Consider a Cartan geometry $\pi: \mathcal{G} \to M$ with Cartan connection $\omega$ modelled on the Klein geometry $(G, H)$.

The Cartan connection is supposed to formalize what it means to "roll without slipping" the homogeneous space $G/H$ on the manifold $M$. I am wondering if the following is one correct way to intuitively think about this.

Let $U \subseteq M$ be an open set such that we have a local section $\sigma: U \to \mathcal{G}$ of the Cartan bundle. Then we can pull down the Cartan connection to get a $\frak{g}$-valued 1-form $\sigma^* \omega$ on $U$.

Suppose we pick a point of contact $o \in G/H$ with a point $x \in M$. Then if we pick a tangent direction $v \in T_x M$ we can imagine rolling the homogeneous space without slipping along the infinitesimal curve corresponding to the tangent vector. The Cartan form $\sigma^* \omega$ assigns to this infinitesimal curve $v$ an infinitesimal transformation $X:= (\sigma^* \omega)(v) \in \frak{g}$. Since $G$ acts on $G/H$, Then I would assume that the infinitesimal transformation $X$ tells us how the point of contact $o \in G/H$ changes as we roll along the infinitesimal curve $v$.

Thus, if the point of contact is initially $o \in G/H$, then our new point of contact with $v(d) \in M$ after rolling for infinitesimal time $d$ along the infinitesimal curve $v$ will be $X(d).o \in G/H$, where the infinitesimal action of $\frak{g}$ on $G/H$ here is induced by the standard left action of $G$ on $G/H$.

I am wondering whether this reasoning is a correct way to think about rolling without slipping the homogeneous space along $M$. If not, is there some other way I can think about it geometrically?

Note that while this statement in terms of infinitesimal transformations is a bit fuzzy and "handwavy" in classical differential geometry where we do not have infinitesimal objects, it is entirely rigorous in synthetic differential geometry, and we can embed the category of smooth manifolds fully and faithfully into whatever smooth topos we use to model SDG.


1 Answer 1


You don't need to be 'handwavy' at all, and, in the standard modern way to understand this, one does not need one to choose local sections and talk about infinitesimal motions. Here's the standard approach:

First, a Cartan connection of type $(G,H)$ on an $n$-manifold $M$ is, as you know, a principal right $H$- bundle $\pi:B\to M$ endowed with a $\frak{g}$-valued $1$-form $\omega:TB\to\frak{g}$ that gives a parallelism of $B$ (i.e., $\omega_b:T_bB\to\frak{g}$ is an isomorphism for each $b\in B$) subject to the following two conditions: (1) $\omega(Y_v) = v$ for all $v\in\frak{h}$, where $Y_v$ is the vector field on $B$ induced by the flow of right action by the $1$-parameter subgroup $\exp(tv)$ in $H$, and (2) $R_h^*(\omega) = \mathrm{Ad}(h^{-1})(\omega)$ for all $h\in H$. The curvature of $\omega$ is $\Omega = \mathrm{d}\omega + \tfrac12[\omega,\omega]$, and it is $\pi$-semibasic. The torsion of $\omega$ is the $\frak{g}/\frak{h}$-valued $2$-form $\Omega\,\text{mod}\,\frak{h}$.

As an example, $\gamma:TG\to\frak{g}$, the canonical left-invariant $1$-form on $G$, is a Cartan connection for the natural coset quotient $q:G\to G/H$. Of course, the curvature of $\gamma$ vanishes identically.

Now, about the rolling interpretation: Given a Cartan connection $(B,\pi,\omega)$ over $M^n$, consider the product $B\times G$ and the modified 'difference' $1$-form $$ \theta =\mathrm{Ad}(g)\bigl(\omega-\gamma\bigr). $$
This $1$-form is invariant under the diagonal right $H$-action on $B\times G$ defined by $(b,g){\cdot}h = (b{\cdot}h,gh)$, and it satisfies $R^*_{a}(\theta) = \mathrm{Ad}(a^{-1})(\theta)$ where $R_a:B\times G\to B\times G$ is the right $G$-action defined by $(b,g)\cdot a = (b,a^{-1}g)$ for $a\in G$. It also satisfies $\iota_{b,g}^*(\theta) = 0$, where $\iota_{b,g}:H\to B\times G$ is the mapping $\iota_{b,g}(h) = (b{\cdot}h, gh)$.

It follows that $\theta$ is the pullback of a well-defined $\frak{g}$-valued $1$-form $\bar\theta$ on the quotient $B\times_HG$ of $B\times G$ by the diagonal right $H$-action. In fact, $\bar\theta$ defines a connection in the usual sense on $B\times_H G$ when this quotient is regarded as a principal right $G$-bundle over $M$, and one has the curvature identity $\bar\Theta = \mathrm{Ad}(g)(\Omega)$. (Hence, even when the torsion of $\omega$ vanishes, the curvature of $\bar\theta$ will not, generally, take values in $\frak{h}$.)

Denote by $\pi:B\times_H G\to M$ the projection to $M$, and denote by $q: B\times_H G\to G/H$ the map defined by $q\bigl([b,g]\bigr) = gH\in G/H$. The pair $(\pi,q)$ defines a double fibration of $B\times_HG$ over the pair of bases $M$ and $G/H$. Any element $[b,g]\in B\times_HG$ defines a pair of points $(\pi(b),gH)$, but, more than this, the horizontal space $\ker(\bar\theta_{[b,g]})\subset T_{[b,g]}B\times_H G$ maps isomorphically onto $T_{\pi(b)}M$ and onto $T_{gH}G/H$, so it defines an isomorphism $\tau_{[b,g]}:T_{\pi(b)}M\to T_{gH}G/H$ that can be regarded as a $1$-jet of a pointed diffeomorphism between $(M,\pi(b))$ and $(G/H, gH)$, i.e., an 'infinitesimal identification' of the two spaces. Thus, $B\times_H G$ can be regarded as providing a family of $1$-jets of identifications of $M$ with the 'model space' $G/H$. (There can be, and, often, there is, more information in $(B\times_HG,\pi,q,\bar\theta)$ than this, but that's enough for the present purposes.)

Finally, the connection $\bar\theta$ on $B\times_H G$ defines a way uniquely to lift any differentiable curve $\alpha:[0,1]\to M$ to a $\bar\theta$-horizontal curve $\hat\alpha:[0,1]\to B\times_HG$ once one specifies $\hat\alpha(0)\in\pi^{-1}\bigl(\alpha(0)\bigr)$. In other words, $\hat\alpha$ provides a 'rolling without slipping' interpretation of the Cartan connection $\omega$. Note that if $\omega$ is flat, then this actually defines the developing map of the simply connected cover of $M$ to $G/H$, as you would expect.

Probably, I should also mention that the (ordinary) holonomy of the connection $\bar\theta$ is what Cartan called the holonomy of (what we now call) a Cartan connection.

  • 1
    $\begingroup$ Thanks for your very detailed answer. I'm still wondering if the infinitesimal description I gave is correct, though? I studied physics so the infinitesimal viewpoint appeals to me. $\endgroup$
    – ಠ_ಠ
    Jan 29, 2016 at 21:15
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    $\begingroup$ Great answer. "There can be, and, often, there is, more information [...]": could you also sketch which further information is always contained in that piece of data? $\endgroup$
    – Qfwfq
    Jan 29, 2016 at 21:20
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    $\begingroup$ @ ಠ_ಠ: Well, I can't say that it's wrong. Your description actually isn't detailed enough for me to decide whether it's 'correct' or not. When the homogeneous geometry $(G,H)$ is of second (or higher) order, the Cartan connection encodes this, but unless your language treats higher order infinitesimals than first order ones (what we call vector fields these days), your description doesn't seem to specify how to treat the second order infinitesimal transformations, as arise, for example, in things like conformal or CR geometry. $\endgroup$ Jan 29, 2016 at 22:04
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    $\begingroup$ @Qfwfq: A full answer to your question gets into technical details, but the main point is that the mapping I described of $B\times_H G$ into $J^1(M,G/H)$ is not usually an embedding (for example, it is not in the conformal and CR cases, and in all of the interesting 'second order' geometries). However, to properly lift this to a mapping of $B\times_HG$ into $J^2(M,G/H)$ (so that one has a well-defined family of second-order identifications of $M$ with the model space), one needs some additional assumption, such as that the torsion of $\omega$ should vanish. $\endgroup$ Jan 30, 2016 at 11:02
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    $\begingroup$ @Qfwfq: The ur-reference is É. Cartan, Les groupes d'holonomie des espaces généralisés (Acta. Math., 48 (1926), 1–42. In this paper, Cartan explains everything about what we now call Cartan connections in general and what we now call 'rolling without slipping'. This is where I learned what I wrote above, though I certainly admit that the modern language helped me figure out what Cartan was saying. I know that there are modern treatments, but I find Cartan's explanation the clearest and cleanest, and, as a bonus, he includes lots of examples and applications. $\endgroup$ Jan 31, 2016 at 22:59

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