Can any metrical or axiomatic geometry be shown to be isomorphic to a Klein geometry? Can any manifold with a (pseudo) Riemannian metric be defined in the Kleinian sense?
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4$\begingroup$ Usually the definition of a Klein geometry requires local homogeneity, so clearly not. $\endgroup$– Ben McKayCommented Feb 2, 2022 at 21:02
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1$\begingroup$ The answer will depend on precisely how constrained your notion of "metrical or axiomatic geometry" is. You could tautologically define things in a way where the answer is yes, I suppose. But you probably won't get agreement on that being the class of "metrical or axiomatic geometry". $\endgroup$– Ryan BudneyCommented Feb 2, 2022 at 23:37
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1$\begingroup$ Here is a kind of answer for dimension at least 3. The case of dimension 2 assuming Desargues' axiom was answered positively by Blaschke. The reference to Blaschke can be found in Sharpe's book. $\endgroup$– KapilCommented Feb 3, 2022 at 3:31
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1$\begingroup$ The answer might also depend on the meaning of "any" ("is every geometry also..." vs. "is some geometry also..."). Many times it is best to avoid the "any" quantifier. $\endgroup$– Jukka KohonenCommented Feb 3, 2022 at 11:15
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