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I'm basically wondering how to make "curved" the first column of the diagram $\require{AMScd}$ \begin{CD} P_1 @>\textrm{inclusion} >> G\\ @V \omega_0 V P_1\cap P_2 V @V\omega V P_2 V\\ P_1/(P_1\cap P_2) @= G/P_2 \end{CD} where $M:=G/P_2$ is a compact homogeneous space, such that the smaller group $P_1\subset G$ still acts transitively on $M$. Above, I regard $M$ as a flat Cartan geometry of type $(G,P_2)$, with Maurer-Cartan form $\omega$.

Simultaneously, I want to regard $M$ as a (non necessarily flat) Cartan geometry of type $(P_1,P_1\cap P_2)$, which is compatible with the flat $(G,P_2)$-type one, in the sense that the first column of $\require{AMScd}$ \begin{CD} \mathcal{G}_\sigma @>\textrm{inclusion} >> G\\ @V \omega_\sigma V P_1\cap P_2 V @V\omega V P_2 V\\ M @= M \end{CD} is no longer flat, i.e., $\mathcal{G}_\sigma$ is a principal sub-bundle of $G$ (and not necessarily a subgroup).

Then I see a nonempty family (it contains at least the flat case) $$ \{(\mathcal{G}_\sigma\to M, \omega_\sigma)\}_{\sigma\in \boldsymbol{\Sigma}} $$ of compatible Cartan geometries of type $(P_1,P_1\cap P_2)$.

MAIN QUESTION: what is the space $\boldsymbol{\Sigma}$ parametrising these compatible "sub-Cartan geometries"?

A rather brutal reasoning led me to believe that $$ \boldsymbol{\Sigma}=\Gamma(\pi), $$ where $\pi:G/N_{P_2}(P_1\cap P_2)\longrightarrow G/P_2=M$. Indeed, the generic fibre of $\pi$ is $P_2/N_{P_2}(P_1\cap P_2)$, which tells how many subgroups exist in $P_2$, isomorphic to $P_1\cap P_2$: hence, a section $\sigma$ of $\pi$ allows me to construct a principal $(P_1\cap P_2)$-subbundle $\mathcal{G}_\sigma$ of $G$.

Finally, suppose that $\boldsymbol{\Sigma}$ admits a geometric description (similar, at least in spirit, to my brute attempt above) in terms of natural structures associated to $M$.

SIDE QUESTION: what is the correct way to "pull-back" the Maurer-Cartan form $\omega$ from $G$ to $\mathcal{G}_\sigma$ thus obtaining $\omega_\sigma$? (In the sense that it should be a pull-back which introduce somehow the curvature.)

If my question is well-posed, then I cannot believe that this matter has not been given attention before: any reference will be appreciated!

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I don't think that this is a real answer to your question, but there is a general concept of extension functors for Cartan geometries. Such a functor extending Cartan geometries of type $(G,P)$ to Cartan geometries of type $(K,L)$ is determined by a homomorphism $i:P\to L$ and a linear map $\alpha:\mathfrak g\to\mathfrak k$ which is equivariant for the actions of $P$ and $L$ over $i$, restricts to $i'$ on $\mathfrak p$ and induces a linear isomorphism $\mathfrak g/\mathfrak p\to \mathfrak k/\mathfrak l$. Given a Cartan geometry $(\mathcal G\to M,\omega)$ you extend the bundle as $\mathcal G\times_P L$ and construct $\omega_{\alpha}$ as an equivariant extension of $\alpha\circ\omega$. The curvature of the resulting geometry is a combination of two terms, one coming from the curvature of the initial geometry, the other from the obstruction against $\alpha$ being a homomorphism of Lie algebras. I am aware of intersting examples of extension functors producing a non-flat geometry out of a flat one, for example the chains of a spherical CR structure define a non-flat path geometry. However, I don't know of an example of obtaining a flat extension from a non-flat geomtry. Indeed, I think that this would need a rather peculiar setup.

These extension functors are discussed in sections 1.5.15 and 1.5.16 of the book on parabolic geometries by J. Slovak and myself ( http://bookstore.ams.org/surv-154/ ).

Another related concept is holonomy reductions of Cartan geometries, but again this only produces locally flat Cartan geometries when starting from a locally flat one.

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  • $\begingroup$ I guess that Proposition 1.5.15 from your book is going to answer my question! I just need time to adapt it to my context. When you say that you don't of any example of a flat geometry extending a non-flat one, you mean that this is impossible, or simply that it is a very special case? $\endgroup$ – Giovanni Moreno Sep 26 '16 at 7:51
  • $\begingroup$ I don't claim that this is impossible, but the situation has to be very special. The point is that the algebraic contribution coming from $\alpha$ exactly has to cancel with the term coming from the original curvature. (In the case of chains, this is impossible since the two terms have different symmetry properties, but of course there can be many other situations.) $\endgroup$ – Andreas Cap Sep 26 '16 at 7:58

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