Following Kobayashi and Nomizu, a connection on a manifold is given by a establishing a notion of horizontal vector in the tangent space of a frame bundle. (Alternative approaches make covariant differentiation foundational.)

An important step in developing Riemannian geometry consists of isolating the Levi-Civita connection as that connection with zero torsion that preserves the metric.

Could an alternative approach to defining the Levi-Civita connection go like this: Given a manifold $M$ with Riemannian metric, construct some natural (family of?) Riemannian metrics on the orthogonal frame bundle of $M$. Then simply define "horizontal" to mean orthogonal (in the sense of the constructed metric) to vertical?

Pedagogically, this might offer a bypass around defining and studying torsion.

About my "family of" hedge. There may be no canonical way to compare the scale of vertical vectors, essentially elements of the Lie algebra of the orthogonal group, with more general vectors.

If this is worked out anywhere, I'd appreciate a reference. If there's some obstruction to this approach, I'd appreciate an explanation.


2 Answers 2


This is not meant to be a real reply, it is more like an extended comment. I have doubts that this is a feasible approach. If you approach the problem naively, finding a metric on the tangent bundle to the orthonormal frame bundle is much more complicated than to just prescribing a complement to the vertical subbundle. If you impose some simple natural conditions on such a metric (e.g. that it coincides with the metric coming from the Killing form on the vertical subbundle, the principal right action is by isometries, and it descends to the given metric on $TM$), then finding such a metric is simply equivalent to finding an $O(n)$-equivariant complement to the vertical subbundle.

On the other hand, I don't believe that you can avoid defining torsion, since this is what makes the Levi-Civita connection natural. Indeed, the set of metric connections (or equivalently, the set of $O(n)$-equivariant complements to the vertical subbundle in the tangent bundle of the orthonormal frame bundle) is in bijective correspondence with the space of sections of $\Lambda^2T^*M\otimes TM$ via mapping a metric connection to its torsion. So it may be possible to hide the torsion, but I don't think you can avoid it.

An approach, in which the torsion does not show up very explicitly is Cartan's construction of the Levi-Civita connection as a family $\gamma^i_j$ of $1$-forms such that $\gamma^j_i=-\gamma^i_j$ and such that $d\theta^i+\sum_j\gamma^i_j\wedge\theta^j=0$ for all $i$, where $\theta^i$ are the components of the soldering form. (Of course, this expression just is the torsion viewed as a function on the orthonormal frame bundle, but there is no need to go into that interpretation.)

  • $\begingroup$ I don't follow. What's wrong with this? The notions of curve length and geodesic can be formulated without reference to the connection. So define an $\epsilon$-frame at a point $p$ as $n$ geodesic length $\epsilon$ curves $(C_i)$ leaving $p$ in mutually orthogonal directions. Write $c_i$ for the other endpoint of $C_i$. Then consider the naive distance between two $\epsilon$-frames as $$d(p,p')+(\sum_i |d(c_i,c'_i) - d(p,p')|)/\epsilon\ .$$ Then consider frames as limits of $\epsilon$-frames, and derive the Riemann from the global metric. $\endgroup$ Nov 22, 2017 at 9:28
  • $\begingroup$ You are looking for an inner product on each tangent space upstairs. Look at the ingredients available to do this: There is the vertical subspace, which is naturally identified with $\mathfrak{o}(n)$ and hence comes with a natural inner product. The quotient of an upstairs tangent space by the vertical subspace is the downstairs tangent space on $M$, so again this comes endowed with an inner product already. Using theses ingredients, the only way to cook up a natural inner product seems to be fixing the orthocomplement of the vertical, and this exactly mean choosing a metric connection. $\endgroup$ Nov 22, 2017 at 9:52
  • $\begingroup$ The inner product upstairs should come from a quadratic form upstairs, which in turn should come from localizing a general-topology-type metric. It is precisely the possibilities of viewing frames, approximately, as extended objects, and not merely as infinitesimal ones, that makes it possible to exploit the metric downstairs. As I've said, no natural scale unifies the quadratic form on the Lie algebra with the downstairs metric, so the upstairs metric won't be unique. But the formula in my previous comment suggests a connection-free definition. $\endgroup$ Nov 24, 2017 at 5:46
  • $\begingroup$ @DavidFeldman, Andreas is correct. The frame bundle does not have a canonical Riemannian metric on it. Choosing one that is invariant under the orthogonal group action is equivalent to choosing a connection. You have to choose which horizontal subspace should be orthogonal to the fiber. $\endgroup$
    – Deane
    Jun 19, 2022 at 17:36

"It is precisely the possibilities of viewing frames, approximately, as extended objects, and not merely as infinitesimal ones."

This is one of the interpretations of parallel transport due to a connection. The frame itself is pointwise (fibrewise) only. To already go to the infinitesimal neighbourhood means to choose a connection (see e.g. the papers of Cartan or Sharpe's book). By using the language of stalks of sheaves, the infinitesimal level already means something like "in an arbitrary small neighbourhood". In differential geometry, this is the meaning of "locally". If you want to go any further than that, this means to globalize in a certain way. By using connections, this extension from local to global is done by integrating out "infinitesimal" differential objects, physically spoken, along paths. The full theory is homotopy theory (or simpler, (co-)homology theory), which clarifies obstructions to extending something from local to global.

1st order differential geometry uses 1st order frames, so in a sense a Taylor approximation at linear level. Higher order differential geometry uses higher order frames, and in a sense a Taylor approximation at higher order level. You can try fancier constructions on higher order bundles and project them down to M, but in any case, you'll have something like a "higher order connection".

You seem to have some variational definition in mind, defining distances via integrals (that's even the case on a Riemannian Mf). This means, that you'll have some kind of Finsler, Weyl, Cartan, or Lichnerowicz geometry - and even there, connections appear (Weyl connection, Cartan connection etc.).

Basically, Andreas Cap has alread told you the minimal given data available, so anything else is an extension of that (e.g. o(n) with inner product, vertical bundle). Finally, you have to reduce to the o(n)-situation to obtain the Levy-Civita connection, and this means to globally reduce a principal bundle to an o(n) subbundle. As written above, you could start with more general and higher order geometries, eg. affine with agl(n) Lie algebra, or projective with pgl(n+1) Lie algebra, or even fancier structures. Somewhere, o(n) or so(n) must occur as a substructure, and its "complement" (be it a quotient or a kernel or an equivariant complementary submodule) provides a notion of torsion. That's what that Cartan point of view tells you for general principal bundles, based on subgroups H -> G and their homogeneous space G/H. The torsion part of the cartan connection belongs to the g/h-component within a 0 -> h -> g -> g/h -> 0 structure of Lie algebras. See the meaning of a "Cartan connection" and its relation to the Levi-Civita connection.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .