Htykuut

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.

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.

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 $psi(x)=(-x,sqrt{1-x^2})$, $xin(-1,1)$, we have $phi[(-1,1)]=psi[(-1,1)]$ but $dphi_0(H^1)=H^1$ and $dpsi_0(H^1)=-H^1$.

Instead, let $phi, psi: Vrightarrow X$ be two parametrizations with $phi(0)=psi(0)=x$. We know that $dphi_0$ and $dpsi_0$ are both isomorphisms from $mathbb{R}^k$ to $T_x(X)$. Assume $vin dphi_0(H^k)$ but $vnotin dpsi_0(H^k)$. Then $dpsi^{-1}_x(v)in mathbb{R}^k-H^k$. Let $alpha$ be a curve in $mathbb{R}^k$ with $alpha(0)=0$ and $alpha'(0)=dpsi^{-1}_x(v)$. 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 $Psicircalpha$ maps $(0,epsilon)$ to an open arc in the ambient space of $X$ that is disjoint from $X$. Finally, consider the map $g=Phi^{-1}circPsicircalpha$ which is an arc in $mathbb{R}^k$ with $g(0)=0$ and
$$
g'(0)=dPhi^{-1}_xcirc dPsi_0(dPsi^{-1}_x(v))=dPhi^{-1}_x(v)in H^k,
$$
by our choice of $v$. The curve $g$ maps a short interval $(0,epsilon)$ into $H^k$ whereas $Psicircalpha$ 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 $dphi_0(H^k)=dpsi_0(H^k)$.

Since you are using G&P (exercise 2.1.7*), here is a solution in that flavor which doesn't use parametrizing local curves.

Let $phi:Uto X$ and $psi:Wto X$ be local parametrizations about a point $xinpartial X$ where $U$ and $W$ are open subsets of $H^k$ with the usual $0mapsto x$ for $phi$ and $psi$. Then, by shrinking neighborhoods if need be, $g=psi^{-1}circphi$ is a diffeomorphism $Ucong W$. Let $G$ be an extension of $g$ to an open subset of $mathbb{R}^k$, $G:U'to W$. By definition $dg_0=dG_0$ (p.59). Since $phi$ and $psi$ map boundary to boundary (exercise 2.1.2), they must map (strict) upper half space to $X^circ$. This gives us that $G$ maps $H^k$ to $H^k$. Now observe that since $G$ is smooth, the limit $lim_{tto 0}frac{G(tv)}{t}$ exists and equals $dG_0(v)$ for all $vinmathbb{R}^k$. So, in particular, for $vin H^k$ since $G$ maps $H^k$ to $H^k$ and $H^k$ is a closed set, we have that
$$dG_0(v)=lim_{tto 0^+}frac{G(tv)}{v}in H^k$$This shows that $dG_0(H^k)subset H^k$. But since the chain rule still works (p.59) we have that $dG_0=dg_0=dpsi^{-1}_xcirc dphi_0$ and therefore we have shown that $dphi_0(H^k)subset dpsi_0(H^k)$. Defining $G$ in the reverse order gives the other inclusion that we seek and thus gives $dphi_0(H^k)= dpsi_0(H^k)$. This guarantees that the definition $H_x(X)=dphi_0(H^k)$ is well defined.