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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.

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    $\begingroup$ the definition of the outward pointing normal appears in page 118 (second edition of the book) $\endgroup$
    – Masacroso
    Sep 14 at 22:06
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    $\begingroup$ 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. $\endgroup$
    – yi li
    Sep 15 at 7:58

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