First, let me explain what I mean by "synthetic" in the title, which is a proof that reasons purely axiomatically and does not explicitly invoke local coordinate charts (either via concrete expansions or Penrose-style abstract tensor notation). For example in Euclidean geometry, one can either prove statements using Euclid's postulates or via vector arithmetic/Cartesian geometry. Euclidean proofs tend to be shorter and more elegant, but the arithmetical proofs tend to be easier to discover.

To better illustrate in the context of geodesic flow, this I shall now sketch out a basic non-synthetic (arithmetical) derivation for the geodesic equation. Given a compact, path-connected Riemannian manifold (M, g) and a pair of designated points A, B, find a path $\gamma : [0, 1] \to M$ to

Minimize: $\int \limits_0^1 g_{\gamma(t)}( \dot{\gamma}(t), \dot{\gamma}(t) ) dt$

Subject to: $\gamma(0) = A, \gamma(1) = B$

Now we construct a variation, $\gamma' : (-\epsilon, \epsilon) \times [0,1] \to M$ such that for all s, t, $\gamma'(0, t) = \gamma(t)$, $\gamma(s,0) = A$ and $\gamma(s, 1) = B$. So, a path $\gamma$ is an extremal if for all variations $\gamma'$,

$\left. \frac{d}{d s} \right|_{s = 0} \int \limits_0^1 g_{\gamma(s, t)}( \frac{d}{d t}\gamma(s, t), \frac{d}{d t}\gamma(s, t) ) dt = 0$

So far so good. Now at this point, the usual strategy is to fix a subinterval $(a, b) \subset [0,1]$ such that for all t in (a,b), $\gamma (t)$ is contained within a single coordinate chart, then expand the functional in terms of this basis and apply the Euler-Lagrange equations. If you do this, then lo-and-behold differentiating the metric gives you a second order term and the Christoffel symbols, which corresponds to the Levi-Civita connection. To extend this to a global result requires showing that the Euler-Lagrange equations agree across a boundary and that the metric is locally convex.

Now my question is can we do this last part synthetically? In other words, just using the basic properties of connections can we prove that this has to be an extremum of the length functional? I really would hope so, otherwise why go through all the trouble of defining connections axiomatically, just pick coordinates and Christoffel symbols and be done with it!

  • 3
    $\begingroup$ I think you're asking for a co-ordinate free approach, right? Check section 1.2 of Cheeger and Ebin's book "Comparison theorems in Riemannian geometry". You can view it on books.google.com $\endgroup$
    – Deane Yang
    Jul 15, 2010 at 19:00
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    $\begingroup$ @Mikola: You seem to be saying "why go to the trouble of developing a coordinate-free viewpoint if you will still sometimes need the other viewpoint?" I am sure that most geometers have good reasons for valuing both viewpoints. (And, yes, the second variation formula can be proved in a coordinate-free way.) $\endgroup$ Jul 15, 2010 at 19:48
  • $\begingroup$ First variation formula, I guess I meant. Anyway, it's often instructive to work out a coordinate-free proof if you already have the other kind, and I suppose it can go the other way, too. $\endgroup$ Jul 15, 2010 at 21:32
  • $\begingroup$ @Tom: I am not saying that it is bad to have two view points; rather I am just trying to make sense of the coordinate free version. I figure if I can't work this out in a coordinate free way, then I probably don't understand that approach very well. I guess I don't understand what the disagreement is? $\endgroup$
    – Mikola
    Jul 15, 2010 at 22:43
  • $\begingroup$ @Deane: Thanks! I will take a look at that book later this evening. $\endgroup$
    – Mikola
    Jul 15, 2010 at 22:44

1 Answer 1


If $(s,t) \mapsto \Gamma(s,t)$ is a family of curves in Riemannian manifold $M$, where $s \in [0,1]$ is the curve parameter and $t \in (-\delta,\delta)$ is the variation parameter, let $S = \partial_s\Gamma, T = \partial_t\Gamma \in T_{\Gamma(s,t)}M$. Note that $[S,T] = 0$. Then the derivative of the energy functional can be computed as follows:

If we assume that the two endpoints are fixed, then $T(0) = T(1) = 0$. Therefore,

$$\frac{d}{dt}\int_0^1 |S|^2\,ds = \int_0^1 \frac{d}{dt}(S\cdot S)\,ds$$ $$= 2\int_0^1 S\cdot\nabla_TS\,ds$$ $$= 2\int_0^1 S\cdot\nabla_ST\,ds$$ $$= 2\int_0^1 \partial_s(S,T) - T\cdot\nabla_S S\,ds$$ $$= -2\int_0^1 T\cdot\nabla_S S\,ds.$$

The second variation formula can be derived using a similar calculation.

This calculation, however, makes no sense on the manifold itself, since the two vector "fields" $S$ and $T$ are defined only on the image of $\Gamma$, which is a possibly degenerate surface. What you need to do is to pull back the tangent bundle of $M$ to a rank $n$ vector bundle over $[0,1]\times(-\delta,\delta)$ and pull back the Riemannian metric and Levi-Civita connection back to an inner product and connection on the vector bundle and view the calculation above as being on that vector bundle. The easiest way to check that this works is, well, write everything with respect to local co-ordinates. Still, I do prefer this approach to doing calculations. Christoffel symbols are to be avoided as much as possible.


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