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Let $(M,g)$ be a Riemannnian manifold and let $f:\Sigma\to M$ be a smooth immersion. Then the vector bundle $f^\ast TM\to\Sigma$ has a natural bundle metric and metric-compatible connection. Can one characterize the situations in which there must exist a section $V$ such that $\nabla V=df$? This is trivially possible if $(M,g)$ is Euclidean space. It feels like it should not be possible in general.

This seems to be equivalent to the existence of a closed 1-form $\omega$ on $\Sigma$ and a normal vector field $w$ along $\Sigma$ such that \begin{align}\nabla \omega-\langle h,w\rangle&=f^\ast g\\ h(\cdot,\omega^\sharp)+\nabla^\perp w&=0\end{align} where $h$ is the second fundamental form. I can't see any immediate conclusions to make.

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Your differential $df\in\Omega^1(\Sigma,f^*TM)$ satisfies the integrability condition $$d^\nabla df=0$$ where $d^\nabla$ is the induced exterior derivative from the (pull-back of) the Levi-Civita connection on $M.$ If $df=\nabla V$ the integrability condition is that the curvature tensor $R$ applied to the vector field $V$ does vanish. As you have guessed this is clearly not possible in general.

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  • $\begingroup$ Thank you, that was essentially my intuition but I failed to realize it explicitly. Could one say anything in the situation that $(M,g)$ is a flat manifold? $\endgroup$
    – Quarto Bendir
    Commented Jul 23, 2020 at 23:29
  • $\begingroup$ Yes, one can say the following: it is always possible locally, but not always possible globally. As a counterexample you can consider a cone $C$ in 3-space with the induced flat metric, and $f$ to be the identity $\Sigma=C\to C.$ If the cone angle is not $2\pi$, i.e., you do not have a degenerate cone (plane), $V$ does not exist globally. $\endgroup$
    – Sebastian
    Commented Jul 24, 2020 at 9:45

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