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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.2010-12-10

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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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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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    ...since you only really wrote about the tan$g$ent bundle. But it is not like it matters much anyway.) :-)2010-12-10