11
$\begingroup$

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.

========

I do see that this is related to A geometric interpretation of the Levi-Civita connection? and Intuition for Levi-Civita connection? .

I have also asked this on Mathematics Stack Exchange without feeling satisfied by the responses there, as you can see from the comments: https://math.stackexchange.com/questions/2529479/levi-civita-connections-from-metrics-on-the-orthogonal-frame-bundle/2530478?noredirect=1#comment5228620_2530478

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Perhaps something like this would work: 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. Haven't worked out the details of this final step. $\endgroup$
    – David Feldman
    Dec 4, 2017 at 2:38
  • $\begingroup$ The catch here is defining the curves $C_i$, which lie in the principal bundle. You are effectively choosing $n$ vectors (the initial velocity of each curve $C_i$) that project down to an orthonormal basis of $T_pM$. However, you say nothing about the vertical components of these vectors, and specifying that is tantamount to specifying a connection. $\endgroup$
    – Deane Yang
    Dec 4, 2017 at 17:53
  • $\begingroup$ Thanks Deane, but no, I mean the curves, geodesic, to lie in $M$. $\endgroup$
    – David Feldman
    Dec 5, 2017 at 1:21
  • $\begingroup$ But if you choose $n$ orthogonal directions and construct $n$ geodesic starting in those directions, you need the Levi-Civita connection to write down the ODE needed to solve for each geodesic. $\endgroup$
    – Deane Yang
    Dec 5, 2017 at 3:02
  • $\begingroup$ But if you pick $n$ nearby points, you don't need the connection to define the geodesics leaving $p$. Now the geodesics might not have length $\epsilon$ and they might not have orthogonal tangent vectors at $p$. But one can restrict to the set where in fact they do and then define my metric on it. And then take the limit. So at least for conceptual understanding, one gets a space and a metric and finally a connection in a fairly direct way. I accept for the sake of argument that the approach may not be suited for calculations. $\endgroup$
    – David Feldman
    Dec 5, 2017 at 3:34

1 Answer 1

Reset to default
3
$\begingroup$

Sure. Take the structure equations of Cartan, with soldering forms $\omega^i$ and connection forms $\omega^i_j$, and then use them as the orthonormal basis of 1-forms for a metric: $ds^2=\sum (\omega^i)^2 + \sum (\omega^i_j)^2$. This metric appears in some work in physics, I think. The same idea works even for a Lorentzian metric down on the manifold: you get a Riemannian metric on the frame bundle.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Marathe (2010, p. 131) calls this the bundle metric, attributed to himself (1972). But it assumes you have the connection, and won’t single out the torsion-free one, right? (Writing $\omega$ and $\theta$ for the connection and soldering forms, he says that the connection is torsion-free iff the torsion 2-form $d^\omega\theta=0$.) $\endgroup$
    – Francois Ziegler
    Dec 4, 2017 at 7:39
  • $\begingroup$ @FrancoisZiegler: yes, I assume that you have the connection forms defined, and this construction applies to any connection on the frame bundle, not necessarily arising as a Levi-Civita connection, and not necessarily compatible with any pseudo-Riemannian metric.. So my answer only points out that it is possible to define such a metric on the frame bundle, but doesn't discover a simpler way to arrive at it than to go through the standard approach. Thanks for the reference. $\endgroup$
    – Ben McKay
    Dec 4, 2017 at 8:52
  • $\begingroup$ @FrancoisZiegler Is the bundle metric the same metric I describe in my comment (without reference to a connection)? $\endgroup$
    – David Feldman
    Dec 4, 2017 at 17:30
  • $\begingroup$ As Ben says, if you start with a connection, it defines a horizontal subbundle, with the right behavior under the group action, of the tangent bundle of the principal bundle. This, along with the bi-invariant metric on $O(n)$, defines a natural Riemannian metric, where the horizontal and vertical subspaces are orthogonal. What's not clear is going in the other direction. To me, it looks like the torsion-free condition is some kind of integrability condition for the horizontal subspace. But that is an extra assumption, so it's just another way to formulate the torsion-free condition. $\endgroup$
    – Deane Yang
    Dec 4, 2017 at 17:58

You must log in to answer this question.

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