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If $M^n$ is a smooth manifold with boundary and $p\in\partial M$, then $T_pM$ is the disjoint union of inward and outward point vectors, and $T_p\partial M$. If $(U,(x^i))$ is a smooth boundary chart at $p$, then the inward pointing vectors are those with positive $x^n$ coordinate, the outward pointing vectors are those with negative $x^n$ coordinate, and $T_p\partial M$ is those with zero $x^n$ coordinate.

What's confusing me is that the coordinate map $(x^i)$ is on $M$, and $T_pM$ is a vector space, not a manifold. So how can you talk of the $x^n$ coordinate of a tangent vector in the tangent space of an abstract manifold?

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  • $\begingroup$ If you have coordinates around a point, these coordinates give you coordinate vectors which form a basis. So the $x^n$ coordinate of a vector $v$ in $T_p M$ would be the $n$-component of $v$. $\endgroup$ – Braindead Aug 9 '14 at 18:59
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It's probably more accurate to say that inward-pointing vectors are those with positive $x^n$ component. In the given boundary chart, each element of $T_pM$ can be written $v = v^i\partial/\partial x^i$ (using the summation convention), and $v$ is inward-pointing if and only if $v^n>0$.

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A tangent vector $v\in T_pM$ does not have any coordinates. However, if we have a smooth boundary chart $\mathbf{X}:U\subset H^n\to M$ with $\mathbf{X}(0)=p$, then the derivative $d\mathbf{X}$ is an isomorphism $\mathbb{R}^n\to T_pM$. Now, in $\mathbb{R}^n$ it is clear enough whether a given tangent vector points up, down, or none of them. In other words, a vector in $\mathbb{R}^n$ does have coordinates.

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