Timeline for Non invertibility of certain integral arising from group action
Current License: CC BY-SA 3.0
44 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| S Nov 6, 2015 at 19:17 | history | bounty ended | CommunityBot | ||
| S Nov 6, 2015 at 19:17 | history | notice removed | CommunityBot | ||
| Nov 3, 2015 at 17:14 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 2, 2015 at 17:49 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| S Nov 1, 2015 at 16:41 | history | suggested | Sylvain JULIEN | CC BY-SA 3.0 |
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| Nov 1, 2015 at 16:38 | review | Suggested edits | |||
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| Nov 1, 2015 at 16:34 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 1, 2015 at 16:15 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Nov 1, 2015 at 16:13 | comment | added | Ali Taghavi | @AndreasCap Now I understand the advantage of your formulation: the action of $G$ on the function under integral(integrad) is: $h.F=\rho(h) F$ but with my formulation it is not the case. So i revise the question. Thanks for your interesting comment. | |
| Nov 1, 2015 at 10:01 | comment | added | Andreas Cap | I mean $V^n$ and not $V$, and the analog of invertibility is that all values consist of linearly independent vectors. For the other part, the question is whether you can interpret the statement as the value at $x$ of $\int_G g\cdot f$ for some action on $G$ on a space of functions or not. (I don't claim that such an interpretation would readily answer your question, but it provides a way how to think about the quesiton.) Concerning motivation, I rather mean motivation why the claimed property should hold, rather than what it could be used for. | |
| Oct 30, 2015 at 18:27 | comment | added | Ali Taghavi | in fact the main motivation of this question is a possible generalization of the Borsuk Ulam Theorem | |
| Oct 30, 2015 at 18:18 | comment | added | Ali Taghavi | ..is almost the same. but I do not know for non abelian. The advantage of your consideration is that it is the usual" average" | |
| Oct 30, 2015 at 18:16 | comment | added | Ali Taghavi | @AndreasCap Thank you for your comment. in the first part of your comment, I think you mean $f:X\to V$ rather to $V^{n}$ so the value of integral lies in $V$, so we could not speak about invertieblity. regarding the motivation of such formula, as I said in the question, we are motivated by representation of cyclic groups of order n on $\mathbb{C}$(when we choose a nth root of unity) and a continuous map $f:X \to C$ is counted as a map to $M_{1}(C)$. the motivation comes from proposition 3 of this note:arxiv.org/abs/1110.0091 But you replace g by g^{-1}. I think for G abelian the result | |
| Oct 30, 2015 at 8:09 | comment | added | Andreas Cap | I don't quite understand the motivation for looking at that integral. If you would put $\int_G\rho(g)f(g^{-1}x)$ then there is a simple interpretation: This would mean that you view $f$ as a function $X\to V^n$, where $V$ is the space on which $G$ is represented and average for the natural action of $G$ on the space of such maps. Thus the result would be $G$-equivariant for that action. One would get an analogous statement for the natural action on maps $X\to L(V,V)$ when considering $\int_G\rho(g)f(g^{-1}x)\rho(g)^{-1}$. | |
| Oct 29, 2015 at 19:05 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| S Oct 29, 2015 at 18:14 | history | bounty started | Ali Taghavi | ||
| S Oct 29, 2015 at 18:14 | history | notice added | Ali Taghavi | Authoritative reference needed | |
| Oct 29, 2015 at 18:14 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| S Aug 10, 2015 at 6:03 | history | bounty ended | CommunityBot | ||
| S Aug 10, 2015 at 6:03 | history | notice removed | CommunityBot | ||
| S Aug 2, 2015 at 4:15 | history | bounty started | Ali Taghavi | ||
| S Aug 2, 2015 at 4:15 | history | notice added | Ali Taghavi | Authoritative reference needed | |
| Oct 12, 2014 at 12:46 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 12, 2014 at 10:02 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 11, 2014 at 18:19 | comment | added | paul garrett | Ah, ok, in light of your edits, I'll remove my earlier comments in a little while... | |
| Oct 11, 2014 at 18:01 | history | edited | Ali Taghavi |
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| Oct 11, 2014 at 17:53 | comment | added | Ali Taghavi | @paulgarrett I forgot to add the assumption "X is Compact" Now I edited the question. So in this case, for G abelian the answer is affirmative. | |
| Oct 11, 2014 at 17:51 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 11, 2014 at 15:22 | comment | added | Ali Taghavi | @paulgarrett thank you for your comment. the answer to my question is affirmative for $G$ abelian, since the only irreducible representation is one dim. | |
| Oct 9, 2014 at 9:55 | comment | added | Fernando Muro | @AliTaghavi No, the complex general linear group has higher homotopy groups, so it cannot have a contractible covering space. | |
| Oct 9, 2014 at 6:42 | comment | added | Ali Taghavi | @FernandoMuro Is $e^{A}:M_{n}(\mathbb{C})\to GL(n,\mathbb{C})$ a covering space? | |
| Oct 9, 2014 at 6:39 | comment | added | Ali Taghavi | @paulgarrett Prof. Garrett Thank you very much for the comment. I apologize if my question is trivial: Could you please explain that Why for translation action, the integral under my question is the fourrier transform? Morover, as I asked in my question, what about compact $G$ and $X$? | |
| Oct 8, 2014 at 21:07 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 8, 2014 at 20:46 | history | edited | Ali Taghavi |
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| Oct 8, 2014 at 20:42 | comment | added | Ali Taghavi | @FernandoMuro Thanks for the comment. i deleted the tag:) | |
| Oct 8, 2014 at 20:41 | comment | added | Fernando Muro | It's just an opinion, I'm not sure about anything. | |
| Oct 8, 2014 at 20:39 | comment | added | Ali Taghavi | @FernandoMuro If you are sure, so i delete the tag. but the covering lifting lemma in algebraic topology is very essential here! | |
| Oct 8, 2014 at 20:37 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 8, 2014 at 20:16 | comment | added | Fernando Muro | The fact that the fundamental group shows up doesn't make the question an algebraic topology one. | |
| Oct 8, 2014 at 20:15 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Oct 8, 2014 at 20:09 | comment | added | Ali Taghavi | @FernandoMuro yes because the (algebraic) topological property of $X$ is essential in this question.9In fact $X$ is simply connected or with finite $\pi_{1}$ | |
| Oct 8, 2014 at 20:07 | comment | added | Fernando Muro | Algebraic topology? | |
| Oct 8, 2014 at 20:01 | history | asked | Ali Taghavi | CC BY-SA 3.0 |