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Why can a diffeomorphism $F: M \to N$ between two smooth manifolds with boundary not take an interior point of $M$ to a boundary point of $N$?

Let $(U, \varphi)$ be a smooth chart for $M$, $(V, \psi)$ a smooth chart for $N$. I believe it is because it would cause $(\psi \circ F \circ \varphi^{-1}): \varphi(U) \to \psi(V)$ to be a diffeomorphism between an open set in $\mathbb{R}^n$ and an open set in $\mathbb{H}^n$ such that its intersection with $\partial \mathbb{H}^n$ is nonempty, and this cannot be. However, I don't know how to prove this. Any help would be appreciated

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A diffeomorphism is in particular a homeomorphism, so a neighbourhood of a point must be mapped to a homeomorphic neighbourhood of the image of the point. In particular, it cannot take interior points to boundary points since they have non-homeomorphic neighbourhoods.

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  • $\begingroup$ Ah, I think I see my confusion. Lee's Smooth Manifolds never explicitly says the nbds of the different kinds of points are not homeomorphic but he says that a point can never be both kinds of points, which is effectively saying the above. Thanks. $\endgroup$ – Emilio Minichiello Sep 20 '18 at 14:31
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    $\begingroup$ @EmilioMinichiello: Javi's answer is correct (and I've upvoted) - but do you know how to prove they have non-homeomorphic neighborhoods? (And I wouldn't say they are non-homeomorphic by definition. They are certainly presented differently, but that doesn't rule out some kind of weird homeomorphism. The usual trick is local homology.) $\endgroup$ – Jason DeVito Sep 20 '18 at 14:32
  • $\begingroup$ I'm on Chapter 2 of Smooth Manifolds, Lee doesn't give us the proof that the nbds are non-homeomorphic. He only says that a point cannot be a boundary point and an interior point simultaneously, and that every smooth chart will map a boundary point to the boundary of $\mathbb{H}^n$. I'm trying to understand how this implies that $F(\text{Int}M) = \text{Int}N$. $\endgroup$ – Emilio Minichiello Sep 20 '18 at 14:39
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The usual proof that neighborhoods that intersect $\partial\mathbb{H}$ are distinct from neighborhoods uses the local homology groups $H^\ast(M,M\setminus\{p\})$ for different choices of $p$. It has the advantage of showing that a homeomorphism must map boundary to boundary and interior to interior.

But since you're asking about diffeomorphisms, we should be able to work less hard. So how about this?

Suppose $f:M\rightarrow N$ is a diffeomorphism and that $p\in \operatorname{int} M$ and $f(p)\in \partial N$. Then $d_p f:T_p M\rightarrow T_{f(p)} N$ must be an isomorphism.

But I claim that is not. To see this, pick a $w\in T_{f(p)}$ which points outside of $N$. I claim there is no $v\in T_p M$ with $d_p f(v) = w$.

To see this, suppose there is such a $v$, choose a smooth path $\gamma_v:(-\epsilon,\epsilon)\rightarrow M$ with $\gamma_v(0) = p$ and $\gamma_v'(0) = v$. Then $f\circ\gamma$ is a smooth curve in $N$. By picking a chart near $f(p)$, we can view all this as happening in $\mathbb{R}^n$.

So, without loss of generality, we have a curve $\gamma$ for which $\gamma(0) = \vec{0}\in \mathbb{R}^n$, $\gamma\subseteq H =\{(x_1,...,x_n)\in \mathbb{R}^n: x_n\leq 0\}$, but $\gamma'(0) \notin H$. That is, the $n$th coordinate of $\gamma'(0)$ is positive.

Since $\gamma$ is smooth, $\gamma'$ is continuous, so the $n$-th coordinate of $\gamma'(t)$ is positive in a neighborhood of $0$. But now the mean value theorem applied to the $n$th coordinate of $\gamma$ implies that $\gamma(t)\notin H$ for small $t$ with $t>0$.

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    $\begingroup$ I like to generalize a little bit, which sometimes helps people see the point. To any shape in $\Bbb R^m$ there is a tangent cone at any given point, consisting of those tangent vectors you can obtain as the derivative of some curve in the shape. Now it is a quick exercise to verify that diffeomorphisms preserve tangent cones. This shows interior and boundary points differ, but also cone points, and that not all cone points are the same... $\endgroup$ – user98602 Sep 20 '18 at 18:47
  • $\begingroup$ PS. I wonder what the odds are we will ever be at the same conference? We are a little too far from each other's orbit to make it too likely. One can hope... $\endgroup$ – user98602 Sep 20 '18 at 22:42
  • $\begingroup$ I do not have something like that, but I'll send you an email to start the chain. Also, I'm going to delete my unrelated comments after you see them (as they are not much help to the OP!) $\endgroup$ – user98602 Sep 21 '18 at 17:20

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