A question on volume form

Question

Let $(M,g)$ be an oriented Riemanian manifold with boundary $\partial M$. Let $f:M\rightarrow \mathbb{R}$ be a boundary defining function, and let $N$ be the outward unit normal vector field along $\partial M$ (or to be more precise, the restriction to $\partial M$ of $-\frac{grad f}{\|grad f\|_{g}}$. So we can think that $N$ is defined on the whole $M$). My question is: it it true that $$df\wedge (N\lrcorner \,\omega_{g})=h\,\omega_{g}$$ for some non-negative function $h$ on $M$? Here $\omega_{g}$ denotes the volume form of $M$.

Answer 1

The interior multiplication is an antiderivation: $X\lrcorner(\omega\wedge\eta)=(X\lrcorner\omega)\wedge\eta+(-1)^k\omega\wedge (X\lrcorner\eta)$, where $k=\deg\omega$. Here, this can be used to show that

$$0=N\lrcorner(0)=N\lrcorner(df\wedge \omega_g)=(N\lrcorner df)\wedge\omega_g-df\wedge(N\lrcorner\omega_g),$$

and so your question is equivalent to show that $N\lrcorner df$ is a non-negative function. But since

$$N\lrcorner df=df(N)=g(\mathrm{grad} f,N)$$

(by definition of the gradient) and that

$$g(\mathrm{grad} f,N)=g\left(\mathrm{grad} f,-\frac{\mathrm{grad} f}{||\mathrm{grad} f||}\right)=-\frac{||\mathrm{grad}f||}{||\mathrm{grad}f||}=-1,$$

it appears that it is in fact a strictly negative function. Also, beware how you extend $N$ on the whole manifold $M$, since you can't divide by $||\mathrm{grad} f||$ at critical points of $f$.