MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '10 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 '10 at 10:22
One invariant construction is here: – Steve Huntsman Aug 1 '10 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 '10 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 '10 at 14:07
up vote 14 down vote accepted

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 $F$.

share|cite|improve this answer
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 '10 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 '10 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 '10 at 3:24

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: What is torsion in differential geometry intuitively?

share|cite|improve this answer
anon -- thanks! The discussion in the thread you gave is very interesting. – algori Aug 1 '10 at 23:51

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.

share|cite|improve this answer

I think that a good way to understand the Levi-Civita connection is to say that is is the Ehresmann connection in $TTM$ obtained from the linearization of the geodesic flow by a natural geometric construction.

I described this construction in my answer to this MO question, but I'll do so again with some improvements.

Dynamic construction.

Let $c(t)$ be an orbit of the geodesic flow in $TM$, consider the vertical subspaces $V(t)$ in $TTM$ along $c(t)$ and bring them back to the tangent space of the cotangent bundle over the point $c(0)$ by using the differential of the flow. You get a family of (Lagrangian) subspaces $l(t) := D\phi_{-t}(V(t))$ in the symplectic vector space $T_{c(0)}TM$.

Now forget you ever had a geodesic flow: all that you need is the curve of subspaces. A bit of differential projective geometry---described below---shows that you also get a second curve $h(t)$ of (Lagrangian subspaces) in $T_{c(0)}(T^*M)$ that is transversal to $l(t)$. The subspace $h(0)$ is the horizontal subspace of the connection and $T_{c(0)}(T^*M) = l(0) \oplus h(0)$ is the decomposition into vertical and horizontal subspaces.

Projective construction.

Now I'll describe as succintly as possible the projective-geometric construction that underlies both the Levi-Civita connection and the Schwartzian derivative. For the detais of what follows see this paper What's new in the description here is that I explicitly use the Springer resolution (Duran and I used implicitly in the paper).

First we need two remarks on the geometry of the Grassmannian $G_n(\mathbb{R}^{2n})$ of $n$-dimensional subspaces in $\mathbb{R}^{2n}$

1. The tangent space of $G_n(\mathbb{R}^{2n})$ at a subspace $\ell$ is canonically identified with the space of linear maps from $\ell$ to $\mathbb{R}^{2n}/\ell$ or, equivalently, with the space $(\mathbb{R}^{2n}/\ell) \otimes \ell^*$. Since $\mathbb{R}^{2n}/\ell$ and $\ell$ have the same dimension, we may distinguish a class of differentiable curves $\gamma$ on the Grassmannian by requiring that at each instant $t$ their velocities are invertible linear maps from $\gamma(t)$ to $\mathbb{R}^{2n}/\gamma(t)$. These curves are called fanning or regular.

Using that the cotangent space of $G_n(\mathbb{R}^{2n})$ at a subspace $\ell$ is canonically isomorphic to $\ell \otimes (\mathbb{R}^{2n}/\ell)^*$, we can lift every fanning curve $\gamma(t)$ to a curve on the cotangent bundle of the Grassmannian by $t \mapsto (\dot{\gamma}(t))^{-1}$.

2. Consider the action of the linear group $GL(2n;\mathbb{R})$ on the Grassmannian $G_n(\mathbb{R}^{2n})$ and lift it to an action on its cotangent bundle. The moment map of this action takes values on the set of nilpotent matrices.

Now consider a fanning curve $\gamma(t)$ on the Grassmannian $G_n(\mathbb{R}^{2n})$ and lift it to the curve $(\dot{\gamma}(t))^{-1}$ on its cotangent bundle. Use the moment map to obtain a curve $F(t)$ of nilpotent matrices. Note that everything we have done is $GL(2n,\mathbb{R})$-equivariant.

Finally we come to the little miracle: the time derivative of $F(t)$ is a curve of reflections $\dot{F}(t)$ (i.e., $\dot{F}(t)^2 = I$) whose -1 eigenspace is the curve of subspaces $\gamma(t)$ and whose $1$-eigenspace defines a "horizontal curve" $h(t)$ equivariantly attached to $\gamma(t)$. This is the construction that yields the Levi-Civita connection (and what is behind the formalisms of Grifone and Foulon for connections of second order ODE's on manifolds).

Differentiate $F(t)$ a second time to find the Schwartzian derivative. Geometrically, it just describes how the curve $h(t)$ moves with respect to $\gamma(t)$. For comparison, recall that the curvature of a connection is obtained by differentiating (i.e., bracketing) horizonal vector fields and projecting onto the vertical bundle.

share|cite|improve this answer

This avoids use of Christoffel symbols and index argument by re-characterizing the torsion free property. The following is one definition for a torsion-free connection. Let $\tau:TM \to M$ be the tangent bundle projection. A torsion free connection is a splitting of $T(TM) = H(TM) \oplus V(TM)$ where $V(TM) = {\rm kernel}(T \tau) \equiv TM \oplus TM$ (see Kolar,Michor,Slovak (1993)). The covariant derivative associated to this splitting is given by $\frac{Dv}{Dt} = {\rm pr_2} \left( {\rm ver}( \frac{dv}{dt} ) \right)$ where ${\rm ver}:TM \to VM$ is the projection onto $VE \equiv TM \oplus TM$ and $\rm pr_2$ is the projection onto the second factor of $TM \oplus TM$. The torsion-free property manifests from the fact that given any surface embedding $m: (s,t) \mapsto m(s,t) \in M$ we observe $\frac{D}{Dt} ( \partial_s m) = \frac{D}{Ds}( \partial_t m)$ by the equality of mixed partials. Consider the map $h^{\uparrow}: TM \to T^{(2)}M$ given by $h^{\uparrow}(v_0) = \left.\frac{d}{dt} \right|_{t=0}(v(t))$ where $v(t)$ is the geodesic with initial condition $v_0 \in TM$. Then (I think) the Levi-Cevita is given by $H(TM) = h^{\uparrow}(TM)$. Clearly, this is a horizontal space, and thus a well-defined connection.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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