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I need to prove that $PGL_2(\mathbb{R})\cong SO_3(\mathbb{R})$. Abstract considerations show that both can be identified with the group of projective motions of a conic curve. But maybe there is more explicit isomorphism (in matrix form, for example)?

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    $\begingroup$ This is not really a research question, so I'm inclined to vote to close. However, before doing that, I should ask you whether you really mean $\mathrm{SO}(3)$ or not? I think you might mean $\mathrm{SO}(2,1)$ instead. $\endgroup$ Apr 22, 2013 at 21:55
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    $\begingroup$ BTW: I would suggest asking this question on math stack exchange, since it is a standard kind of question for that site. $\endgroup$ Apr 22, 2013 at 22:00
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    $\begingroup$ Dear Robert: For a non-degenerate quadratic space $(V,q)$ over a field $k$, usually ${\rm{SO}}(q)$ denotes the algebraic $k$-group classifying automorphisms of $(V,q)$ (over extensions of $k$). If $q$ is the standard split quadratic form $q_n$ on $k^n$ ($x_1 x_2 + x_3 x_4 + \dots + x_{n-1}x_n$ for even $n$, $x_0^2 + q_{n-1}$ for odd $n > 1$), it is common for algebraists to write ${\rm{SO}}_n$ to denote ${\rm{SO}}(q_n)$. So for $k = \mathbf{R}$, the Lie group ${\rm{SO}}_n(\mathbf{R})$ is not the same as ${\rm{SO}}(n)$. And ${\rm{PGL}}_2 = {\rm{SO}}_3$ as algebraic groups (over any $k$)! $\endgroup$
    – user28172
    Apr 22, 2013 at 22:31
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    $\begingroup$ @nosr: Thanks for pointing this out; I wasn't familiar with this convention among algebraists (which would be very confusing for geometers if they were to try to adopt it). Since the title of the question used '$\mathrm{SO}(3)$', I assumed that it was interchangeable (in the OP's mind) with $\mathrm{SO}_3(\mathbb{R})$ (a notation I did not use in my comment). Would $\mathrm{SO}(3)$ also mean $\mathrm{SO}_3(\mathbb{R})$ to you? How would you denote $\mathrm{SO}(2,1)$? Also, your definition confuses me a bit because it seems to be what I would denote as $\mathrm{O}_3$. How do you get the '$S$'? $\endgroup$ Apr 23, 2013 at 0:36
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    $\begingroup$ I think nosr neglected to include the condition that the automorphisms have Dickson invariant 1 (equivalent to determinant 1 when 2 is invertible). $\endgroup$
    – S. Carnahan
    Apr 23, 2013 at 2:24

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Put the bilinear form $\langle, \rangle$ on $2 \times 2$ real matrices by setting $\langle A,B \rangle = {\rm tr}(AB).$ The space of matrices breaks with respect to this form as the orthogonal direct sum of the space of scalar matrices and the $3$-dimensional subspace of matrices of trace zero. Now ${\rm GL}(2,\mathbb{R})$ acts by conjugation on the the matrices of trace zero, and preserves this bilinear form in that action. Furthermore, scalar matrices (and nothing more) in ${\rm GL}(2,\mathbb{R})$ are in the kernel of this action, so the action is really one of ${\rm PGL}(2,\mathbb{R}).$ Every matrix in ${\rm GL}(2,\mathbb{R})$ has the eigenvalue $1$ in this action- a scalar matrix certainly does and any non-scalar matrix $A$ fixes the matrices of trace zero in ${\rm span}(I,A).$ Every element of ${\rm PGL}(2,\mathbb{R})$ acts with determinant $1$ in this action, as diagonal elements clearly do. This gives an embedding of ${\rm PGL}(2,\mathbb{R})$ in the special orthogonal group determined by this form,and dimension shows that it is surjection.

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    $\begingroup$ Maybe worth noting that the choice of bilinear form is not ad hoc: aside from being the obvious one to choose, it descends to a multiple of the Killing form on $\mathfrak{sl}(2, \mathbb R) \cong \mathfrak{pgl}(2, \mathbb R)$. $\endgroup$
    – LSpice
    Sep 26, 2019 at 15:50

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