Timeline for A particular embedding of a Lie group in Euclidean space
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Feb 20, 2018 at 8:05 | comment | added | Ali Taghavi | @Ycor So I think two notion s of "isometry" are equivalent. Right? That is distance preserving isometries and infinitiesimally isometries are the same. | |
| Feb 19, 2018 at 12:25 | comment | added | YCor | @AliTaghavi I said in my previous comment that I forgot to assume that the speed is 1. An affine map $t\mapsto tu+v$ is isometric (in any possible sense) if and only if $\|u\|=1$. | |
| Feb 19, 2018 at 11:12 | comment | added | Ali Taghavi | @Ycor I think I am missingsome thing. You wrot e in your 2 previous comment that e an arbitrary affine map is isometric in distance. Is it correct? | |
| Feb 19, 2018 at 9:10 | comment | added | YCor | @AliTaghavi Yes by affine I mean $t\mapsto tu+v$, for $u,v\in\mathbf{R}^d$. Of course yes, I need $\|u\|=1$ for it to be an isometry. | |
| Feb 19, 2018 at 8:53 | comment | added | Ali Taghavi | @Ycor an affine map is distance preserving? By affine do you mean $tA+B$? Are you sure it preserve the distance? | |
| Feb 19, 2018 at 8:24 | comment | added | YCor | @AliTaghavi just any embedding of $\mathbf{R}$ into $\mathbf{R}^d$. It's isometric for the distance iff it is an affine map. Same for locally isometric (in the sense of small distance). But it's infinitesimally isometric as soon as its derivative has norm 1 everywhere, which holds much more often. (If we post-compose an affine map $\mathbf{R}$ into $\mathbf{R}^d$ with the quotient map from $\mathbf{R}^d$ to a torus, we obtain a locally isometric map that is not distance-isometric.) | |
| Feb 19, 2018 at 8:19 | comment | added | Ali Taghavi | @Ycor I always thought two concepts of isometry are equivalent in the following sense: Assume $f:M \to N$ is a diffeomorphism. Then f preserve the distance iff $Df$ preserve the inner product.But I think that you said it is not the case.Am Yes? What is a counter example? | |
| Feb 18, 2018 at 5:42 | comment | added | Ali Taghavi | @YCor Yes I mean associative algebra, since I was motivated by the fact that every Lie algebra can be embedded in an inner Lie algebra, an associative algebra with inner Lie bracket. As for invariant metric, I mean left invariant. | |
| Feb 18, 2018 at 0:26 | comment | added | YCor | @LSpice yes in (2) I meant "does not cover at all any". In general I don't think we can say better than having a common cover with a linear Lie group: the direct product of the $\widetilde{SL_2}$ example and the Heisenberg/cyclic examples has no cover and does not cover a linear Lie group. | |
| Feb 17, 2018 at 23:40 | comment | added | LSpice | @YCor, indeed, I appear to have been quoting a theorem without remembering it. There's some statement describing the farthest a Lie group can be from being linear, though, right? (Also, I think your wording in (2) must be backwards, right? I did not claim that an arbitrary Lie group is covered by a linear Lie group, but rather that it covers one.) | |
| Feb 17, 2018 at 23:27 | comment | added | YCor | @AliTaghavi by algebra do you mean associative algebra? (there are interesting structures, called pre-Lie algebras or left-symmetric algebras, whose brackets are Lie brackets). Also, I guess that by "invariant metric" you mean "left-invariant metric"? | |
| Feb 17, 2018 at 23:26 | comment | added | Ali Taghavi | @YCor I mean the second one, according to Nash embedding theorem, | |
| Feb 17, 2018 at 23:25 | comment | added | YCor | @AliTaghavi unfortunately there are two non-equivalent definitions of isometric embedding: isometric for the distance, or infinitesimally isometric (i.e. the pull-back of the Riemannian metric of the target space equals the Riemannian metric of the left-hand space). Which one do you have in mind (one can even imagine intermediate notions, such as "locally isometric", i.e. preserving small distances). | |
| Feb 17, 2018 at 23:23 | comment | added | YCor | @LSpice no (1) the universal covering of $SL_2(R)$ is not finite cover of a linear Lie group (2) the quotient of the Heisenberg group by an infinite cyclic central subgroup has no cover at all that is a linear Lie group. Also there are two definitions of linear: has an injective continuous representations, vs has one with closed image. These are equivalent, but it's a theorem (maybe of Mostow, I don't remember right now). | |
| Feb 17, 2018 at 23:02 | comment | added | Ali Taghavi | @LSpice However it is not an isometric embedding/ | |
| Feb 17, 2018 at 23:00 | comment | added | Ali Taghavi | @LSpice It is interesting to think about non linear ones. BTW I was not aware of the fact that every Lie group is a cover of a linear group. Thanks for this point and for your comment. | |
| Feb 17, 2018 at 22:53 | answer | added | Igor Rivin | timeline score: 1 | |
| Feb 17, 2018 at 22:50 | comment | added | LSpice | Every (real) Lie group is a finite cover of a linear Lie group (i.e., a closed subgroup of some $\mathrm{GL}_m(\mathbb R)$), and this can obviously be done for linear Lie groups; so maybe it's enough to show that this condition is preserved under covers? (Or, if it's false, then a non-linear group like a spin or metaplectic group is the place to look.) | |
| Feb 17, 2018 at 22:41 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 17, 2018 at 22:35 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 17, 2018 at 22:04 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 17, 2018 at 21:52 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 17, 2018 at 21:41 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 17, 2018 at 21:32 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Feb 17, 2018 at 21:21 | history | asked | Ali Taghavi | CC BY-SA 3.0 |