# Definition of outward-pointing unit normal vector field

I am to trying to understand the Divergence Theorem. This say:

Theorem. Let $$(M,g)$$ be a oriented Riemannian manifold with boundary. For any compactly supported smooth vector field $$X$$ on $$M$$,

$$\begin{equation} \int_{M}(\text{div} X) dV_{g} = \int_{\partial M} \langle X , N \rangle_{g} dV _\widetilde{g}, \end{equation}$$

where $$N$$ is the outward-pointing unit normal vector field along $$\partial M$$ and $$\widetilde{g}$$ is the induced Riemannian metric on $$\partial M$$.

I don't know what is the outward-pointing unit normal vector field. In fact, up to this point I haven't even had a need to know what it means an unit normal vector field along $$\partial M$$. Intuitively, I think that means that

$$\begin{equation} \langle v , N_{p} \rangle_{g} = 0 \;\; \forall v \in T_{p} \partial M, \;\; \forall p \in \partial M, \end{equation}$$

and

$$\begin{equation} \langle N_{p} , N_{p} \rangle_{g} = 1 \;\; \forall p \in \partial M. \end{equation}$$

I think so, but honestly I have never seen the formal definition of unit normal vector field, much less than outward-pointing.

What is the correct meaning of this?

P.S. I saw the above theorem in John Lee's Introduction to Smooth manifolds, but couldn't find the definition I needed.

• the definition of the outward pointing normal appears in page 118 (second edition of the book) Sep 14 at 22:06
• there is an exercise need to prove the existence of outward normal v.f along the boundary see Problem 8.4.The rough idea is construct it inside coordinate chart,then gluing them together. Sep 15 at 7:58