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Let $\gamma:[0,\ell] \rightarrow M$ be a normalized geodesic on a manifold $M$.

How can I construct a parallel vector field $w(t)$ along $\gamma$ with the following properties:

  • $\langle \gamma'(t), w(t) \rangle =0$

  • $|w(t)|=1$

I am trying to take a vector $w$ that is perpendicular to $\gamma'(0)$ and taking the parallel transport $w(t)$ along $\gamma$ and using the fact that the parallel transport is an isometry to conclude $\langle \gamma'(t),w(t)\rangle=0$ but this doesn't seem to work.

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1 Answer 1

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Your proposed procedure works, thanks to the fact that $\gamma$ is a geodesic. By metric-compatibility we have $$\frac d{dt}\langle \gamma',w \rangle = \langle \nabla_{\gamma'}\gamma',w\rangle+\langle\gamma',\nabla_{\gamma'}w\rangle.$$ The first term vanishes because $\gamma$ is a geodesic, and the second term vanishes because $w$ is parallel; so $\langle \gamma',w\rangle$ is constant along $\gamma$. Since we choose $w$ to make $\langle \gamma',w \rangle$ vanish at $t=0$, it therefore vanishes everywhere. The other part ($|w|=1$) follows from $|w(0)|=1$ and the fact that parallel transport is an isometry, as you seem to have deduced.

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  • $\begingroup$ Ah I see, thanks very much ! $\endgroup$
    – Tuo
    Mar 30, 2018 at 2:33

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