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Jan 17, 2019 at 5:19 vote accept Ali Taghavi
Dec 21, 2018 at 16:15 answer added Tobias Diez timeline score: 6
Dec 21, 2018 at 6:40 comment added Thomas Richard (The first arrow should be $M\mapsto \frac{1}{|\det M|^{1/2}}M$ and it works only on $GL(2,\mathbb{R})^+$.)
Dec 20, 2018 at 10:14 comment added Thomas Richard The first map is just the the map $M\mapsto \tfrac{1}{|\det(M)|}M$. Maybe the map $f:SL(2,\mathbb{R})\to\mathbb{H}^2$ is easier to see, consider the action $(A,p)\mapsto A\cdot p$ of $SL(2,\mathbb{R})$ by isometries on $\mathbb{H}^2$. $f$ is then $A\mapsto A\cdot p_0$ where $p_0$ is some reference point. The isotropy group of $p_0$ is then some copy of $SO(2,\mathbb{R})$ inside $SL(2,\mathbb{R})$.
Dec 20, 2018 at 7:08 comment added Ali Taghavi @Thomas Thank you for your comment. What do you mean by the quotiont in your first comment?Further more what is the map in the first arrows, is it the Grahm Schmitd map? If yes is not the range (SO2)?
Dec 19, 2018 at 10:07 comment added Thomas Richard This at least gives a submersion, the remaining question is wether it is a Riemannien submersion for some metric of $GL(2,\mathbb{R})$.
Dec 19, 2018 at 8:19 comment added Thomas Richard Have you tried considering the natural maps $GL(2,\mathbb{R})\to SL(2,\mathbb{R})\to SL(2,\mathbb{R})/SO(2,\mathbb{R})\sim \mathbb{H}^2$ and see if the metric on $\mathbb{H}^2$ comes from something interesting ?
Dec 19, 2018 at 5:58 history asked Ali Taghavi CC BY-SA 4.0