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Given a fiber bundle $\pi:E\to M$, a curve $\gamma:[0,1]\to M$, and a point $p \in \pi^{-1}(\gamma(0))$, a connection on the bundle allows us to uniquely lift $\gamma$ to a horizontal curve in E through $p$. In almost all situations I have encountered, the horizontal lift does not depend on the orientation of $\gamma$. To be precise, the two curves $t\to \gamma(t)$ and $t\to\gamma(1-t)$ have the same horizontal lift through $p$.

I have a fiber bundle for which I would like to have a type of parallel transport which depends on which direction one is moving in the base. So my question is: what is the best way to formulate a connection which is orientation dependent, and so enables this type of parallel transport?

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    $\begingroup$ Given a curve $\gamma$ on $M$ and fixing a point $x$ in fibre of $\gamma(0)$, there exists a curve that starts at $x$. So, lift has starting point as $x$. Suppose you choose $t\mapsto \gamma(1-t)$, you fix a point $y$ in fibre of $\gamma(1)=\gamma(1-0)$, you get a lift whose starting point is $y$. This does not say horizantal lift of $\gamma(t)$ and $\gamma(1-t)$ are same if you are looking from orientation perspective. What is that I am misunderstanding in your question? $\endgroup$
    – Praphulla Koushik
    Commented Jan 31, 2019 at 8:05
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    $\begingroup$ I agree with @PraphullaKoushik. Simply put: he lifts of the two curves $t\mapsto \gamma(t)$ and $t\mapsto \gamma(1-t)$ are not the same (their images are), so the usual lift is already orientation dependent. $\endgroup$
    – Michael Bächtold
    Commented Feb 2, 2019 at 8:19
  • $\begingroup$ I had written that as an answer and then I was not sure as question is not clear.. so deleted my answer... So, left it as a comment... $\endgroup$
    – Praphulla Koushik
    Commented Feb 2, 2019 at 9:44

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If you look at the Ehresmann connection in T(TM) (or T(TM\0)) associated with a second order differential equation or a Finsler metric (which is not necessarily reversible), then this is orientation dependent. This is not a fancy notion, just a straight-forward generalization (or just geometrization) of the Levi-Civita connection for Riemannian metrics, when you interpret it as an Ehresmann connection: https://mathoverflow.net/a/256484/21123

Related to these connections in $TM$ (or $PTM$, $STM$) is the notion of non-linear connection and that may be what you are looking for. Instead of decomposing $T_eE$ at every point into the vertical subspace $V_eE$ and a horizontal subspace $H_eE$ you decompose it into $V_eE$ and a cone in $T_eE$ such that at each non-zero point of the cone the vertical subspace and the tangent to the cone form a linear decomposition of $T_eE$. The cone is not necessarily symmetric about the origin and you can capture non-reversibility in this way.

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  • $\begingroup$ How does this not violate Picard–Lindelöf theorem? $\endgroup$
    – Vít Tuček
    Commented Jan 31, 2019 at 10:38
  • $\begingroup$ Because the initial conditions for a geodesic and for the one with its orientation reversed are not the same. $\endgroup$
    – alvarezpaiva
    Commented Jan 31, 2019 at 13:25
  • $\begingroup$ It IS cheating: the connection is not on a bundle over M (but over TM). If not, it is impossible with anything that could be described as an Ehresman connection: a horizontal curve remains horizontal in whatever reparametrization. $\endgroup$
    – alvarezpaiva
    Commented Jan 31, 2019 at 13:28
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    $\begingroup$ I see. That's along the lines I was thinking. So perhaps in general one could try connection on $J^1E$? $\endgroup$
    – Vít Tuček
    Commented Jan 31, 2019 at 14:27
  • $\begingroup$ I am sure I am misunderstanding something in the question... I was thinking "Given a curve $\gamma$ on $M$ and fixing a point $x$ in fibre of $\gamma(0)$, there exists a curve that starts at $x$. So, lift has starting point as $x$. Suppose you choose $t\mapsto \gamma(1-t)$, you fix a point $y$ in fibre of $\gamma(1)=\gamma(1-0)$, you get a lift whose starting point is $y$. This does not say horizantal lift of $\gamma(t)$ and $\gamma(1-t)$ are same if you are looking from orientation perspective." Can you (if possible) tell me what I am misunderstanding $\endgroup$
    – Praphulla Koushik
    Commented Jan 31, 2019 at 15:51
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Given a curve $\gamma$ on $M$ and fixing a point $x$ in the fibre of $\gamma(0)$, there exists a curve that starts at $x$. So, this lift has starting point as $x$. Suppose you choose $t\mapsto \gamma(1-t)$, you fix a point $y$ in the fibre of $\gamma(1)=\gamma(1-0)$, you get a lift whose starting point is $y$. This does not say horizontal lift of $\gamma(t)$ and $\gamma(1-t)$ are same if you are looking from orientation perspective.

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