Proving the decomposition is independent of the choice of chart. Let $M$ be a manifold with boundary, and let $p\in \partial M.$ So there is a decomposition of $T_pM$ as follows: For a chart $\phi : U\to \tilde{U} \subset \mathbb{H}^n,$ with coordinate functions $(x^1,\dots, x^n),$ we say that $X\in T_pM$ points into $M$ if $Xx^n > 0,$ out of $M$ if $Xx^n<0,$ and parallel to $\partial M$ if $Xx^n = 0.$ How do I prove this decomposition is independent of the choice of chart? And that the set of vectors that point parallel to $\partial M$ form an $n-1$-dimensional subspace of $T_pM?$
Can someone please help me with this? Thank you
 A: Take another chart $(y^1,\dots,y^n):V\rightarrow \tilde{V}$. Look at the induced transition diffeomorphism $\Phi =y \circ x^{-1} \in C^\infty(\tilde{U},\tilde{V})$. To not get confused by notation, let's denote its variable inputs by $\Phi(s_1,\dots,s_n)$. Then we know that by change of coordinates formula (remember $X[x^i]$ is the component in the $\partial_{x^i}|_p$ direction of $X$ meaning $X=\sum_i X[x^i]\partial_{x^i}|_p$)
$$X[y^j]=\sum_i \frac{\partial \Phi_j}{\partial s_i}(x(p)) X[x^i]$$
Denote $q=x(p)\in \partial \mathbb{H}^n$ for simplicity. looking at the case $j=n$, let's inspect each term:
for all $i<n$
$$ \frac{\partial \Phi_n}{\partial s_i}(q)=\lim_{h\rightarrow 0^+} \frac{\Phi_n(q+h\hat{e}_i)-\Phi_n(q)}{h}=0$$
The reason being is (topological!) boundary is preserved under homeomorphisms, so $q+h\hat{e}_i,q \in \partial \mathbb{H}^n$ must still have zero $n$th coordinate after applying $\Phi$.
In addition, the case $i=n$ yields a positive derivative, since $\Phi_n(q+h\hat{e}_i)\geq 0$ for all $h>0$. It can't be zero because then the derivative matrix of $\Phi$ at $q$ won't be invertible, as it has a row of $0$s.
If so, we have $X[y^n]=\frac{\partial \Phi_n}{\partial s_n}(q) X[x^n]$
where $\frac{\partial \Phi_n}{\partial s_n}(q)> 0$ and thus the sign is preserved no matter the chart, as required.
In regard to the subspace, we have just seen that it is $\left \{ X\in T_pM|\ X[x^n]=0 \right \}$. this is basically the vector space of all tangent vectors whose $n$th component according to the basis $\partial_{x^1}|_p,\dots,\partial_{x^n}|_p$ is zero. Thus, it is spanned by $\partial_{x^1}|_p,\dots,\partial_{x^{\color{red}{n-1}}}|_p$.
