Let $M$ be a smooth manifold with boundary and $p \in \partial M$. In Lee's Introduction to Smooth Manifolds he gives the following definition:
If $p \in \partial M$, a vector $v \in T_pM \setminus T_p \partial M$ is said to be inward-pointing if for some $\epsilon > 0$ there exists a smooth curve $\gamma: [0, \epsilon) \rightarrow M$ such that $\gamma(0) = p$ and $\gamma'(0) = v$, and it is outward-pointing if there exists such a curve whose domain is $(-\epsilon, 0]$.
Intuitively I understand what an inward and outward pointing vector is from calculus and Euclidean spaces. However, I am having some trouble mapping that intuitive picture to the definition above. In particular, how does the condition $\gamma'(0) = v$ suggest that the vector is pointing inwards, and how does taking the domain to be $(-\epsilon, 0]$ suggest that the vector is pointing outwards?
Best Answer
Before talking of a boundary point, let's look at the more general picture of an interior point $p$. A tangent vector $v \in T_pM$ can be thought of as a vector tangent to a curve $\gamma \colon (-\varepsilon,\varepsilon)$ at $p=\gamma(0)$, with $\gamma'(0)=v$. This means that when passing through $p$, $\gamma$ follows the direction defined by $v$. Actually, this tells two things on the path $\gamma$:
Now, imagine that $p$ lies on the boundary of some domain, for instance, the boundary of a flat earth. There are three kinds of tangent vectors at that point:
The same thing occurs for a more general manifold with boundary. The tangent space $T_pM$ at $p\in \partial M$ splits into three different components $T_pM^{\text{out}}$, $T_pM^{\text{in}}$ and $T_p\partial M$. $T_p\partial M \subset T_pM$ is an hyperplane, and the two other components are the two connected components of $T_pM \setminus T_p\partial M$.