# Is there a notion of a connection for which the horizontal lift of a curve depends on its orientation?

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?

• 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? – Praphulla Koushik Jan 31 '19 at 8:05
• 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. – Michael Bächtold Feb 2 '19 at 8:19
• 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... – Praphulla Koushik Feb 2 '19 at 9:44

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.

• How does this not violate Picard–Lindelöf theorem? – Vít Tuček Jan 31 '19 at 10:38
• Because the initial conditions for a geodesic and for the one with its orientation reversed are not the same. – alvarezpaiva Jan 31 '19 at 13:25
• 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. – alvarezpaiva Jan 31 '19 at 13:28
• I see. That's along the lines I was thinking. So perhaps in general one could try connection on $J^1E$? – Vít Tuček Jan 31 '19 at 14:27
• 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 – Praphulla Koushik Jan 31 '19 at 15:51

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.