Uncertain about the statement "equivalent condition for an isometry of Riemannian manifolds" in Lee's Intro to Riemannian Manifolds

Question

In Professor Lee's Introduction to Riemannian Manifolds, second edition on page 12, the first paragraph on Isometries reads

Suppose $(M,g)$ and $(\tilde{M},\tilde{g})$ are Riemannian manifolds with or without boundary. An isometry from $(M,g)$ to $(\tilde{M},\tilde{g})$ is a diffeomorphism $\phi\colon M\to\tilde{M}$ such that $\phi^*\tilde{g}=g$. Unwinding the definitions shows that this is equivalent to the requirement that $\phi$ be a smooth bijection and each $d\phi_p\colon T_pM\to T_{\phi(p)}\tilde{M}$ be a linear isometry.

Only one part of a proof eludes me. That is, given a smooth bijection $\phi$ such that each $d\phi_p\colon T_pM\to T_{\phi(p)}\tilde{M}$ is a linear isometry, I haven't been able to prove that $\phi^{-1}$ is smooth if $M$ has nonempty boundary. When $M$
has no boundary, I can use his Introduction to Smooth Manifolds, second edition (ISM) Proposition 4.8(a) and ISM Exercise 4.9 to get that $\phi$ is a local diffeomorphism and ISM Proposition 4.6(f) to get that $\phi$ is a diffeomorphism. ISM Proposition 4.8(a) relies on the Inverse Function Theorem for Manifolds (ISM Theorem 4.5) which Professor Lee warns us "can fail for a map whose domain has nonempty boundary."

Is a map with invertible differential that maps boundary to boundary a local diffeomorphism? might be part of an answer if I could prove that $\phi$ mapped the boundary of $M$ into the boundary of $\tilde{M}$. (I can already prove that it maps the interior of $M$ into the interior of $\tilde{M}$ using ISM Problem 4-2.) Of course, such a requirement is necessary for $\phi$ to be a diffeomorphism by the Diffeomorphism Invariance of the Boundary Theorem (ISM Theorem 2.18), but I haven't been able to prove that $\phi$ maps boundary to boundary either.

So, how can $\phi^{-1}$ be proven to be smooth, or is the statement in the book incorrect?

Have you considered mapping of $[0,1)$ onto $S^1$ given by $e^{2\pi i x}$? I think it's smooth, and definitely not diffeomorphism. The inner product is irrelevant to that part (you could make it preserve metric if you wanted), so it seems false. — Esgeriath, Jun 13 '23 at 23:18
I did consider that. However, since it has sine and cosine in it, there will be points where $d\phi_x$ is the zero map and therefore not invertible. So I think it is not a viable counterexample. — Jeff Rubin, Jun 14 '23 at 02:13
@JeffRubin It is of course a local diffeomorphism. I think you’d better double-check that vague reasoning. — Ted Shifrin, Jun 14 '23 at 03:28
It can't be a local diffeomorphism: since it is bijective it would then be a diffeomorphism, which it is not. — Jeff Rubin, Jun 14 '23 at 04:45
The derivative is an isomorphism at every point. The issue is point-set topology, not the inverse function theorem. — Ted Shifrin, Jun 14 '23 at 04:49
Using terminology from Lee ISM: Let $\phi\colon[0,1)\to S^1$ be defined by $\phi(t)=(\cos 2\pi t,\sin 2\pi t)$. Then $d\phi_t\colon T_t([0,1))\to T_{(\cos 2\pi t,\sin 2\pi t)}S^1$ is a linear map whose matrix with respect to coordinate bases is the Jacobian of the representative $\hat{\phi}$ with respect to smooth charts for $[0,1)$ and $S^1$, namely $([0,1),\mathrm{id}{[0,1)})$ and $(U_i^\pm\cap S^1,\phi_i^\pm)$. In particular, $J{1/2}(\phi_1^-\circ\phi)=\sin(2\pi/2)=0$, so no, the differential is not an isomorphism at $t=1/2$. — Jeff Rubin, Jun 14 '23 at 05:03
That should have been $2\pi\sin(2\pi/2)=0$ in the last sentence. — Jeff Rubin, Jun 14 '23 at 05:09
The issue with your reasoning is that $\phi(\frac 1 2)$ is not in set $U_1^-$, since it's $x$ coordinate is zero, and not negative. You have to consider domains of maps — Esgeriath, Jun 14 '23 at 06:40
From ISM page 5: $U^-_1={(x^1,x^2)\in\mathbb{R}^2\colon x^1<0}$, so $\phi(1/2)=(\cos\pi,\sin\pi)=(-1,0)\in U^-_1\cap S^1$, so $\phi(1/2)$ is most certainly in the domain of the smooth chart $(U^-_1\cap S^1,\phi^-_1)$ for $S^1$. — Jeff Rubin, Jun 14 '23 at 13:40
And the chart is $\phi_1^-(x^1,x^2)=x^2$, and the derivative of $\sin[2\pi x)$ is $2\pi\color{red}{\cos}(2\pi x)$. It helps to visualize the mapping directly, too. — Ted Shifrin, Jun 14 '23 at 14:18
Oh! My apologies. You’re right! So it is a counterexample after all! If you write it as an answer, I’ll be happy to accept it. — Jeff Rubin, Jun 14 '23 at 15:13
@Esgeriath Why don’t you write it up? It was your suggestion. — Ted Shifrin, Jun 14 '23 at 15:32
@JeffRubin: You're absolutely right; these conditions are equivalent only when $M$ has empty boundary. I've added a correction to my online list. — Jack Lee, Jun 14 '23 at 19:10

score 3 · Accepted Answer · answered Jun 14 '23 at 16:32

Given this asumptions, $\phi^{-1}$ cannot be proven to be smooth.

Consider $[0, 2\pi)$ and $S^1$ with standard smooth structures. Manifold $[0, 2\pi)$ has nonempty boundary $\partial [0, 2\pi) = \{0\}$.

The function $\phi\colon[0, 2 \pi) \to S^1$ given by $\phi(t) = (\cos(t), \sin(t))$ is smooth bijection, but not diffeomorphism - inverse function is not continuous at $(1, 0)$.

The part about $d\phi_t$ being linear isometry does not invalidate this example. This is kind of independent condition. But to be more formal, we can endow both spaces with metric tensors by setting $\frac\partial{\partial x^1}$ and $ d\phi\frac\partial{\partial x^1} $ to be one-element orthonormal basis in corresponding manifolds. Notice how vectors $d\phi_t \frac\partial{\partial x^1}$ are always nonzero, and thus definition is correct. This deifnition gives us that $d\phi_t\colon T_t [0, 2\pi) \to T_{\phi(t)}S^1$ is isometry for each $t$.

Uncertain about the statement "equivalent condition for an isometry of Riemannian manifolds" in Lee's Intro to Riemannian Manifolds

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