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I was trying to understand the notion of a connection. I have heard in seminars that a connection is more or less a differential equation. I read the definition of Kozsul connection and I am trying to assimilate it. So far I cannot see why a connection is a differential equation. Please help me with some clarification.

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I'd rather say it's a differential operator than an equation. But you can build equations with it. – Raskolnikov Dec 10 '10 at 17:53

I'm going to be bold and go against the wisdom of the rest of the posters here.

A connection can be regarded as a differential equation, through the notion of parallel transport and geodesics. In the parallel transport formulation, a connection can be associated with the following map: give a point $p$ in your manifold $M$, a tangent vector $v \in T_pM$ in the tangent space at $p$, and a curve $\gamma:[0,1] \to M$ such that $\gamma(0) = p$, you have

$$ (p,v,\gamma) \mapsto \tilde{v}$$

where $\tilde{v}$ is a vector field on $M$ defined along $\gamma$. That is, $\tilde{v}(\gamma(s)) \in T_{\gamma(s)}M$ for every $s\in [0,1]$. The specification of $\tilde{v}$ is given by the ordinary differential equation

$$ \nabla_{\dot{\gamma}}\tilde{v} = 0, \qquad \tilde{v}(0) = v$$

or, more explicitly

$$ \frac{d}{ds} \tilde{v}(\gamma(s)) = 0, \qquad \tilde{v}(\gamma(0)) = v$$

In terms of the geodesic formulation, in local coordinates, a specification of a parallel transport is equivalent to the specification of the geodesics for that connection. More precisely, fixing a coordinate system $\{x_1, \ldots, x_n\}$ for your manifold $M$, the connection can be represented by its Christoffel symbols (relative to the coordinate system) $\Gamma_{ij}^k$, such that the geodesics satisfy Newton's Equations

$$ \frac{d^2 x_i}{dt^2} = -\sum_{k,j} \Gamma^i_{jk} \frac{d x_j}{dt} \frac{d x_k}{dt} $$

and in this way the connection precisely is associated to a particular set of second order differential equations, such that for each prescribed initial data $x_i(0) = x_{i,0}$ and $dx_i/dt (0) = y_{i,0}$ you have a corresponding geodesic.

In a more abstract setting, a connection on the tangent space $TM$ of a manifold $M$ can be associated to a vector field $v$ define over the total space $TM$ (so that $v$ is a section of $TTM$) so that the natural projection $\pi_*v(x) = x$ where $\pi$ is the projection map from $TM$ to $M$. The vector $v$ defines what is known as the geodesic spray of the connection, which can be written as a first order ordinary differential equation on $TM$ (or a second order ordinary differential equation on $M$). So yes, it is possible to associate a connection to a differential equation. (The reverse association, however, is more delicate.)

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None of the other posters denied that, Willie. – Raskolnikov Dec 10 '10 at 18:57
Oh? What about Eric's post below? The point is that it is no less valid to think of a connection as given you an ODE on $TM$ as it is as a differential operator; depending on the applications it may even be preferred. – Willie Wong Dec 10 '10 at 19:03
The OP was asking about Koszul connections, which I have always seen defined as a differential operator. You can view them as a set of differential equations as you point out (and as I did as well) but it is sloppier in my opinion to start with the geodesic equation since it is not invariant as stated (you need to say how the Christoffel symbols transform, which looks weird unless you see them as coming from a connection seen as an operator). – Eric O. Korman Dec 10 '10 at 19:24
@Eric: I agree that for many purposes that the differential operator point of view may be more natural, but it is not true that you must go through the Christoffel symbols as you stated. In some situations it may be more natural to even treat the Koszul connection as a one form with values in the vertical bundle. (See the book of Kolar, Michor, and Slovak for example.) My point is that depending on situation, different points of view may be more appropriate. (Also, it seems to me that your answer does not specifically address the Koszul connection either, if you want to pick bones... – Willie Wong Dec 10 '10 at 19:54
...since you only really wrote about the tangent bundle. But it is not like it matters much anyway.) :-) – Willie Wong Dec 10 '10 at 19:55

A connection is not a differential equation-- it is a differential operator. Where the differential aspect comes from is that it is a derivation. Remember in calculus the product rule says that $\frac{d}{dx} fg = \frac{df}{dx} g + f\frac{dg}{dx}$. Well for a vector field $\sigma$, a function $f$, and Koszul connection $\nabla$, the property $$ \nabla_V (f\sigma) = df(V) \sigma + f\nabla_V \sigma $$ is satisfied. This is what makes it differential.

What makes it a "connection" is that with $\nabla$ you can define parallel translation. If you have a curve $c$, you can say that a vector field $\sigma(t)$ defined on the curve is parallel if $\nabla_{c'(t)} \sigma = 0$. Think of this as $\frac{d}{dt} \sigma = 0$. Now take a path $c$ starting at a point $x$. Then by basic theorems of differential equations, for any vector at $x$ there is a unique extension to a parallel vector field along $c$. This allows you to connect the tangent space at $x$ with the tangent space at any point on $c$.

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