Timeline for A particular Lie algebra $L_{n}$ and (various) lie groups whose Lie algebra is isomorphic to $L_{n}$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 14, 2020 at 23:17 | history | bounty ended | Ali Taghavi | ||
| Dec 12, 2015 at 17:35 | comment | added | Qiaochu Yuan | @Ali: there is no need to think about this fiber bundle. Take $G = GL_n$ and take $X$ to be the $G$-set of nonzero vectors in $\mathbb{R}^n$, on which $G$ acts transitively. All I am doing is changing coordinates. | |
| Dec 12, 2015 at 12:35 | comment | added | Ali Taghavi | @QiaochuYuan i realy do not know how to apply your comment to the questions. | |
| Dec 12, 2015 at 8:58 | comment | added | Ali Taghavi | @QiaochuYuan yes this elementary. But how you apply this to my question. We have a projection map from $\{(v,g)\in Gl(n)\times S^{n-1}\mid gv=v\}\to S^{n-1}$. Two questions: is it a locally trivial fiber bundle?Can one construct a principal fiber bundle. In the latter we do not specify a certain group action. Could you please more explain? | |
| Dec 12, 2015 at 8:26 | comment | added | Qiaochu Yuan | @Ali: it's much simpler than that. If any group $G$ acts transitively on any set $X$, then the stabilizers of every $x \in X$ are conjugate: explicitly, if the stabilizer of $x$ is $H$, then the stabilizer of $gx$ is $gHg^{-1}$. This is elementary group theory. | |
| Dec 12, 2015 at 7:51 | comment | added | Ali Taghavi | @QiaochuYuan Yes, the projection is that.The motivation for this question is your statment "This is conjucate to the same group but where v replaced..."So the fibers are the same group and topological space(up to isomorphism and homemorphism) So is there a local trivialization for a fiber or principal fiber bundle structure? | |
| Dec 12, 2015 at 7:45 | comment | added | Qiaochu Yuan | @Ali: I mean, I'm not sure what kind of answer you're expecting. Are you asking whether the map $(v, g) \to v$ is in fact a fiber bundle? | |
| Dec 12, 2015 at 7:41 | comment | added | Ali Taghavi | @QiaochuYuan is my other question on the fiber bundle structure, obvious?Thanks again for your answer on my main question. | |
| Dec 12, 2015 at 7:38 | comment | added | Ali Taghavi | @YCor Thank you for your interesting comment which completes the answer. What about my other question in the previous comment? | |
| Dec 11, 2015 at 13:19 | comment | added | YCor | In $GL_n$ the corresponding Lie group is $\mathbf{R}^{n-1}\rtimes GL_{n-1}^+(\mathbf{R})$ (which has index 2 in the non-connected $\mathbf{R}^{n-1}\rtimes GL_{n-1}(\mathbf{R})$). This group has a trivial center; its $\pi_1$ is the same as in the maximal compact subgroup $S0(n-1)$ which has $\pi_1$ of order $1$ if $n=2$, cyclic infinite if $n=3$, and of order 2 if $n\ge 4$. This allows to list those connected Lie groups having $L_n$ as Lie algebra. | |
| Dec 11, 2015 at 7:40 | comment | added | Ali Taghavi | motivated by your answer, can one consider $\{ (v,g)\in S^{n-1}\times Gl(n)\mid g.v=v \}$, as total space of a natural fiber or principal fiber bundle over $S^{n-1}$? | |
| Dec 11, 2015 at 7:29 | vote | accept | Ali Taghavi | ||
| Dec 11, 2015 at 6:42 | comment | added | Qiaochu Yuan | @Ali: $X \mapsto - X^T$ is a Lie algebra automorphism of $\mathfrak{gl}_n$, which is the Lie algebra version of $g \mapsto (g^T)^{-1}$ being a Lie group automorphism of $GL_n$. | |
| Dec 11, 2015 at 6:40 | comment | added | Ali Taghavi | Thank you very much for your answer. I think you mean $v^{tr}X=0$ ? So a related question: is every Lie subalgebra $L$ of $M_{n}(\mathbb{R})$ isomorphic to its transpose $L^{tr}$ while the transpose operator is not Lie algebra automorphism? | |
| Dec 11, 2015 at 6:32 | history | answered | Qiaochu Yuan | CC BY-SA 3.0 |