Let $(M^{n+1},g)$ be a Riemannian manifold and let $\Sigma^n \hookrightarrow M$ be a smooth, closed, embedded submanifold. Let $\Omega$ be the volume form of $\Sigma$. It is well-known that a volume form is parallel, i.e. $$ \nabla^{\Sigma}\Omega \equiv 0 $$ on $\Sigma$. Can we always extend $\Omega$ to an $n$-form $\tilde{\Omega}$ defined on a neighborhood $U$ of $\Sigma$ such that $$ \nabla \tilde{\Omega} \equiv 0 $$ on $U$?
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$\begingroup$ @XipanXiao: Why is that necessarily parallel on that neighborhood? $\endgroup$– Ted ShifrinCommented Jul 1, 2015 at 17:34
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$\begingroup$ @TedShifrin: en you're right I didn't notice that at all...deleted. $\endgroup$– Xipan XiaoCommented Jul 1, 2015 at 18:12
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This seems to be false, even if one assumes that $\Sigma$ is isometrically embedded. Counterexample: Let $M$ be the $2$-sphere and $\Sigma$ some great circle in $M$; the volume form $\Omega$ of $\Sigma$ is some parallel one-form on $\Sigma$. $\Omega$ can't possibly extend even locally to some neighborhood of $M$, because $M$ doesn't admit even locally parallel nonzero one-forms.
(One argument for this last statement, in case it is not immediately clear: If $\omega$ were a nonzero parallel one-form on some neighborhood in $M$, it would have some corresponding nonzero parallel local vector field $X$ (such that $\langle X, Y \rangle = \omega(Y)$ for every vector field $Y$), and you could define a parallel local frame for $M$ by taking some constant oriented rotation of $X$; but of course $M$ is not locally flat anywhere.)
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$\begingroup$ Thanks mollyerin; this is convincing. Do you know what curvature information represents the obstruction? i.e. Can I construct some sort of "most parallel" extension $\tilde{\Omega}$ and then bound its covariant derivative by some sort of curvature information? $\endgroup$– ThompsonCommented Jul 2, 2015 at 12:30
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$\begingroup$ @Thompson I'm not sure how you might go about trying to do this. One method would be to follow Xipan Xiao's (now deleted) comment above, taking a unit normal $\nu$ to $\Sigma$, extending it (perhaps via the exponential map) to a v.f. $N$ on a neighborhood of $\Sigma$, and looking at the interior product of $N$ with the volume form $\omega$ of $M^{n+1}$. You can estimate the covariant derivative of this $n$-form in terms of $\nabla N$, which perhaps you can then estimate in terms of the shape operator of $\Sigma$ and the curvature operator of $M$ (though I haven't thought it through precisely). $\endgroup$ Commented Jul 3, 2015 at 5:53