Is the Jacobian a divergence of some vector field? (manifolds)

Question

Let $M,N$ be $d$-dimensional oriented Riemannian manifolds, and let $f:M \to N$ be smooth. Assume $M$ is compact with boundary.

Does there exist a vector field $X \in \Gamma(TM)$ s.t $\text{div} \, X=\det df$?

If $M$ has no boundary this is impossible in general since $$ \int_M \det df \, dV=\int_M \text{div} \, X \, dV=\int_{\partial M} \langle X ,n \rangle \, dV=0.$$

Such a vector field always exists in the Euclidean case, when $M=N=\mathbb{R}^n$, see e.g theorem 8.33 in "Direct methods in the calculus of variations" - by Dacorogna (second edition).

score 3 · Accepted Answer · answered Sep 21 '17 at 17:07

3

If $M$ is a compact $d$-dimensional manifold with nonempty boundary, then $H^d(M) = 0$, and so any $d$-form is exact. Thus, $\det df\,dV = d\eta$ for some ($d-1$)-form $\eta$. Consider the $1$-form $\omega = \star\eta$. Using the metric, $\omega$ corresponds to a vector field $X$ and $\text{div}\,X\,dV = d{\star}\omega = \pm d\eta = \pm \det df\,dV$, as required (up to a sign).

answered Sep 21 '17 at 17:07

Ted Shifrin

115,160

Just to be sure: your argument showed every function is a divergence, right? There is no use of the specific structure of the Jacobian. – Asaf Shachar Sep 22 '17 at 06:26
Indeed. That's correct. – Ted Shifrin Sep 22 '17 at 06:28

Is the Jacobian a divergence of some vector field? (manifolds)

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