Contact distributions on $(G_2,P)$-type Cartan geometries in dimension 5

Question

Up to topology, the 5D homogeneous space $$ G_2/P $$ of the (real form of the) 14D exceptional Lie group $G_2$ is the 5D jet space $$ M:=J^1(2,1)=\{(x,y,u,p,q)\} $$ of scalar functions in two independent variables.

WHY? Because $G_2/P=S^2\times S^3\approx\mathbb{P}T^*S^3\approx\mathbb{P}T^*\mathbb{R}^3=\mathbb{R}^2\times(\mathbb{R}\times\mathbb{R}^{2\ast})=J^2(2,1)$, where $\mathbb{R}^2$ is the plane of independent variables, $\mathbb{R}$ the line of the dependent one, and $\mathbb{R}^{2\ast}$ the plane of the derivatives of the latter w.r.t. the former.

Now $M$ comes equipped with a flag of distributions, viz $$ (M,V,C)\, , $$ where $V=\langle\partial_p,\partial_q\rangle$ is the 2D vertical and $C=V\oplus\langle \partial_x+p\partial_u, \partial_y+q\partial_u\rangle$ the 4D contact one.

Now there is a result (I've learned it from P. Nurowski) saying that

(THEOREM) There is a one-to-one correspondence between $(G_2,P)$-type Cartan geometries and generic 2D distributions in dimension 5.

Since - as textbooks recite - a $(G_2,P)$-type Cartan geometry is obtained by rolling without slipping the homogeneous space $G_2/P\approx M$, I guess that a generic 2D distribution in dimension five is obtained by "rolling without slipping" the vertical distribution $V$ of the homogeneous model $M$. This should be the essence of the theorem.

My point is simple: if I can "roll" the vertical distribution $V$, why I cannot "roll" the contact distribution $C$ as well?

QUESTION: Is it true that a $(G_2,P)$-type Cartan geometry in dimension five is the same as a manifold equipped with a $(2,4)$-type flag of distributions $(\widetilde{V},\widetilde{C})$? Under which circumstances $\widetilde{C}$ is contact? (This makes sense, being so at least for the flat case $M$.)

Maybe the question is ill-posed, due to my still poor grasp on Cartan geometries, but I hope that its essence will be clear: the "flat model", i.e., $M$, is a contact manifold equipped with a distinguished 2D contact-subdistribution (no doubts about that), and I'm just wondering how much of this structure descends to the "curved cases".

"I'm just a simple man trying to make my way in the universe of Cartan geometries" - J. Fett.

Answer 1

As Ben wrote, the question appears to conflate two different parabolic geometries of type $\newcommand{bfD}{{\bf D}}\newcommand{bfE}{{\bf E}}\newcommand{bfH}{{\bf H}}G_2$:

Let $\Bbb V$ be the standard (i.e., $7$-dimensional irreducible) representation of $\mathfrak{g}_2$ (either the split real or the complex form); recall that $G_2$ is the stabilizer of cross product map $\times : \Bbb V \times \Bbb V \to \Bbb V$. The inclusion $G_2 \hookrightarrow SO(3, 4)$ ($G_2 \hookrightarrow SO(7, \Bbb C)$) determines an indefinite, nondegenerate, symmetric bilinear form $H$ on $\Bbb V$.

The "first" parabolic subgroup $P_1$ (corresponding to a cross on the first node of the Dynkin diagram of $G_2$ in the usual Bourbaki ordering) is the stabilizer of an isotropic $1$-dimensional subspace of $\Bbb V$. The cone $\mathcal C$ of nonzero isotropic vectors inherits an invariant filtration of tangent distributions, whose fibers at $Y \in \mathcal C$ are $$\ker (Z \mapsto Y \times Z) \subset \operatorname{im} (Z \mapsto Y \times Z) \subset T_Y \mathcal C$$ of dimensions $3, 4, 6$ (to apply the cross product, we implicitly use here the identifications determined by the canonical isomorphism $T_Y \Bbb V \leftrightarrow \Bbb V$). By linearity, this descends to a filtration $$\bfD \subset \bfD' \subset T\Bbb Q_5,$$ of dimensions $2, 3, 5$, on the null quadric $\Bbb{Q}_5 := \Bbb P(\mathcal C) \subset \Bbb P(\Bbb V)$, which is diffeomorphic to $(\Bbb S^2 \times \Bbb S^3) / \Bbb Z_2$, where $\Bbb Z_2$ acts by the antipodal map on both factors. Since the ingredients are $G_2$-invariant, so is $\bfD$ under the induced action on $T\Bbb Q_5$, and as one expects, it turns out that $[\bfD, \bfD] = \bfD'$ and $[\bfD', \bfD] = T\Bbb Q_5$. (NB there are no $G_2$-invariant linear or hyperplane distributions on $T\Bbb Q_5$.)

On the other hand, consider differential equations of the form $z' = F(x, y, y', y'', z)$. Any function $F(x, y, p, q, z)$ determines a total derivative $D_x := \partial_x + p \partial_y + q \partial_p + F \partial_z$ on the corresponding partial jet space $J^{2, 0}(\Bbb R, \Bbb R) \cong \Bbb R^5_{xypqz}$. Suitably regarded, the vertical fibers of the jet truncation map $J^{2, 0}(\Bbb R, \Bbb R) \to J^{1, 0}(\Bbb R, \Bbb R)$ are spanned by $\partial_q$. Computing directly shows that the distribution $$\bfD_F := \langle D_x, \partial_q \rangle$$ is generic iff $F_{qq}$ vanishes nowhere; conversely, a theorem of (I believe) Monge states than any generic $2$-plane distribution on a $5$-manifold is locally equivalent to $\bfD_F$ for some function $F$. If $F(x, y, p, q, z) := q^2$, then the resulting distribution has infinitesimal symmetry algebra isomorphic to $\mathfrak{g}_2$, so by a general fact about parabolic geometries the distribution $(J^{2, 0}(\Bbb R, \Bbb R), \bfD_F)$ corresponding to the differential equation $z' = (y'')^2$ is everywhere locally diffeomorphic to the homogeneous model distribution $(\Bbb Q_5, \bfD)$ above. In particular, it follows from this that neither of the distributions $V$ and $\widetilde{C}$ are invariant under the action of the infinitesimal symmetry algebra $\mathfrak{g}_2$ of $(J^{2, 0}(\Bbb R, \Bbb R), \bfD_F)$.

The correct statement of the theorem you mention is that there is an equivalence of categories between generic $2$-plane distributions on $5$-manifolds and normal, regular parabolic geometries of type $(G_2, P_1)$. (See the end of Subsubsection 4.3.2 in Cap & Slovak's book, Parabolic Geometries.)

On the other hand, we can consider the action of $G_2$ on the space of isotropic $2$-planes in $\Bbb V$. This action has two orbits, according to whether the cross product $\times$ restricts to the zero map on each $2$-plane. (Bryant calls the $2$-planes on which the restriction is zero special in his highly enjoyable lecture notes Elie Cartan and Geometric Duality [pdf], which treats the correspondence space construction for $G_2 / P_1 \leftarrow G_2 / (P_1 \cap P_2) \to G_2 / P_2$, as well as the analogous construction for $A_2$ and $B_2 \cong C_2$. NB that this article seems contains a few typos, replacing $\Bbb N_5$, introduced in a moment, with $\Bbb Q_5$, which is the essential apparent confusion in the question here.) The isotropy subgroup of a point in the $5$-dimensional space $\Bbb N_5$ of special $2$-planes is the "second" parabolic $P_2 \subset G_2$. Analogously to the situation for the first parabolic, we can view $\Bbb N_5$ as a subset of $\Bbb P^{13} = \Bbb P(\mathfrak{g}_2)$, but its geometry is apparently much more complicated than that of $\Bbb Q_5$: In the complex case, $\Bbb N_5$ is a variety of degree $18$, and its complete intersection with three hyperplanes in a general configuration is a K3 surface of genus $10$, but NB other geometric descriptions of this space (which look less daunting to non-algebraic geometers like myself) are available, too. Apparently this is worked out in the paper of Borcea cited below, but I can't find an ungated copy. See also the accessible historical survey paper of Agricola, also cited below.

Now, $\Bbb N_5$ inherits an invariant contact distribution $\bfH$. (Surely this can be written down with some much effort in terms of the cross product on $\Bbb V$, but to my knowledge this hasn't been done anywhere.) Moreover, the representation $P_2$ induces on each fiber of $\bfH$ turns out to be a trivial extension of a representation of $GL(2, \Bbb F) \subset P_2$, and this representation is isomorphic to $S^3 \Bbb F^2$ (this representation is conformally symplectic and so determines equivalently a nondegenerate cone in each fiber of $\bfH$). All of the $G_2$-invariant structure on $\Bbb N_5$ can be recovered from these objects, corresponding to the fact that a (normal, regular) parabolic geometry of type $(G_2, P_2)$ is a $5$-manifold $M$ equipped with a "$G_2$ contact structure", which is a contact structure $\bfH \subset TM$ together with an auxiliary rank-$2$ vector bundle $\bfE \to M$ and a vector bundle isomorphism $S^3 \bfE \stackrel{\cong}{\to} \bfH$ such that the Levi bracket $\bfH \times \bfH \to TM / \bfH$ (the tensorial map induced by the Lie bracket) is invariant under the induced action of $\mathfrak{sl}(\bfE)$. See Subsubsection 4.2.8 of Cap & Slovak's book. I don't know of any sensible analog of the Monge (quasi-)normal form $z' = F(\cdots)$ for $G_2$ contact structures, but I would be pleased to hear about one.

There are connections between these two types of parabolic geometry beyond the mentioned correspondence space construction. See this preprint of Leistner, Nurowski, & Sagerschnig.

I. Agricola, Old and New on the Exceptional Group $G_2$ [pdf], Notices Amer. Math. Soc. 55(8) (2008), 922-929.

C. Borcea, Smooth global complete intersections in certain compact homogeneous complex manifolds, J. Reine Angew. Math. 344 (1983), 65–70.

A. Cap, J. Slovak, Parabolic geometries I: Background and general theory. Math. Surveys Monogr. 154, Amer. Math. Soc., Providence, RI, 628pp.

T. Leistner, P. Nurowski, K. Sagerschnig, New relations between $G_2$-geometries in dimensions $5$ and $7$, Internat. J. Math. 28(13) (2017). arXiv:1601.03979

Answer 2

You are confusing (me and) the two different 5-dimensional homogeneous spaces of $G_2$, both of which are $X=G_2/P$ but for different parabolic subgroups $P \subset G_2$. One has a $G_2$-invariant nondegenerate rank 2 distribution, contained in a rank 3. The other has a $G_2$-invariant contact structure. See my paper on characteristic classes for lots of examples of invariant distributions on the models: http://arxiv.org/abs/0704.2555. The only $G_2$-invariant distributions on $G_2/P_2$ (in the notation of that paper) are the rank 2 and rank 3 (and the tangent bundle and its zero section). The only $G_2$-invariant distribution on $G_2/P_1$ is the contact structure.

A little more about rolling: take two surfaces $S_1, S_2$ embedded into $\mathbb{R}^3$. Draw a curve along one of them, and roll the other along that curve. (We need some geometric hypothesis to keep them from getting ``stuck'', but we ignore this.) At each moment in time, they are tangent at some pair of points, giving an isometry of their tangent planes. So the rolling is described as a curve in the manifold $M$ of tuples $(x_1,x_2,u)$ where $x_i \in S_i$ and $u \colon T_{x_1} S_1 \to T_{x_2} S_2$ is a linear isometry. Conversely, there is a 2-plane field $V$ on $M$ (i.e. a rank 2 subbundle $V \subset TM$) so that each path $x_1(t)$ with given initial values of $x_2(0), u(0)$ determines a unique path $(x_1(t),x_2(t),u(t))$ tangent to $V$, and this path is the rolling of one surface on the other. This $V$ depends only on the Riemannian metrics of the two surfaces, not on how they are embedded into $\mathbb{R}^3$. So we have a map $(S_1,S_2) \mapsto (M,V)$ from pairs of surfaces with Riemannian metric to 5-manifolds with 2-plane field. A 2-plane field $V$ on a 5-manifold is nondegenerate if any two local linearly independent sections $X,Y$ have $X, Y, [X,Y], [X,[X,Y]], [Y,[X,Y]]$ linearly independent. The 2-plane field $V$ on $M$ we have constructed by rolling is nodegenerate just at the points $(x_1,x_2,u)$ so that $S_1$ and $S_2$ have distinct Gauss curvature at $x_1$ and $x_2$. There is no rolling of $M$ involved in this picture.

So we have a map $(S_1,S_2) \mapsto (M,V)$ to nondegenerate 2-plane fields on 5-manifolds (cutting out points $(x_1,x_2,u)$ of $M$ where Gauss curvatures agree at $x_1$ and $x_2$). Two great unsolved questions: (1) what is the image and (2) what are the fibers? Well known methods should suffice to solve both problems. Not every nondegenerate 2-plane field $V$ arises, even locally, this way, as you can see by counting, once you know that Cartan proved that every nondegenerate 2-plane field has finite dimensional symmetry group. There are examples of pairs of pairs, i.e. $(S_1,S_2)$ and $(S_1',S_2')$ pairs of surfaces, so that the associated $(M,V)$ nondegenerate 2-plane fields are isomorphic, even though $(S_1',S_2')$ are not isometric to $(S_1,S_2)$ even after constant rescaling. But it seems likely that a typical choice of $(S_1,S_2)$ pair gives rise to an $(M,V)$ which is not obtained from any other pairs of surfaces $(S_1',S_2')$ except for rescaling the Riemannian metrics of both $S_1$ and $S_2$ by the same positive constant.