5
$\begingroup$

I read somewhere that minimum energy paths are geodesics under the Levi-Civita connection, on a Riemannian energy landscape.

The Levi-Civita connection is the "unique connection on the tangent bundle of a manifold that preserves the Riemannian metric and is torsion-free".

What exactly is the intuition of geodesics under the Levi-Civita connection? Why would torsion-free connections give geodesics with least energy?

$\endgroup$
5
  • 1
    $\begingroup$ What exactly are you calling a torsion-free geodesic? In general, the word "torsion-free" refers to the connexion, not the geodesic. $\endgroup$
    – Didier
    Mar 30, 2021 at 13:27
  • $\begingroup$ @Didier right, edited. $\endgroup$
    – 900edges
    Mar 30, 2021 at 13:27
  • 1
    $\begingroup$ Some clarifications: The energy of a curve (path) is defined using the Riemannian metric (and not the Levi-Civita connection). When you compute the variational formula for energy, the Levi-Civita connection appears naturally. The fact that the Levi-Civita connection preserves the metric is used in this. The critical points of energy are constant speed geodesics. But constant speed geodesics are not necessarily energy minimizing. They are locally energy minimizing (any sufficiently short piece of the curve is energy minimizing)., $\endgroup$
    – Deane
    Mar 30, 2021 at 15:44
  • $\begingroup$ @Deane interesting, so does this mean the Levi-Civita connection can be derived by minimizing the energy of a curve with the constraint that the speed is constant? Or that this is a equivalent definition of the L-C connection? $\endgroup$
    – 900edges
    Mar 30, 2021 at 15:52
  • $\begingroup$ I don't think so. The variational equation uses only the Levi-Civita connection for the covariant derivative of the velocity vector of a curve with respect to itself. This, I believe, is not enough to define the connection for two arbitrary vector fields. $\endgroup$
    – Deane
    Mar 30, 2021 at 16:21

1 Answer 1

6
$\begingroup$

If $\nabla$ is any connection and $f$ a function, its Hessian with respect to $\nabla$ is $\mathrm{Hess}^{\nabla}f = \nabla \mathrm{d}f$, and one can see, after a messy calculation, that: $$ \mathrm{Hess}^{\nabla}f(X,Y) - \mathrm{Hess}^{\nabla}f(Y,X) = \pm\mathrm{d}f\left([X,Y] - (\nabla_XY - \nabla_YX) \right) $$ (where the $\pm$ sign is here because I don't remember the exact sign, but the computations are not that hard, just messy.) Hence, Hessians are symmetric if and only if the connection is torsion-free. This is the main motivation to consider torsion-free connections: in the euclidean space, Hessians are symmetric!

Moreover, the fundamental theorem of Riemannian geometry tells us that on a Riemannian manifold, there is a unique connexion that is torsion-free and lets the metric invariant, that is: $$ \forall X,Y,Z, \left(\nabla_Zg\right)(X,Y) = Z\cdot g\left(X,Y \right) - g\left(\nabla_ZX,Y\right) - g\left(X,\nabla_ZY\right) = 0. $$ (compare with the euclidean case, where $\langle X,Y\rangle ' = \langle X',Y\rangle + \langle X, Y' \rangle$.) This theorem thus says that given any Riemannian metric $g$, there is a connection that is better than others: Hessians are symmetric and the metric is invariant under the action. We call it the Levi-Civita connexion.

If a connection is chosen, a geodesic is a parametrized curve satisfying the equation of geodesics : $\nabla_{\gamma'}\gamma' = 0$. Thus a curve $\gamma$ is a geodesic with respect to the connection, and can be a geodesic for some connection $\nabla^1$ but not for another connecion $\nabla^2$. Therefore, your question does not really have sense: we do not say that a connexion gives the least energy of a geodesic. I think you got confused, believing that being a geodesic is an intrinsic notion, but it really depends on the connection you consider.

Now, suppose $(M,g)$ is a Riemannian manifold endowed with its Levi-Civita connexion. Then if $\gamma : [a,b] \to M$ is a curve, we define its energy to be: $$ E(\gamma) = \frac{1}{2}\int_a^b \|\gamma'\|^2 $$ and one can show that, in the space of all curves $\{\gamma : [a,b] \to M\}$ with same end points, a curve $\gamma$ is a point where the energy functional is extremal if and only if $\nabla_{\gamma'}\gamma'=0$, that is if and only if $\gamma$ is a solution of the equation of geodesics. Hence, a minimizer of the energy functional is a geodesic.

$\endgroup$
11
  • 1
    $\begingroup$ Quibble: $\nabla_{\gamma'}\gamma' = 0$ implies the parameterized curve is not just a geodesic but a constant speed geodesic. $\endgroup$
    – Deane
    Mar 30, 2021 at 15:45
  • $\begingroup$ Thanks! So your definition of a geodesic as an energy-minimizer gives different geodesics (and different minimum energies) based on which $\nabla$ we use. I'm still unsure about why the Levi-Civita connection in particular provides a smaller minimum energy compared to other choices of connection. $\endgroup$
    – 900edges
    Mar 30, 2021 at 16:09
  • $\begingroup$ The Levi-Civita does not provide a minimum energy compared to other connection: this, I think is another question that is unrelated. Also, I did not define geodesics as energy minimizing curves. I defined the as solution to the equation of geodesics. $\endgroup$
    – Didier
    Mar 30, 2021 at 17:24
  • $\begingroup$ @Didier well, geodesics by definition minimize energy under the given connection. My interpretation of the statement "minimum energy paths are geodesics under the Levi-Civita connection" has the emphasis on "under the L-C connetion" rather than "are geodesics". So I was wondering why (or whether) the energy-minimizer over geodesics is the one under the L-C connection. $\endgroup$
    – 900edges
    Mar 30, 2021 at 17:29
  • $\begingroup$ +1 by the way for the explanation about geodesics being energy-minimizing paths. $\endgroup$
    – 900edges
    Mar 30, 2021 at 17:33

You must log in to answer this question.

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