MathOverflow Asked by Quarto Bendir on December 1, 2021
Let $(M,g)$ be a Riemannnian manifold and let $f:Sigmato M$ be a smooth immersion. Then the vector bundle $f^ast TMtoSigma$ 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,wrangle&=f^ast g\ h(cdot,omega^sharp)+nabla^perp w&=0end{align}
where $h$ is the second fundamental form. I can’t see any immediate conclusions to make.
Your differential $dfinOmega^1(Sigma,f^*TM)$ 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.
Answered by Sebastian on December 1, 2021
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