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Let $f:X\to Y$ be a vector bundle with a chosen Ehresmann connection. Let $\gamma:[0,1]\to Y$ be a curve downstairs and let $\underset{\gamma:t_0\to t}{\mathrm{tra}}:\alpha^{-1}(\gamma t_0)\to \gamma ^{-1}(\gamma t)$ be (a portion of) parallel transport along $\gamma$. The covariant derivative along $\gamma$ is an arrow $$\nabla_\gamma:\Gamma(\gamma^\ast f)\to \Gamma(\gamma ^\ast f)$$ defined by $$\nabla_\gamma h(t_0)=\lim_{t\to t_0}\frac{\underset{\gamma:t_0 \to t}{\mathrm{tra}}^{-1}(h(t))-h(t_0)}{t}.$$ I think a section of $\gamma^\ast f$ is precisely a lift of $\gamma$.

  1. What's $\nabla_\gamma$ geometrically doing to a lift of $\gamma$? How's it modifying these curves?
  2. Suppose now $\pi:\mathrm TX\to X$ is a tangent bundle of a premanifold $X$, also with a connection. Say a section of $\gamma^\ast \pi$ is constant w.r.t the connection if it's a parallel transport along $\gamma$. Is $\gamma$ being geodesic equivalent to $\dot \gamma$ being constant in this sense? Intuitively I feel this must be true since the parallel transport tells which curves upstairs are "constant", and these are what we want geodesics to be (I think).
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    It doesn't make sense to speak of geodesics in your context. A geodesic is a curve whose velocity is "constant". You need a connection on $TY$ in order to define that, not on some arbitrary bundle $X$ over $Y$. Then the velocity of $\gamma$ is a lift of $\gamma$ to $TY$ (a section of the pullback bundle $\gamma^{*}(TY)$) and if the velocity is constant (the covariant derivative is zero) then the curve is (defined to be) a geodesic. – 2017-01-30
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    @levap sure. I forgot to replace $f$ with a tangent bundle in the second question. I'll fix that. I think my first question makes sense. – 2017-01-30

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