Timeline for Examples of group $G=N \rtimes H$ where $N$ and $H$ are as below
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2018 at 8:09 | comment | added | Venkataramana | @Mambo: yes. The simple connectedness follows from the homotopy exact sequence since $N \rightarrow {\mathbb R}^2$ is a locally trivial fibration with fibre $\mathbb R$. | |
| Dec 15, 2018 at 6:19 | comment | added | Mambo | Is every central(non-split) extension with kernel $\mathbb{R}$ and quotient $\mathbb{R^2}$ simply connected nilpotent Lie group? | |
| Dec 14, 2018 at 1:40 | comment | added | Venkataramana | @YCor: Yes, you are right. I do not need $N$ to be a semi-direct product. There is an exact sequence $$ 0 \rightarrow {\mathbb R} \rightarrow N \rightarrow {\mathbb R}^2 \rightarrow 1$$. The group $GL(2,{\mathbb R})$ acts on the exact sequence. | |
| Dec 13, 2018 at 19:33 | comment | added | YCor | $N$ is not a semidirect product $\mathbf{R}\rtimes\mathbf{R}^2$; I don't know if you mean to think of it as a central (not split) extension with kernel $\mathbf{R}$ and quotient $\mathbf{R}^2$, or as a semidirect product with kernel $\mathbf{R}^2$ and quotient $\mathbf{R}$. Anyway, it works. | |
| Dec 13, 2018 at 11:54 | comment | added | Venkataramana | OK. You replace $SL(2,{\mathbb R})$ by $H=GL(2,{\mathbb R})$ except that on $\mathbb R$ the group $H$ acts via determinant, and hence does not fix anything except identity. It acts on ${\mathbb R}^2$ preserving the symplectic form up to sclars, and hence still acts on $N$. | |
| Dec 13, 2018 at 8:51 | comment | added | Mambo | The above action is symplectic automorphism. Apply inversion further to get desired action. | |
| Dec 13, 2018 at 8:37 | comment | added | Mambo | If the action of $H$ on $N$ is as follows: for $h \in H, (t,v) \in N, h.(t,v) = (t,hv)$ then $H$ fixes $\mathbb{R} \rtimes \{(0,0)\}$ ! What exactly is your action? | |
| Dec 13, 2018 at 8:20 | comment | added | მამუკა ჯიბლაძე | Is it obvious that $H$ acts without fixed points? | |
| Dec 13, 2018 at 8:20 | vote | accept | Mambo | ||
| Dec 13, 2018 at 6:12 | history | answered | Venkataramana | CC BY-SA 4.0 |