Is the Lie quadric $Q^3$ isomorphic to the Lagrangian Grassmannian $LG(2,4)$?

On the paper

http://arxiv.org/abs/1009.1364

(published on Proc. London Math. Soc.) I've found an interesting statement:

The Lie quadric $Q^3$, i.e., the space of all points, lines and circles of $\mathbb{R}^2$, is isomorphic to the Lagrangian Grassmannian $LG(2,4)\subset G(2,4)$ of all 2D subspaces of $\mathbb{R}^4$ which are isotropic with respect to the canonical symplectic form $\eta=e^1\wedge e^3+e^2\wedge e^4$, where $\{e^i\}$ is a dual basis of $\mathbb{R}^4$.

But this fact is not proved clearly, and I'm beginning to wonder whether it is actually true.

Just looking at the dimensions, everything looks all-right: $LG(2,4)$ contains an open subset of symmetric $2\times 2$ matrices, so it must be 3D; to parameterise all circles in $\mathbb{R}^2$ (comprising zero and infinite radii) I need 3 numbers (2 for the position of the center and 1 for the radius).

Then, the actual proof should go as follows: Plucker-embed $LG(2,4)$ into $\mathbb{P}(\mathrm{ker}\,\eta)\cong\mathbb{RP}^4$, where $\mathrm{ker}\,\eta\subset\Lambda^2\mathbb{R}^4$, and observe that its image is the 3D quadric $Q\subset \mathbb{RP}^4$ of isotropic elements with respect to the obviously defined (conformal) symmetric form $$\Lambda^2\mathbb{R}^4\otimes\Lambda^2\mathbb{R}^4\longrightarrow \Lambda^4\mathbb{R}^4\cong\mathbb{R}.$$ On the other hand, $Q^3$ is the 3D quadric of isotropic lines in $\mathbb{R}^{2,3}$, but I cannot manage to make the last step and answer myself the following

QUESTION: is it true that $Q=Q^3$?

In the affirmative case, it should be such a basic fact that even several alternative proofs should have be given in literature - but I wasn't able to find any. For example, I'd like to see a very elementary one, i.e., not involving multilinear algebra, but just incidence/tangency of lines, planes and circles. Any reference will be appreciated! Thanks.

• Thanks for linking to the freely available copy (as well as pointing out the 'official' version). – David Roberts May 12 '14 at 10:33

This, of course, is "classical". Here is a simple way to see the identification between the space of all circles (including point circles) on the two-sphere and the Lagrangian Grassmannian in $\mathbb{R}^4$ taken from Section 5 of the paper Finsler surfaces with prescribed geodesics by Gautier Berck and myself.
Consider $S^3$ as the set unit quaternions and let $\pi : S^3 \rightarrow S^2$ be the Hopf fibration $\pi(q) := qi\bar{q}$. Every Lagrangian subspace in $\mathbb{R}^4$ intersects the sphere in a (Legendrian) great circle and the image of this great circle under the Hopf fibration is a circle on the two-sphere. Conversely, every circle on the two-sphere can be "lifted" to a Legendrian great circle on the three-sphere (the lift is actually the Frenet frame of the circle in the group of unit quaternions or, equivalently, in $SU(2)$).
• I don't know the exact reference, but this fact is in Lie's work. He proved that the group of contact transformations of the plane that carry circles to circles (points are 'ideal circles' in the classical language) is a $10$-dimensional group isomorphic to what we now call $\mathrm{SO}(2,3)$, which is, in turn, 'isomorphic' to what we now call $\mathrm{Sp}(2,\mathbb{R})$. Of course, he didn't make a careful distinction between real and complex, and he didn't worry too much about 'points at infinity' (i.e., global questions). I'm sure I could find the exact reference, if you want to know it. – Robert Bryant May 10 '14 at 9:53
• @alvarezpaiva your explanation is "almost" what I needed. I can see that intersecting $S^3$ with a Lagrangian plane yields a circle on $S^2$ via Hopf fibration, but then - I guess - it still needs to be stereographically projected to $\mathbb{R}^2$ to become an actual circle on a plane. Still, I don't see where 'ideal circles' come from (I think your construction cannot deal with them: what is the Frenet frame of a point?) But I have to read carefully your paper, and Lie's work, kindly suggested by R. Bryant, and will take a few days... meanwhile, I'll be waiting for more hints from MO users! – Giovanni Moreno May 12 '14 at 7:35