Can a volume form on a submanifold be extended to a parallel form in a neighbourhood?

Question

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 $U$?

Answer 1

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.)