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The origin question: Let $\Omega \subset \mathbb{H}^2$ be a domain of the hyperbolic plane $\mathbb{H}^2$. Let $u: \Omega \to \mathbb{H}^2$ be injective and an isometry from $\Omega$ to its image. Does there exist a Mobius transformation $\gamma\in \text{PSL}(2,\mathbb{R})$ such that $u=\gamma\mid_\Omega$?

The modified question: Let $\Omega \subset \mathbb{H}^2$ be a connected domain of the hyperbolic plane $\mathbb{H}^2$. Let $u: \Omega \to \mathbb{H}^2$ be an orientation-reserving $C^1$ isometry from $\Omega$ to its image. Does there exist a Mobius transformation $\gamma\in \text{PSL}(2,\mathbb{R})$ such that $u=\gamma\mid_\Omega$?

Thanks for all comments and answers. I have found the answer from "Dierkes, Ulrich; Hildebrandt, Stefan; Tromba, Anthony J. Global analysis of minimal surfaces", on page 273, Lemma 1, which reads as follows:

Lemma: Let $f: U \to \mathbb{H}^2$ be a $C^1$ isometry on an open connected subset $U$ of the hyperbolic plane. Then $$ f(w)=\frac{Aw+B}{Cw+D}, \, A,B,C,D \in \mathbb{R}, $$ and $AD-BC=1$.

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  • $\begingroup$ Isn't "injective" redundant in "injective and an isometry"? $\endgroup$
    – LSpice
    Nov 25, 2022 at 0:26
  • $\begingroup$ You should edit the question and explain what do you mean by an "isometry" (so that the word "injective" is not redundant and also make it clear if your definition includes "orientation-preserving") and does the word "domain" mean "open and connected" (as customary in complex analysis). $\endgroup$ Nov 25, 2022 at 5:30
  • $\begingroup$ If your "domain" isn't connected, then the answer would be no. $\endgroup$ Nov 25, 2022 at 16:12
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    $\begingroup$ As I noted below, “half” of the isometries of the hyperbolic plane are orientation reversing. However, Mobius transformations are orientation preserving. So the lemma you cite is missing a hypothesis (or a conclusion). If you add “negative complex conjugation” to the group $\mathrm{PSL}$ then you will get all isometries. $\endgroup$
    – Sam Nead
    Nov 26, 2022 at 8:42

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Assuming that $\Omega$ is open, non-empty, and connected, then yes. In fact, an isometry determines, and is determined by, how it transforms any one point and any one "frame" (orthonormal basis) at that point.

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    $\begingroup$ An isometry is determined by how it tramsforms at a point. However, it doesn't mean a local isometry can be extended to a global isometry. It only says that if such an extension exists, then it is unique. $\endgroup$
    – gaoqiang
    Apr 16, 2023 at 0:13
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The hyperbolic plane has this property, as does the Euclidean plane.

If $K$ is any subset of $ \mathbb{H}^2$, and $u : K \to\mathbb{H}^2$ is an isometry, then there is an extension of $u$ which is an isometry of $\mathbb{H}^2$ onto itself. I use "isometry" in the sense: $d(u(x),u(y)) = d(x,y)$ for all $x,y \in E$, where $d$ is the hyperbolic distance in $\mathbb{H}^2$.

So the OP reduces to: is an isometry of $\mathbb{H}^2$ onto itself necessarily a Möbius transformation (or the conjugate of a Möbius transformation)?

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  • $\begingroup$ It's maybe too early in the morning. I'll think about what I said. $\endgroup$ Nov 25, 2022 at 17:01
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    $\begingroup$ Regarding your reformulation (in the last line): Mobius transformations are orientation preserving, so they make up an index two subgroup of the full isometry group. $\endgroup$
    – Sam Nead
    Nov 26, 2022 at 8:43
  • $\begingroup$ @SamNead thanks, fixed. $\endgroup$ Nov 20, 2023 at 22:28

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