Confirming whether a vector field is inward/outward pointing

Question

Suppose $M \subseteq \mathbb{R}^{n}$ is an $m-$dimensional manifold with boundary, and let $x \in \partial{M}$. Let $f: U \cap M \rightarrow V \cap H^{m}$ be some boundary chart such that $x \in U$. Then the tangent space of $x$ is given by $$T_{x}M=Df^{-1}(f(x))(\mathbb{R}^{m}).$$

Consider a smooth vector field $h:M\rightarrow \mathbb{R}^{n}$, which requires that for all $x \in M$, $h(x) \in T_{x}M$.

1. What does it mean to say that $h$ is inward pointing? Would it be correct to confirm that $$[Df(x)\cdot h(x)]_{m}>0$$ where $[y]_{m}$ denotes the $m-$th component of a vector $y$?
2. How can one check this in practice, so that we can apply results such as the Poincare-Hopf theorem? On page 5 in this text http://webapps.towson.edu/cbe/economics/workingpapers/2016-05.pdf it says that for $x \in \partial M$, it is enough to confirm that $x+\alpha h(x) \in int(M)$ for $\alpha$ small. I don't see this this though. It's clear that this would imply that for $\alpha$ small, $x+\alpha h(x)$ lies in $f$'s domain, so that $[f(x+\alpha h(x)]_{m}>0$, but I do not know how to conclude the necessary $[Df(x)\cdot h(x)]_{n}>0$ from here.

Answer 1

The answer to question 1. depends entirely on the conventions you use. If you define the half space $H^m$ as the set of those $y\in\mathbb R^m$ such that $[y]_m\geq 0$ (which is one of the usual conventions), then your description is correct. In fact, you can always think about the half space itself in which the tangent space in $y$ is just a copy of $\mathbb R^m$ centered in $y$. Then "inward pointing" and "outward pointing" get their intuitive meaning, and all that just gets transported to $M$ via $f$.

The criterion you are trying to use in 2. does not really make sense in the given form. You are trying to use the affine line through $x$ in direction $h(x)$ but this is not adapted to $M$. (You certainly need the assumption that $\alpha\geq 0$ for the criterion to make any sense. Then it does work in this form in the case of an open subset of $H^m$ as discussed above.) In a general version, you would have to use curves in $M$, for example integral curves of $h$. The technical details about this depend a bit on the setting you use, e.g. whether you assume that $h$ is defined on an open neighborhood of $M$. A correct version of the condition would be that to consider smooth curves $c$ in $\mathbb R^n$ with $c(0)=x$ and $c'(0)=h(x)$. Then $h$ is invard pointing if there are such curves for which $c([0,\epsilon))\subset M$ for some small $\epsilon$ but there are no such curves for which $c((-\epsilon,\epsilon))\subset M$.

Edit (to address your comment): The criterion you try to impose does not work in the case of a general submanifold. It does work for an open subset $U\subset H^m$: For $x\in\partial U$, you have $[x]_m=0$ and then vor a vector $v\in\mathbb R^m$ and $t\in\mathbb R$ you get $[x+tv]_m=t[v]_m$. Since $U$ is open in $H^m$ this lies in $U$ for sufficiently small positve $t$ if and only if $[v]_m\geq 0$ and if $[v]_m=0$, then it also lies in $U$ for sufficiently small negative $t$. Using elementary analysis, this is equivalent for the analogous condition for arbtirary smooth curves. If $c$ is a smooth curve in $H^m$ such that $c(0)=x$ then $c(t)$ lies in $U$ for sufficiently small positive $t$ iff $[c'(0)]_m\geq 0$ (i.e. iff $c'(0)$ is invard pointing) and if it is $=0$ then the same holds for sufficiently small negative $t$. In this form, the condition extends to arbritrary submanifolds, since everthing is transported to $H^m$ via $f$. This is technically a bit more demanding, but the ideas are the same.