MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question

closed as off topic by Robert Bryant, Misha, Qiaochu Yuan, Angelo, Chris Gerig Apr 23 '13 at 4:30

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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. – Robert Bryant Apr 22 '13 at 21:55
BTW: I would suggest asking this question on math stack exchange, since it is a standard kind of question for that site. – Robert Bryant Apr 22 '13 at 22:00
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 $k$)! – user28172 Apr 22 '13 at 22:31
I think nosr neglected to include the condition that the automorphisms have Dickson invariant 1 (equivalent to determinant 1 when 2 is invertible). – S. Carnahan Apr 23 '13 at 2:24
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. ... – Emerton Apr 23 '13 at 4:07
up vote 4 down vote accepted

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.