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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.

Diane Vanderwaif
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3 Answers3

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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 $\phi:U\to X$ and $\psi:W\to X$ be local parametrizations about a point $x\in\partial X$ where $U$ and $W$ are open subsets of $H^k$ with the usual $0\mapsto x$ for $\phi$ and $\psi$. Then, by shrinking neighborhoods if need be, $g=\psi^{-1}\circ\phi$ is a diffeomorphism $U\cong W$. 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 $\lim_{t\to 0}\frac{G(tv)}{t}$ exists and equals $dG_0(v)$ for all $v\in\mathbb{R}^k$. So, in particular, for $v\in H^k$ since $G$ maps $H^k$ to $H^k$ and $H^k$ is a closed set, we have that $$dG_0(v)=\lim_{t\to 0^+}\frac{G(tv)}{v}\in H^k$$This shows that $dG_0(H^k)\subset H^k$. But since the chain rule still works (p.59) we have that $dG_0=dg_0=d\psi^{-1}_x\circ d\phi_0$ and therefore we have shown that $d\phi_0(H^k)\subset d\psi_0(H^k)$. Defining $G$ in the reverse order gives the other inclusion that we seek and thus gives $d\phi_0(H^k)= d\psi_0(H^k)$. This guarantees that the definition $H_x(X)=d\phi_0(H^k)$ is well defined.

J. Moeller
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  • Could you please provide a solution for exercise 2.1.2 Prove that if $f: X \rightarrow Y$ is a diffeomorphism of manifolds with boundary, then $\partial f$ maps $\partial X$ diffeomorphically onto $\partial Y$. – Intuition Nov 25 '18 at 13:19
  • I have just posted a question about it. – Intuition Nov 25 '18 at 13:21
  • Do we have to prove 2.1.2 or as stated from here https://math.stackexchange.com/questions/554156/the-boundary-of-an-n-manifold-is-an-n-1-manifold in the first answer , it is clear? – Intuition Nov 25 '18 at 19:42
  • my question is that a definition that boundary points are mapped to boundary points as stated in the link I have given in my previous comment? – Intuition Nov 25 '18 at 19:53
  • Could you please if you have time look at this question ? https://math.stackexchange.com/questions/981033/prove-that-nzy-is-trivial-if-and-only-if-there-exists-a-set-of-k-independ – Intuition Dec 01 '18 at 16:36
  • The question I mentioned in the above comment contain a 50 bounty that I can donate with no more than 20 hours ..... I want to give them to you as you helped me alot. – Intuition Dec 02 '18 at 03:59
  • I think the functions that cut out Z are not specically described in terms of $g_{i}$ ....... adjustment of the domains and codomains ? – Intuition Dec 02 '18 at 14:49
  • What is the importance of the normal bundle being trivial is not stated clearly – Intuition Dec 02 '18 at 14:50
  • from where the zeros of the functions $g_{i}$ can we get ? why we are sure that there exist such values? – Intuition Dec 02 '18 at 15:13
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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 $d\phi_0(H^k)=d\psi_0(H^k)$.

Emilho
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  • It is hard to see this an Answer to the Question. Of course the assumption that $X$ is a manifold with boundary, and that $x$ is a point on the boundary, is important to the Question (about defining a "half space" $H_x(X)$). Observing that without those assumptions "the assertion would not be true" seems hardly worth the discussion you've given here. Since the Question is somewhat old (almost three years) and has an Accepted Answer, discussion of what seems to be a basic departure from the Question as it was asked is of doubtful value. – hardmath Sep 08 '17 at 18:46
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    The reason I posted my answer is that I was not satisfied with the one already given. Citation "dϕo(Rk)=dωo(Rk)dϕo(Rk)=dωo(Rk) implies dϕo(Hk)=dωo(Hk)", is not correct. This argument could just as well be applied to the boundaryless manifold case, which is why I show a counterexample. The discussion was still relevant for me, and I'm convinced it will still be for others. – Emilho Sep 08 '17 at 18:57
  • One of the reasons posts are deleted is that instead of attempting to answer, the post is an attempt to reply to or comment on another post. That sort of "crosstalk" is not allowed by the StackExchange guidelines, but if you earn $50$ points in reputation, then you can Comment on the posts of others. – hardmath Sep 08 '17 at 20:07
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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.

XiaoXiao Zhen
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    I don't think this is a solution as presented. This is a proof that tanget space is well defined, but not a proof that upper half space is well defined. All you have shown is that $dϕ_o (R^k)=dω_o (R^k)$. What if one of these maps flips $R^k$ and the other leaves it alone? $dϕ_o (R^k)=dω_o (R^k)$ doesn't alone imply $dϕ_o (H^k)=dω_o (H^k)$ – J. Moeller May 24 '18 at 14:30