Let $M$ a differentiable manifold and $H\subset G$ a Lie group with a closed subgroup such that $G/H$ is connected. A $H\subset G$-Cartan connection on $M$ can be defined by

  • A principal $G$-bundle on $M$, we call it $Q$
  • A $G$-principal connection on $Q$, we call its connection $1$-form $\omega$ and its curvature $2$-form $\Omega$
  • A reduction of the structure group of $Q$ to $H$, under the form of a subbundle $P$ which is $H$-principal and on which $\omega$ in non-degenerate

I am trying to get an idea for the torsion of a Cartan connection associated to a $H\hookrightarrow G$-geometry. There are great answers about the torsion of linear connections and G-structures in What is torsion in differential geometry intuitively? in particular Mathieu ANEL's answer gives an interpretation as the translation component of the "Cartan curvature" form.

I am looking for a "geometrical" interpretation of the vanishing of torsion. Vanishing of the total Cartan curvature is equivalent to the Cartan geometry being "locally flat", that is, locally isomorphic to the reference homogeneous space [1,2]. That said I see on the ncatlab that vanishing of the torsion should be enough of flatness to have local trivialisations (it seems too strong but I may be missing a subtlety).

I tried to see what vanishing torsion implied on parallel transport. As written by Robert Bryant if $\mathfrak h$ is not stable under $\operatorname{Ad}(G)$ then vanishing torsion does not mean that $\Omega$ has value in $\mathfrak h$ on all of the principal $G$-bundle $Q$. That said, $Q$ comes with a reduction to $H$ which defines a section $\sigma$ of the fibre bundle $Q/H$.

Vanishing torsion means that $\Omega$ takes value in $\mathfrak h$ on $P$. It follows that for any $x$ in $M$, if we identify the fibre $Q_x/H$ with $G/H$ using $\sigma(x)$ as origin ($[eH]\in G/H$; there is $H$ worth of ambiguity) the representation of the curvature on $Q/H$ (a vertical vector field-valued $2$-form) takes value in $\mathfrak h$ at the origin, which means that it vanishes on $\sigma(x)$.

Are there any more geometrical conclusions to be drawn? I first mistakenly hoped that restricted holonomy would have to preserve $\sigma$ but according to the Ambrose-Singer theorem the infinitesimal holonomy at $p$ is generated by the values of the curvature at all points that can be connected to $p$ by a parallel path so that vanishing torsion does not seem to be sufficient.

[1] : Sharpe, R.W, Differential Geometry. Cartan's Generalization of Klein's Erlangen Program, Graduate Texts in Mathematics

[2] : Čap, Andreas; Slovák, Jan, Parabolic geometries I. Background and general theory, Mathematical Surveys and Monographs


1 Answer 1


In my Introduction to Cartan geometries, p.30, I define a near parallelogram, a 5-sided figure, of constant vector fields flows, say of two elements $A,B$ in the Lie algebra of the model, to be a figure given by flowing along the flow associated to $A$, then that of $B$, then $-A$, then $-B$, and then $-[A,B]+\dots$ where the $\dots$ are Baker-Campbell-Hausdorff terms needed to make the figure close up in the model.

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I show that the lowest order term of the error in closing up is the curvature, in the total space of the Cartan geometry. This just follows, once we have the structure equations, from the usual computations of Lie brackets using the Cartan formula. In section 13, p. 34, I show that the torsion is the lowest order term of the error in closing up of the projection to the underlying manifold.

In the special case where $[A,B]=0$, for example in the horizontal directions of a reductive homogeneous space, the picture reduces to the usual picture of a parallelogram in the model, nearly closing up in any geometry with that model.

Andreas Cap recently pointed out to me that nobody ever seems to make much serious use of these intuitive pictures of curvature or torsion in differential geometry. (I hope I am not misrepresenting what he was saying.) Perhaps the intuition from these pictures is overrated.

  • $\begingroup$ I don't have any strong intuition as to why the parallelogram should close (of course it does in the Euclidean space) and what means its failure to do so... In the end, we are using the picture for the Lie bracket of vector fields, right? Thus I think that the content of the picture is related to the interpretation of torsion as obstruction to the existence of coordinates such that the connection symbols vanish at a given point (if $(\nabla X_i)|_p = 0$ then $[X_i, X_j]|_p = -T(X_i, X_j)|_p$). $\endgroup$
    – jpdm
    Commented Nov 20, 2023 at 15:07
  • $\begingroup$ Correct: this is about brackets of vector fields. You can map the total space of any Cartan geometry to the affine connection bundle of the underlying manifold (i.e. the bundle whose sections are affine connections), in an obvious construction, and the torsion of each of these connections is just the torsion of the Cartan geometry. $\endgroup$
    – Ben McKay
    Commented Nov 21, 2023 at 10:39

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