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Let $\mathbb{H}$ be the Poincare half plane with the hyperbolic metric. Let $Gl(2,\mathbb{R})$ be equipped with a left invariant metric?

Is there a Riemannian submersion from $Gl(2,\mathbb{R})$ to $\mathbb{H}$? If yes, what is a precise formula for such a Riemannian submersion?

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    $\begingroup$ 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 ? $\endgroup$ – Thomas Richard Dec 19 '18 at 8:19
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    $\begingroup$ This at least gives a submersion, the remaining question is wether it is a Riemannien submersion for some metric of $GL(2,\mathbb{R})$. $\endgroup$ – Thomas Richard Dec 19 '18 at 10:07
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    $\begingroup$ @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)? $\endgroup$ – Ali Taghavi Dec 20 '18 at 7:08
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    $\begingroup$ 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 $SL(2,\mathbb{R})$. $\endgroup$ – Thomas Richard Dec 20 '18 at 10:14
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    $\begingroup$ (The first arrow should be $M\mapsto \frac{1}{|\det M|^{1/2}}M$ and it works only on $GL(2,\mathbb{R})^+$.) $\endgroup$ – Thomas Richard Dec 21 '18 at 6:40
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Define the hyperbolic metric $g$ on $\mathbb{H}^2$ by $$ g = \frac{1}{2y^2} (d x^2 + d y^2). $$ The Iwasawa decomposition yields an isomorpism $SL(2)/SO(2)=\mathbb{H}^2$ which is given by $$ \pmatrix{a_{11} & a_{12}\\ a_{21} & a_{22}} \mapsto \frac{1}{(a_{21}^2 + a_{22}^2)}(a_{11} a_{21} + a_{12} a_{22}, 1). $$ Moreover, if you define a metric $h$ on $SL(2)$ by $$ h_A (V, W) = Tr(A^{-1} V A^{-1} W), $$ then the projection $SL(2) \to \mathbb{H}^2$ is a Riemannian submersion. Some time ago, I've written some notes on this (and related things concerning the Poincare upper half plane). They are not in a very good shape but may be helpful nonetheless (they were meant as a first experiment with sage).

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  • $\begingroup$ Thank you very much for your answer. I read it to understand its detail. $\endgroup$ – Ali Taghavi Dec 22 '18 at 19:21
  • $\begingroup$ Is it really an inner product?$h_A$?may be you mean $W^{tr}?in this case why the form is symmetric? $\endgroup$ – Ali Taghavi Dec 27 '18 at 14:05
  • $\begingroup$ $h_A$ is the left translation of the Killing form of $\mathfrak{sl}$ (up to some constant). Since $\mathfrak{sl}$ is semisimple, the Killing form is non-degenerate. $\endgroup$ – Tobias Diez Dec 27 '18 at 17:28

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