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I'm having trouble dealing with the following question : what is the isomety group of $\mathbf{PSL}_2(\mathbb{R})$ viewed as a Lie group with its Killing form ? For the record, its Killing form is the push forward of the determinant on $\mathbf{SL}_2(\mathbb{R})$. I'm almost sure that the component of the identity is $\mathbf{PSL}_2(\mathbb{R}) \times \mathbf{PSL}_2(\mathbb{R})$ acting by right and left multiplication, but i think i don't know enough theory to show it.

Thank you for looking at my problem

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I think we had a similar question on Mathoverflow a while ago. Maybe you will find information there. –  Mariano Suárez-Alvarez Jun 4 '12 at 17:12
    
i checked but i didn't find an answer –  Selim Ghazouani Jun 4 '12 at 17:42
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Robert Bryant gave an actual algorithm to compute isometry groups of homogeneous spaces here mathoverflow.net/questions/75742/…, at least. In your case, you have to take $H=(e)$. –  Mariano Suárez-Alvarez Jun 4 '12 at 18:05
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