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I already asked the same question here, but received no answer. I was reading this interesting article by Givental

Givental, Alexander. "The Pythagorean theorem: What is it about?" Amer. Math. Monthly 113 (2006), no. 3, 261–265

And my attention was caught by this passage:

"This explains why the Pythagorean theorem is a genuinely Euclidean phenomenon (and not only in the historical sense of the word): among all Riemannian metrics of constant curvature only the Euclidean one admits nontrivial conformal isometries."

So that's the question: there exist Riemannian metrics of non constant curvature admitting nontrivial conformal isometries? If so, the Pythagorean theorem holds on them?

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    $\begingroup$ I don't have a proof, but my belief is that whether the Pythagorean Theorem holds is equivalent to whether there is any nontrivial holonomy when a tangent vector is parallel translated around a closed curve. If so, this would prove that the Pythagorean holds if and only if the metric is flat. $\endgroup$
    – Deane Yang
    Commented May 22, 2023 at 3:50

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I think (see (??) below) there is no connected complete Riemannian manifold $(M,g)$ which is not flat and such that there is an isometry $f:(M,g)\to(M,\lambda^2g)$, where $\lambda\neq0,1$. I will assume $\lambda>1$, if not take the inverse function.

If there were such an isometry $f$, then for any $p\in M$ the sequence $f^np$ would be Cauchy in the Riemannian distance $d_g$, because $d_g(f^n(p),f^{n+1}(p))=\frac{d_g(p,f(p))}{\lambda^n}$.

As $M$ is not flat there are non zero sectional curvatures so we can find a small geodesic triangle $T$ with sum of angles $S\neq\pi$. The sum of angles $S_n$ of the triangle $T_n=f^n(T)$ is just $S_n=S$, because $f$ is conformal. However, if $M$ is complete, then the triangles $T_n$ converge to some point $q$. But $g$-sectional curvatures are bounded in a small neighborhood of $q$, and the diameters of $T_n$ converge to $0$, so $S_n$ should converge to $\pi$, (?? my previous argument had a fallacy I think, I was assuming that a subsurface was totally geodesic. The fact that the sums of angles of geodesic triangles convergent to a fixed point must converge to $\pi$ is probably still true. By the CAT inequalities, using that sectional curvatures are bounded near $q$, one can see that $S_n$ cannot converge to anything above $\pi$, I don't know if there is a similar result to see that they cannot stay below some constant $<\pi$ either).

In the non-complete case there are counterexamples: consider a bump function $f:\mathbb{R}^2\to\mathbb{R}$ which is supported in the ball centered at $(1,0)$ and of radius $0.1$. Now consider the function $g:\mathbb{R}^2\setminus\{0\}\to\mathbb{R}$ given by $g(x)=\sum_{n\in\mathbb{Z}}2^nf\left(\frac{x}{2^n}\right)$.

Then the graph of $g$, as a Riemannian submanifold of $\mathbb{R}^3$, is not flat, but it is invariant by the isometry $f:(\mathbb{R}^3,g)\to(\mathbb{R}^3,g/4);x\mapsto2x$ (where $g$ usual metric of $\mathbb{R}^3$).

In any case, the usual Pythagorean theorem with right triangles cannot always hold if there are non zero sectional curvatures: in the book "Metric spaces of non positive curvature", by Bridson and Haefliger, in the proof of theorem 1A.6, it is mentioned that if $M$ has sectional curvature $\kappa\in\mathbb{R}$ in some plane, then we can find small right triangles with squared sides $\varepsilon^2,\varepsilon^2,2\varepsilon^2-\frac{K}{6}\varepsilon^4+O(\varepsilon^5)$, thus the Pythagorean theorem does not hold.

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  • $\begingroup$ Thank you for your detailed answer, I really appreciate that! So in the non-complete case is it possible to affirm that the Pythagorean theorem holds in $ M $ locally, and so that $ M $ is locally isometric to $ \mathbb{R^2} $? $\endgroup$
    – user967210
    Commented May 24, 2023 at 10:17
  • $\begingroup$ What do you mean exactly by the Pythagorean theorem holding locally? Do you mean in the sense that every point has a nhood such that for every right geodesic triangle with lengths $a\geq b\geq c$ we have $a^2=b^2+c^2$? $\endgroup$
    – Saúl RM
    Commented May 24, 2023 at 11:50
  • $\begingroup$ Yeah, I mean that $\endgroup$
    – user967210
    Commented May 24, 2023 at 12:41
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    $\begingroup$ And when do you say it should be possible to affirm that? Because in my non complete counterexample it need not hold. $\endgroup$
    – Saúl RM
    Commented May 24, 2023 at 13:18
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    $\begingroup$ I see. So the new question is whether Pythagoras' theorem (the one with right triangles) holds in any non flat manifold. I am pretty sure it doesn't (again it will be enough to consider a subsurface with nonzero sectional curvature), I will write it in another answer when I can $\endgroup$
    – Saúl RM
    Commented May 24, 2023 at 16:00

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