Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm wondering what else one can do with Koszul's formula

$$2\langle\nabla_XA,B\rangle = X\langle A,B\rangle-B\langle X,A\rangle + A\langle X,B\rangle - \langle A,[X,B]\rangle + \langle[B,X],A\rangle - \langle B,[A,X]\rangle$$

beyond proving existence and uniqueness of the Levi-Civita connection. I haven't yet seen anybody using it for anything else, which would be quite curious.

Here's a pretty and simple example. I don't know if it is known...

Let $\nabla^{LC}$ be the Levi-Civita connection and $\nabla$ be some other metric connection with torsion $T$. Then

$$2\langle\nabla_XA,B\rangle - 2\langle\nabla_X^{LC}A,B\rangle= \langle T(A,X),B\rangle -\langle T(A,B),X\rangle - \langle T(B,X),A\rangle$$

An application of this would be to compute the LC-connection for the metric $\langle K\cdot,K\cdot\rangle$ in terms of the endomorphism $K$ and the LC-connection for $\langle \cdot,\cdot\rangle$. The computation starts with the connection $K^{-1}\nabla^{LC} K$, which is metric for $\langle K\cdot,K\cdot\rangle$...

I can't guarantee correct letters and signs :-)

share|improve this question
Riemannian geometry IS the study of the Levi-Civita connection and the Koszul formula is an explicit expression for the connection. So ultimately anything you do in Riemannian geometry is bound to be using it at some level. If, on the other hand, you mean what other formulae you can get by formally manipulating the Koszul formula, then there are plenty, but I would not necessarily call them "applications". Killing's equation comes to mind, as well as the O'Neill formulae for submersions, the basic formulae for homogeneous geometry,... Pick any book on differential geometry: e.g., do Carmo. –  José Figueroa-O'Farrill Mar 7 '11 at 11:18
Indeed. Except when the glory breaks down and coordinates get introduced (or used implicitly). (If not, wait till play with Laplacians begins and frames introduced...) BTW, one application of my application would be the coordinate representation of the LC connection. PLEASE NOTE: I'm far away from the library and made it there only a few times this century - and it was open a few of these occasions. But I've seen and possibly remember quite a few of these books, e.g. none on Ricci flow without ungeometricly resorting to coordinate stuff when things get most basic –  Martin Gisser Mar 7 '11 at 15:09
So let me get this straight: you make a distinction between the Koszul formula and the formula for the Christoffel symbols? I don't think there is any conceptual difference between the two. I suppose it's a question of how much pain are you willing to endure in order to keep everything coordinate-free. Analysts as a general rule seem to have a low pain threshold :) –  José Figueroa-O'Farrill Mar 7 '11 at 17:40
Hm, yes I'm actually trying to avoid pain. I might perhaps have an infinitude of indices (symbolic or not) or try to keep track of the tensor product rule (after all, it's tensor calculus) and not get lost (or waste sparse IQ) in the debauch of indices. Why else Koszul connections? Of course Christoffel stuff and Koszul covariant derivatives (plus Cartan's calculus, Penrose spaghetti, etc. pp.) are all about the same thing, but at least there should be some motivation to stay in one concept (plus, avoid coordinates (it could be a ringed space) (or the coordinates being unknown or irrelevant)). –  Martin Gisser Mar 7 '11 at 18:05
Quite a borderline comment but I find it funny. There is (at least) one thing I don't know how to prove without coordinates. Whena metric evolves by the Ricci flow, the evolution of the curvature operator is given by : $\partial_t R=\Delta R + R^2+R^# $, where $R^#$ can be defined in a coordinate free way using the Lie algebra structure of $\Lambda^2 TM $. A substantial part of the proof can be done without coordinates (as in Toppong lectures) but I never saw anyone that goes from the expression ypu found in Topping book to the Lie algebraic expression without coordinates. –  Thomas Richard Mar 10 '11 at 7:51

2 Answers 2

Koszul's formula simply expresses the Levi-Civita connection explicitly in terms of the Riemannian metric. It is quite useful any time you want to eliminate the connection from a formula and write the formula in terms of the metric only. José cited some nice examples. I haven't checked, but I bet the book by Cheeger and Ebin discusses these examples and maybe more quite explicitly.

But it is no different from writing a formula in local co-ordinates and replacing all appearances of Christoffel symbols by their formula in terms of partial derivatives of the Riemannian metric. This is often an equally useful thing to do when, for example, you want to apply PDE techniques or theorems that are stated in terms of co-ordinates to a problem in Riemannian geometry.

share|improve this answer
Deane, 1) my example even involves 2 connections... 2a) I find it disadvantageous to have to learn at least 3 calculi for the whole picture. 2b) I found me developing my own calculus to sanely get into Hilbert space via Stokes theorem (for doing geometric PDE geometrically, like Bochner theorems) Indeed José has an excellent answer. There's sure much more. Alas I had to suffer through O'Neill's formulae etc. in Christoffelian (by a famous German differential ring theorist, no less). –  Martin Gisser Mar 8 '11 at 3:05
Martin, I also prefer to understand everything from a single approach. And to some extent I have succeeded in developing a single way to do everything I need to do in Riemannian geometry. But it appears that each of us ends up developing our own preferred approach and even notation. But even then my approach does not accomplish absolutely everything I want, just most of it. Depending on the context, I still sometimes find either the differential form or local co-ordinate approach the easiest way to do what I want. So I find it quite handy to be able to use them, too. –  Deane Yang Mar 8 '11 at 3:22
P.S.: Said German differential ring theorist's lectures had at least a complete proof of the fundamental lemma of Riem. Geom., completely proving tensoriality on the r.h.s. P.P.S.: My answer-comment to Jose is hidden (anyhow syntacticly mangled) - Killing fields indeed serving an example of Koszul vs. not-Koszul. P.P.P.S.: Something learned today. Thanks, sirs. –  Martin Gisser Mar 8 '11 at 3:24
What's the "fundamental lemma of Riemannian geometry"? –  Deane Yang Mar 8 '11 at 4:09
@Deane: Connections with torsion are popular in physics. They also arise naturally in pure geometry. If one considers Hermitian non-Kahler manifolds, then the Chern connection (compatible with the Hermitian metric and the complex structure) has torsion. In fact, its torsion vanishes if and only if the manifold is indeed Kahler. There is an analogous "Obata" connection in quaternionic geometry. When one has an additional structure on a Riemannian manifold (a special tensor, usually a form), then it is more natural to study connections which make that form parallel. Usually these have torsion. –  Spiro Karigiannis Mar 26 '11 at 16:08

I've meanwhile found out that my innocent example ...

1.) ... amounts to a rediscovery of Schouten's contorsion tensor. (See e.g. this note).

This concept is important in torsion gravity (which isn't as exotic as it may sound - except for some charlatanry derived from it...) My example equation seems to express the equivalence principle (cf. link eq. (255)). See also Rodrigues & Oliveira: The many faces of Maxwell, Dirac and Einstein equations.

As I've already hinted here, contorsion stuff (or what I would term the contorsion operator) can also occur in computations with torsion-free connections.

2.) ... leads to a generalization of the fundamental "lemma", a.k.a. Levi-Civita theorem: For any given vector-valued 2-form $T$ there exists a unique metric connection with torsion $T$.

share|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.