6
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – David Feldman Nov 22 '17 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. $\endgroup$ – Andreas Cap Nov 22 '17 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. $\endgroup$ – David Feldman Nov 24 '17 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.