# Vector field along an immersion whose covariant derivative is the differential

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.

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.
• 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? – Quarto Bendir Jul 23 '20 at 23:29
• 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. – Sebastian Jul 24 '20 at 9:45