$\DeclareMathOperator\LG{LG}$In the paper [The - Conformal geometry of surfaces in the Lagrangian—Grassmannian and second order PDE](http://arxiv.org/abs/1009.1364) (published on [Proc. London Math. Soc.](https://doi.org/10.1112/plms/pdr023)), 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 $\LG(2,4)\subset \operatorname G(2,4)$ 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: Plücker-embed $\LG(2,4)$ into $\mathbb{P}(\ker \eta)\cong\mathbb{RP}^4$, where $\ker \eta\subset{\bigwedge}^2\mathbb{R}^4$, 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
$$
{\bigwedge}^2\mathbb{R}^4\otimes{\bigwedge}^2\mathbb{R}^4\longrightarrow {\bigwedge}^4\mathbb{R}^4\cong\mathbb{R}.
$$
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!



Is the Lie quadric $Q^3$ isomorphic to the Lagrangian Grassmannian $\operatorname{LG}(2,4)$?
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](https://arxiv.org/abs/1002.0243) 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 $SU(2)$).


The details are in the paper.


