W_1\oplus W_2)\cong\Lambda^*(W_1)\otimes\Lambda^*(W_2)$ allows one to view $\Lambda^*(W)$ as "$2^W$" of sorts, so there must be more to it. I mean, It might be that there is some more tricky invariance wrt $V$ (whereas action on the $W$ side might be "ordinary"). What do you think? | |
| Sep 30, 2015 at 4:35 | comment | added | Qiaochu Yuan | I think the answer is morally no. If you additionally require that $V^{\otimes (-)}$ sends direct sums to tensor products and that $V^{\otimes 1} \cong V$, then using Schur-Weyl duality you can conclude that if $\dim W = 2$ then $GL(W)$ can never act faithfully on $V^{\otimes W}$ (in a way compatible with the natural action of $GL(V)$), regardless of the size of $V$. I think the lesson of Schur-Weyl duality here is that unlike the situation with taking iterated direct sums, the exponent when taking iterated tensor products can be upgraded to at best a set but not really a vector space. | |
| Sep 30, 2015 at 4:13 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 3.0 |