Timeline for A new Lie group associated to a given Lie group
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
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| Sep 13, 2019 at 1:53 | comment | added | Ali Taghavi | @YCor Thank you very much Yves for your attention to my question and your very interesting comments. I try to understand its details. | |
| Sep 12, 2019 at 7:16 | comment | added | Ali Taghavi | @user44191 i sincerely thank you very much for your very intriguing beautiful and clear answer in the chat discussion. now i completely understand. | |
| Sep 12, 2019 at 3:16 | comment | added | user44191 | Let us continue this discussion in chat. | |
| Sep 12, 2019 at 3:12 | comment | added | user44191 | The equation is incorrect (I forgot to un-exponentiate and divide), but can be corrected: $[a, b] = \lim_{s, t \rightarrow 0} \frac{exp^{-1}(\text{exp(sa)} exp(tb) exp(-sa) exp(-tb))}{st}$, where we take $s, t$ small enough for the exponential map to be bijective. The idea is that multiplying on the left by $exp(sa)$ (and each of the others) should be equivalent to following the left-invariant vector field, and therefore $f$ should pass through the exponentiations. | |
| Sep 12, 2019 at 1:48 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 12, 2019 at 1:37 | comment | added | Ali Taghavi | The right side (befor we take limit) are member of G. | |
| Sep 12, 2019 at 1:35 | comment | added | Ali Taghavi | @user44191 moreover is the equality you wrote correct? | |
| Sep 12, 2019 at 1:31 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Sep 12, 2019 at 1:24 | comment | added | Ali Taghavi | @user44191 that f preserves the space $\mathcal{g}$ of left invariant vector fields does not mean that $f^* X=X\quad \forall X \in \mathfrak{g}$. Did not you used the later in your comment? | |
| Sep 11, 2019 at 17:48 | comment | added | user44191 | A quick argument that I haven't checked explicitly: $[a, b] = \lim_{s, t \rightarrow 0} \text{exp}(sa) \text{exp}(tb) \text{exp}(-sa) \text{exp}(-tb)$; as $f$ preserves left-invariant vector fields, it preserves the entire right-hand side of the equation, so $df([a, b]) = [df(a), df(b)]$, so $df$ is an automorphism of $\mathfrak{g}$, which implies that $f$ must be an automorphism on $G_0$. $f$ doesn't have to be a group automorphism otherwise (e.g. take $G$ discrete), so I was still slightly wrong above, but if $G$ is connected, I should be right. | |
| Sep 10, 2019 at 22:45 | comment | added | Ali Taghavi | @user44191 Sorry if my question is elementary Let $f$ fix the identity. Why it must be a group homomorphism?9Not that $f$ is assumed to be a diffeomorphism whose action keep invariant the lie algebra of left invariant vector field) | |
| Sep 10, 2019 at 21:35 | comment | added | user44191 | @YCor I think my answer is incorrect (as shown by the example in the OP), though I was aiming at the "purely abstract" method you have. I should have said that the subgroup of $H$ that fixes the identity is the orientation-preserving subgroup $Aut^+(G)$, which leads to an answer of $Aut^+(G) \ltimes G_{left}$, if I'm not mistaken again. | |
| Sep 9, 2019 at 22:02 | comment | added | Ali Taghavi | @YCor BTW after we fix an orientable atlas, the volum form i mentioned does not depend on chart. | |
| Sep 9, 2019 at 21:48 | comment | added | Ali Taghavi | @YCor (and even analytic, i think) But I do not see why? Even translations( which belongs to H, they are not group homomorphism) i really do not understand why the H I described has two interpretations described by you and user44191 | |
| Sep 9, 2019 at 21:38 | comment | added | YCor | It's the group of differentiable group automorphisms. Which turns out to also be the group of continuous automorphisms. | |
| Sep 9, 2019 at 21:23 | comment | added | Ali Taghavi | @user44191 What is the definition of $Aut_{Lie }G$? | |
| Sep 9, 2019 at 21:21 | comment | added | Ali Taghavi | @YCor Yes you are right I should write " we fix an orientation for G since it is orientable. | |
| Sep 9, 2019 at 19:55 | comment | added | YCor | @AliTaghavi It isn't. Every manifold admits a Riemannian metric, including non-orientable ones... also the isometry group of many Riemannian manifolds, including positive-dimensional Euclidean spaces, does not preserve the orientation. Your argument fails because you're just working in a chart and using the orientation of the chart. | |
| Sep 9, 2019 at 19:21 | comment | added | Ali Taghavi | @YCor It is realy true that every Riemannian metric gives an orientation. If $g_{ij}$ be the tensor metric then $\sqrt{det (g_{ij})} dx_1\wedge \ldots\wedge dx_n$ is the volum form associated to our metric. It is independent of coordinate chart. So this volum form determine a unique orientation. | |
| Sep 9, 2019 at 18:10 | comment | added | YCor | @user44191 is right, and the argument is purely abstract: in a group $G$, denoting by $G_{\mathrm{left}}$ the subgroup of the group $S(G)$ of permutations of $G$ consisting of left translations, then its normalizer in $S(G)$ is the semidirect product $\mathrm{Aut}(G)\ltimes G_{\mathrm{left}}$ (exercise). Restricting to the subgroup of orientation-preserving diffeomorphisms gives the conclusion. | |
| Sep 9, 2019 at 18:06 | comment | added | YCor | The second sentence has two verbs so should be syntactically fixed. It also suggest that a Riemannian metric induces an orientation, which is false. | |
| Sep 9, 2019 at 17:03 | comment | added | user44191 | Isn't $H$ just the orientation-preserving subgroup of $Aut_{Lie}(G)$? | |
| Sep 9, 2019 at 16:20 | comment | added | user44191 | I assume you are including $\mathfrak{g}$ into vector fields on $G$ by taking the left-invariant extension? | |
| Sep 9, 2019 at 15:41 | history | edited | LSpice | CC BY-SA 4.0 |
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| Sep 9, 2019 at 15:24 | history | edited | Dima Pasechnik | CC BY-SA 4.0 |
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| Sep 9, 2019 at 14:08 | history | asked | Ali Taghavi | CC BY-SA 4.0 |