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

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.

• A $(G_2,P)$-geometry with torsion can fail to induce a generic 2-dimensional distribution. In one of my papers (I am so old!) I worked out the translation invariant parabolic geometries. There are many, for every model, and all of the distributions they induce are bracket closed. – Ben McKay Mar 6 '16 at 8:52


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

• You gave a sophisticated answer to a lousy question: chapeau! – Giovanni Moreno Mar 13 '16 at 22:06
• You're welcome, Giovanni, I hope you found it useful, too! It was good to meet you in Vienna earlier this year. – Travis Mar 13 '16 at 22:12

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.

• Every complex simple Lie group has a unique homogeneous contact manifold, the orbit of a highest weight vector in the adjoint form. (See the papers of Landsberg and Manivel.) So $G_2$ doesn't preserve any contact structure on any but one of its homogeneous spaces, at least after you complexify. – Ben McKay Mar 6 '16 at 9:00
• I confess that after "A little more about rolling" I had a hard time understanding - though it is interesting and surely we'll get back to it in September. Concerning my specific question: if I got it, $G_2$ has two 5D homogeneous spaces, one is equipped with a contact distribution, and another with a $(2,3)$-type flag of distributions (the 2D one being "nondegenerate" in the sense of yours, or $(2,3,5)$ as others like to call it). Question: the former is $S^3\times S^2$; how the latter looks like? What is a good and quick reference to learn about $G_2$ and its homogeneous spaces? – Giovanni Moreno Mar 8 '16 at 6:56
• The second one is the projectivized null cone in the irreducible 7-dimensional representation of the split form of $G_2$, so a 5-sphere. It is also the adjoint variety (projectivized orbit of a highest weight vector in the adjoint form). Its complex form is a quadric hypersurface in projective space, so a 5-sphere. I don't know a good place to look this up. Maybe when I get to Poland, we can write a book. – Ben McKay Mar 8 '16 at 14:56
• Book-writing is a good entertainment for people over sixties - so let's discuss it in two decades - but a review paper why not? So, the two homogeneous 5-folds of $G_2$ are $S^2\times S^3$ and $S^5$, which are both contact, but only the former is contact homogeneous. The latter should be equipped with a $(2,3)$-type flag of distributions, but I fail to see it: is it evident? In fact, Sagershnig's paper says that the projectivised null cone in the imaginary octonions is $S^2\times S^3$, i.e., the opposite of what you say! That's why I'd like to see how a 2D and a 3D distribution on $S^5$ arise. – Giovanni Moreno Mar 8 '16 at 17:44
• An & Nurowski have given explicit examples of nonhomothetic pairs $(S_1, S_2)$ and $(S_1', S_2)$ of surfaces for which the corresponding plane fields $(M, V)$ are distinct: The usual choice rolling surface realization of the flat model of the geometry (i.e., model with local symmetry group $G_2$) is a pair of spheres, one of which has radius three times that of the other. On the other hand, there are exactly three surfaces $S$ up to homothety that admit a Killing vector such that $(\Bbb R^2, S)$ is also flat, and hence induce a flat distribution; see arxiv.org/pdf/1210.3536.pdf . – Travis Mar 12 '16 at 20:25