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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3$\begingroup$ 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. $\endgroup$– Robert BryantCommented Apr 22, 2013 at 21:55
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1$\begingroup$ BTW: I would suggest asking this question on math stack exchange, since it is a standard kind of question for that site. $\endgroup$– Robert BryantCommented Apr 22, 2013 at 22:00
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2$\begingroup$ 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 $k$)! $\endgroup$– user28172Commented Apr 22, 2013 at 22:31
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2$\begingroup$ 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 ♦Commented Apr 23, 2013 at 2:24
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2$\begingroup$ Dear Tim, There is a three dimensional irrep. of $PGL(2)$, given by the symmetric square of the standard rep. of $GL(2)$. This gives a map from $PGL(2)$ to $GL(3)$, which must be an injection, since $PGL(2)$ is a simple adjoint group. It suffices to show that the image preserves a non-degen. quadratic form (since then we will get an embedding of $PGL(2)$ into $SO(3)$, which will be an isomorphism for dimension reasons). For this, consider $Sym^2$ of the three dimensional rep'n. A simple calculation with weights shows that this rep'n contains a unique copy of the trivial rep'n. ... $\endgroup$– EmertonCommented Apr 23, 2013 at 4:07
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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.
answered Apr 23, 2013 at 2:57
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1$\begingroup$ 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 $\mathfrak{sl}(2, \mathbb R) \cong \mathfrak{pgl}(2, \mathbb R)$. $\endgroup$– LSpiceCommented Sep 26, 2019 at 15:50