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I have a function $f\colon SO(3)\to U$ with $U\subset \mathbb{R}^n$, for some $n\in\mathbb{N}$, that is a diffeomorphism. The space $SO(3)$ is equipped with the geodesic distance as metric and $U$ is equipped with the standard Euclidean metric on $\mathbb{R}^n$.

I want to show that $f$ is bi-Lipschitz (i.e. $f$ and $f^{-1}$ are Lipschitz-continuous). I know already from Lee (Introduction to Smooth Manifolds, Prop. C.29) that if we would have an ambient open space of $SO(3)$ such that the function is still $C^1$ and $SO(3)$ is convex, then I could follow that $f$ is bi-Lipschitz.

Since this is not the case, is there any theorem that could prove the above? If yes, is there any literature where I can find this?

Thank you for your help!

Arctic Char
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CostaZach
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    I have a dumb question: isn't a diffeomorphism between compact spaces automatically bi-lipschitz? – Jason DeVito - on hiatus Nov 09 '20 at 16:05
  • I also think it should be true but what do you mean by automatically? Because it is not a standard result. The most results require an open set. And $SO(3)\subset\mathbb{R}^{3\times 3}$ is not open and $SO(3)$ doesn't even contain open sets in $\mathbb{R}^{3\times 3}$. – CostaZach Nov 09 '20 at 16:25
  • By "automatically", I mean "with no other assumptions." Specifically, I was asking if the question "if $X$ and $Y$ are compact metric spaces and $f:X\rightarrow Y$ is a homeomorphism, then $f$ is bi-Lipschitz" is true. But I've now checked out wikipedia, which reminded me about the homeomorphism $[0,1]\rightarrow [0,1]$ given by $x\mapsto x^3$. Googling also led me to an MSE question which may answer your question.https://math.stackexchange.com/questions/2338799/c1-function-on-compact-set-is-lipschitz – Jason DeVito - on hiatus Nov 09 '20 at 16:46
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    I agree with @JasonDeVito : a diffeomorphism is a bijection which is smooth, with smooth inverse and with $\mathrm{d}f$ invertible everywhere. Moreover, as $SO(3)$ is compact, $df$ is continuous over a compact and thus have bounded norm, and $f$ is Lipschitz. Same thing for $f^{-1}$. – Didier Nov 09 '20 at 19:24
  • Sry, I don't see it. Because all the proofs are know use fundamental theorem of calculus and that the line $x+t(y-x)$ is in $SO(3)$. But this is not the case for $SO(3)$, since it is not convex. So how can I bound something if it not even in the space itself? – CostaZach Nov 10 '20 at 18:11
  • https://math.stackexchange.com/questions/563925/when-is-a-lipschitz-homeomorphism-of-metric-spaces-bi-lipschitz/565973#565973 – Moishe Kohan Nov 18 '20 at 21:27

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