Prove that$ H_x (X)$ does not depend on the choice of local parametrization.

Question

Suppose that $X$ is a manifold with boundary and $x∈∂X$. Let $ϕ:U→X$ be a local parametrization with $ϕ (0)=x$ where $U$ is an open subset of $H^k$. Then $dϕ_0:R^k→T_x (X)$ is an isomorphism. Define the upper half space $H_x (X)$ in $T_x (X)$ to be the image of $H^k$ under $dϕ_0, H_x (X)=dϕ_0 (H^k )$. Prove that$ H_x (X)$ does not depend on the choice of local parametrization.

I tried to construct another local parametrization say $\omega:V→X$ where $V$ is also an open subset of $H^k$ and $\omega (0)=x$. Since both $U$ and $V$ are subset of $H^k$, $U \cap V$ is also subset of $H^k$. I found a hint that tell me to consider $ \phi (U) \cap \omega(V)$ but I can't see how this can help me.

Answer 1

This problem is very similar to one of mine. So I just take my solution and modify a little bit. Hope it helps.

Let $ω:V→X$ be a local parametrization with $ω(0)=x$ and $V$ is also a an open subset of $H^k$. If we shrink both $U$ and $V$ small enough, we will have $ϕ(U)=ω(V)$. Then we have the map $g=ω^{-1} o ϕ:U→V$ is diffeomorphism. From this if we write $ϕ=ω o g$ then take derivative both sides , we have $dϕ_o=dω_o o dg_o$. This mean that $im(dω_o )⊂im(dϕ_o )$

Repeat the above process but swap $ϕ$ and $ω$ to each other, we will have $im(dϕ_o )⊂im(dω_o )$. So $im(dω_o )=im(dϕ_o )$, thus $dϕ_o (R^k)=dω_o (R^k)$ implies $dϕ_o (H^k)=dω_o (H^k)$. From the definition of $H_x (X)$, we can conclude that $H_x (X)$ does not depend on the choice of local parametrization.

Answer 2

I think a point of this exercise is that a boundary in some extent orients the tangentplane. The assertion would not be true for a boundaryless manifold, as take for example the sphere $S^1$. Choosing two parametrizations around the north pole $\phi(x)=(x,\sqrt{1-x^2})$ and $\psi(x)=(-x,\sqrt{1-x^2})$, $x\in(-1,1)$, we have $\phi[(-1,1)]=\psi[(-1,1)]$ but $d\phi_0(H^1)=H^1$ and $d\psi_0(H^1)=-H^1$.

Instead, let $\phi, \psi: V\rightarrow X$ be two parametrizations with $\phi(0)=\psi(0)=x$. We know that $d\phi_0$ and $d\psi_0$ are both isomorphisms from $\mathbb{R}^k$ to $T_x(X)$. Assume $v\in d\phi_0(H^k)$ but $v\notin d\psi_0(H^k)$. Then $d\psi^{-1}_x(v)\in \mathbb{R}^k-H^k$. Let $\alpha$ be a curve in $\mathbb{R}^k$ with $\alpha(0)=0$ and $\alpha'(0)=d\psi^{-1}_x(v)$. By definition, $\phi$ and $\psi$ can be extended to smooth functions $\Phi, \Psi$ on open nbhds of $0$ in $\mathbb{R}^k$, on which they are both still diffeomorphisms.

Note that $\alpha$ maps a short interval $(0,\epsilon)$ into $\mathbb{R}^k-H^k$, so $\Psi\circ\alpha$ maps $(0,\epsilon)$ to an open arc in the ambient space of $X$ that is disjoint from $X$. Finally, consider the map $g=\Phi^{-1}\circ\Psi\circ\alpha$ which is an arc in $\mathbb{R}^k$ with $g(0)=0$ and $$ g'(0)=d\Phi^{-1}_x\circ d\Psi_0(d\Psi^{-1}_x(v))=d\Phi^{-1}_x(v)\in H^k, $$ by our choice of $v$. The curve $g$ maps a short interval $(0,\epsilon)$ into $H^k$ whereas $\Psi\circ\alpha$ maps the same interval outside of $X$, but this would imply that the extension map $\Phi$ maps points in $H^k$ outside of $X$! This cannot be true since $\Phi|_{H^k}=\phi$ maps $H^k$ into $X$. We conclude that $d\phi_0(H^k)=d\psi_0(H^k)$.