Is the Lie quadric $Q^3$ isomorphic to the Lagrangian Grassmannian $\operatorname{LG}(2,4)$? $\DeclareMathOperator\LG{LG}$In the paper The - Conformal geometry of surfaces in the Lagrangian—Grassmannian and second order PDE (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 $\LG(2,4)\subset \operatorname G(2,4)$ 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!
 A: 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 $SU(2)$).
The details are in the paper.
A: Replying to this somewhat late but I think there is important context to this. The Lagrangian Grassmannian and the quadric are both examples of R-spaces (symmetric R-spaces in particular) as such they are given by $ G/P $ for $G$ a semisimple Lie Group and $P$ some parabolic subgroup. So this question boils down to $ \mathfrak{sp}(4,\mathbb{R}) \cong \mathfrak{so}(2,3) $ which is one of the exceptional Lie isomorphisms. (Note that the stabiliser of an isotropic plane in $\mathbb{R}^4$ is sent to that of a null line in $\mathbb{R}^{2,3}$.) All the projective quadrics up to dimension 4 have such an identification to some Grassmannian (isotropic, isotropic for a hermitian form, Lagrangian or normal) over the reals, complex numbers or quaternions.
From a more practical perspective take the Klein correspondence:
$G_2(\mathbb{R}^4)$ embeds into $\mathbb{P}(\bigwedge^2 \mathbb{R}^4)$ as the light cone. Then a symplectic form $\omega$ on $\mathbb{R}^4$ is an element of $\bigwedge^2 (\mathbb{R}^4)^* \cong \left(\bigwedge^2 \mathbb{R}^4\right)^*$. A plane $V$ is isotropic iff $\bigwedge^2 V$ is in the annihilator of $\omega$. Then if $\omega$ is nondegenerate the annihilator must have signature $(2,3)$ or $(3,2)$ so its intersection with the Klein quadric is the Lie quadric you require.
