A geometric interpretation of the Levi-Civita connection?

Let $M$ be a Riemannian manifold. There exists a unique torsion-free connection in the (co)tangent bundle of $M$ such that the metric of $M$ is covariantly constant. This connection is called the Levi-Civita connection and its existence and uniqueness are usually proven by a direct calculation in coordinates. See e.g. Milnor, Morse theory, chapter 2, \S 8. This is short and easy but not very illuminating.

According to C. Ehresmann, a connection in a fiber bundle $p:E\to B$ (where $E$ and $B$ are smooth manifolds and $p$ is a smooth fibration) is just a complementary subbundle of the vertical bundle $\ker dp$ in $T^*E$. If $G$ is the structure group of the bundle and $P\to B$ is the corresponding $G$-principal bundle, then to give a connection whose holonomy takes values in $G$ is the same as to give a $G$-equivariant connection on $P$.

If $p:E\to B$ is a rank $r$ vector bundle with a metric, then one can assume that the structure group is $O(r)$; the corresponding principal bundle $P\to B$ will in fact be the bundle of all orthogonal $r$-frames in $E$. One can then construct an $O(r)$-equivariant connection by taking any metric on $P$, averaging so as to get an $O(r)$-equivariant metric and then taking the orthogonal complement of the vertical bundle.

Notice that in general one can have several $O(r)$-equivariant connections: take $P$ to be the total space constant $U(1)$-bundle on the circle; $P$ is a 2-torus and every rational foliation of $P$ that is non-constant in the "circle" direction gives a $U(1)$-equivariant connection. (All these connections are gauge equivalent but different.)

So I would like to ask: given a Riemannian manifold $M$, is there a way to interpret the Levi-Civita connection as a subbundle of the frame bundle of the tangent bundle of $M$ so that its existence and uniqueness become clear without any calculations in coordinates?

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If you want a manifestly invariant construction of the connection, can't you use a variational formulation in terms of infinitesimal geodesics? – Per Vognsen Aug 1 2010 at 8:32
No, a connection making the metric tensor parallel is not unique; you also need symmetry: $\nabla_X Y-\nabla_Y X=[X,Y]$. And this symmetry refers to a special structure that exists only in tangent bundles. So I doubt that an extra abstraction can make things simple - you'll need to go back to that tangent structure at some point. On the other hand, it is easy to prove the thing in an invariant language (and many textbooks do so), using Lie bracket manipulations rather than coordinates. – Sergei Ivanov Aug 1 2010 at 10:22
One invariant construction is here: books.google.com/… – Steve Huntsman Aug 1 2010 at 13:18
To add on Sergei's comment, that symmetry property is usually called "torsion-free" property of Levi-Civita connection. The proof that uses Lie bracket manipulations is usually through the Koszul formula, it is in, for example, Barrett O'Neill's Semi-Riemannian Geometry on page 61. – Willie Wong Aug 1 2010 at 13:24
Steve, the construction you refer to appears to be using local co-ordinates and Christoffel symbols and not "invariant" (at least by my definition of the word) – Deane Yang Aug 1 2010 at 14:07

To understand the existence and uniqueness of the LC connection, it is not possible to sidestep some algebra, namely the fact (with a 1-line proof) that a tensor $a_{ijk}$ symmetric in $i,j$ and skew in $j,k$ is necessarily zero. The geometrical interpretation is this: once one has the $O(n)$ subbundle $P$ of the frame bundle $F$ defined by the metric, there exists (at each point) a unique subspace transverse to the fibre that is tangent both to $P$ and to a coordinate-induced section $\{\partial/\partial x_1,\ldots,\partial/\partial x_n\}$ of $F$.

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 Simon -- thanks. This sounds interesting, but could you please clarify a couple of points: namely which ordinates do you use to induce a "coordinate-induced" section? Different choices of coordinates at a point lead to very different sections passing through given orthonormal frame. – – algori Aug 1 2010 at 21:48 Yes, but any such section $s$ that passes through $p\in P$ is unique to first order. If we set $a_{ijk} = \Gamma_{ij}^r g_{rk}$ then $s$ is tangent to $P$ at $p$ iff $a_{ijk}+a_{ikj}=0$, which forces the Christoffel symbols to vanish at the point in question. – Simon Salamon Aug 2 2010 at 9:11 I like this answer very much. I can see roughly how it connects to the proofs I know for the uniqueness of the Levi-Civita connection, but haven't figured out the precise details. I encourage students, however, to figure it out. It looks like a great exercise in tearing down the formalism and building it back up. – Deane Yang Aug 11 2010 at 3:24

The Levi-Civita connection is locally described by the Christoffel symbols. How does one obtain these in a natural fashion: write the Euler-Lagrange for the length functional. The extremals of this functional are the geodesics and once you write the Euler-Lagrange equations you obtain the Christoffel symbols. For details see Example 5.1.8 from my book.

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Is the following description correct?

The metric determines the geodesics: pull a string tight enough and it will be a geodesic.

These in turn determine a class of connections, determined up to torsion: twist the string while parallel transporting a tangent vector along it, and you are changing the connection keeping the same geodesics.

Now choose a connection, parallel transport an infinitesimal vector along a geodesic curve $\gamma$. The tip of the vector will draw a curve $\gamma '$. The zero torsion connection in the class, i.e. the Levi-Civita connection, is the one minimizing the lenght of $\gamma '$.

BTW, this question is related: http://mathoverflow.net/questions/20493

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anon -- thanks! The discussion in the thread you gave is very interesting. – algori Aug 1 2010 at 23:51