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TLDR: What condition should a parallel transport (assignment from paths in the base of a vector bundle to linear isomorphisms between fibers) satisfy to ensure the associated covariant derivative is linear in the tangent vector argument?

A covariant derivative $\nabla$ on a vector bundle $E\to M$ is an operator $\nabla\colon TM\times_M J^1E\to E$ satisfying a few conditions. Let $f\in C^\infty(M),$ $X\in\Gamma(TM)$ and $s\in\Gamma(E).$

  • a Leibniz condition for the section: $\nabla_X(f s)=X(f)s+f\nabla_X(s)$
  • a linearity condition for the tangent vector: $\nabla_{aX+bY}s = a\nabla_Xs + b\nabla_Ys$

There are other formulations of connection. One is as an Ehresmann connection, which specifies a horizontal subbundle of a fiber bundle. An Ehresmann connection is equivalent to a covariant derivative if the choice of horizontal subspace depends linearly on $s$, in a way that is described on wikipedia section on "Vector bundles and covariant derivatives" in the Ehresmann connection article. And see the related question How do connection 1-form and Ehresmann version of connections relate to each other?

Another formulation for a connection is in terms of parallel transport $\Gamma$. To each path in the base $\gamma\colon I\to M$, we assign an isomorphism of fibers $\Gamma^t_0(\gamma)\colon E_{\gamma(0)}\overset{\cong}\to E_{\gamma(t)}.$ The assignment should obey some functoriality and smoothness conditions. One expects to be able to recover the covariant derivative from the parallel transport. Indeed, the process is described on Wikipedia in the section "Recovering the connection from the parallel transport" in the parallel transport article.

Define

$$ \nabla_{\gamma}s|_{t=0}=\lim_{t\to 0}\frac{\Gamma^t_0(\gamma)^{-1}s(\gamma(t))-s(\gamma(0))}{t}. $$

Transport the neghboring vector back to the beginning fiber, take the Newton quotient there, and take the limit. From this it is easy to derive the Leibniz law for covariant derivatives.

But how do we see that $\nabla_{aX+bY} = a\nabla_X + b\nabla_Y$? Surely some additional criterion must be stipulated for $\Gamma,$ as there exist non-linear connections. Something analogous to the linearity condition for Ehresmann connections. What is this condition?

ziggurism
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  • It's not just that the fiber isomorphism $E_{\gamma(0)}\to E_{\gamma(t)}$ be linear, is it? – ziggurism Dec 04 '17 at 02:27
  • Related questions: http://math.stackexchange.com/q/669828/16490, http://math.stackexchange.com/q/2112576/16490, http://math.stackexchange.com/q/1407800/16490 – ziggurism Nov 02 '18 at 02:46
  • I guess this is a circular answer: I believe that your definition of the connection from parallel transport defines a horizontal distribution on the associated frame bundle, where a curve in the principal bundle is parallel transport if and only if it is an integral curve of the distribution. Then the connection is linear if and only if the distribution satisfies the usual required conditions for linearity. What you presumably want is an answer involving only the curves directly and not the associated distribution. – Deane Nov 02 '18 at 19:53
  • @Deane, you may be right, that the easiest way to state it is in terms of its associated covariant derivative, since it is ultimately a condition that only applies to its infinitesimal action. – ziggurism Nov 04 '18 at 20:57
  • Actually, I think there should be a way to express it directly in terms of some kind of limit of lifts to the frame bundle of closed curves starting and ending at a fixed point in the manifold. If I had the time, I'd try to work this out. I like questions like this. It all helps elucidate the relationship between a connection, its parallel transport, and its curvature. – Deane Nov 05 '18 at 00:07
  • @Deane General Theory of Lie Groupoids by MacKenzie 2005, defines a path connection as a functor $\nu\mapsto\bar{\nu}$ on the path groupoid satisfying five axioms including the additivity condition, which is that if $\nu_1, \nu_2, \nu_3$ are paths such that $\dfrac{d\nu_1}{dt}(t_0)+\dfrac{d\nu_2}{dt}(t_0)=\dfrac{d\nu_3}{dt}(t_0),$ then $\dfrac{d\bar{\nu}_1}{dt}(t_0)+\dfrac{d\bar{\nu}_2}{dt}(t_0)=\dfrac{d\bar{\nu}_3}{dt}(t_0).$ He shows that such path connections are in bijection with what he calls infinitesimal connections. This is halfway through ... – ziggurism Nov 05 '18 at 16:48
  • ... a text on Lie groupoids and algebroids which is dense and technical and unfamiliar, but I think probably his infinitesimal connections are the same as or at least equivalent in some groupoid sense to Koszul connections, and that this is the linearity criterion I am looking for. I wonder why it's just sums instead of full linear combinations though. Maybe scalar multiples of tangent vectors of curves just correspond to reparametrizations... – ziggurism Nov 05 '18 at 16:48

1 Answers1

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The term "linear connection" refers to linearity in the other slot. So the point here is that $\nabla_X(s_1+s_2)=\nabla_X s_1+\nabla_X s_2$. This is indeed equivalent to the parallel transport being linear as a map between fibers (and thus only makes sense on vector bundles).

What you discuss in the post is linear dependence on the direction into which you differentiate. This is not a specific feature of linear connections, it is also true for connections on general fiber bundles and indeed for arbitrary tangent maps.

Andreas Cap
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  • While it is true that linearity in the tangent vector slot is common to many classes of connections (eg also connection on principal bundle), it is not common to all of them. For example, the general Ehresman connection does not assume linearity. And the Lie derivative does not obey it. – ziggurism Dec 04 '17 at 13:18
  • In the comments I put forth a guess, maybe linearity in the tangent vector is a consequence of linearity of fiber isomorphisms. You point out that linearity of the fiber isomorphisms is responsible for additivity in the sections. It must also be present for the Leibniz law. Therefore it cannot be the source of linearity in the tangent vector, that guess was wrong. Thanks for making that clear. – ziggurism Dec 04 '17 at 13:21
  • As stated in my reply an Ehresmann connection is linear (even over smooth functions) in the tangent vector slot and also the Lie derivative is linear (over the reals) in the vector field slot. – Andreas Cap Dec 05 '17 at 08:15