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A homogeneous metric on a space $X$ is one for which the isometry group acts transitively on its points (for all $x,y\in X$, there is an isometry $\phi$ of $X$ such that $\phi(x)=y$).

If we repeatedly puncture a sphere, can we always give the resulting space a homogeneous metric?

The usual Euclidean metric on the sphere is homogeneous. Puncturing it once yields the plane, on which the usual Euclidean metric is again homogeneous. Puncturing the plane yields a space homeomorphic to an infinitely long cylinder, which has a homogeneous metric induced by restricting the Euclidean metric on $\mathbb{R}^3$ to $\{(x,y)|x^2+y^2 = 1\}\times \mathbb{R}$.

We then get to the twice-punctured plane. Does the ability to apply a homogeneous metric end here, or is it also possible for this space?

Ted Shifrin
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volcanrb
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    Homogeneous :) Nice question to ponder. – Ted Shifrin Jun 14 '24 at 19:37
  • The answer should be yes, as the universal covering space is the upper half-plane $\Bbb H$, and the group of deck transformations is a subgroup of $PSL(2,\Bbb Z)$, which is a subgroup of the group of Möbius transformations. These are the isometries of the hyperbolic metric on $\Bbb H$. (For more on this, see this post.) – Ted Shifrin Jun 14 '24 at 19:43
  • @Ted: but hyperbolic surfaces aren't usually homogeneous, right? – Qiaochu Yuan Jun 14 '24 at 20:27
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    Indeed, @Qiaochu, not in the compact case, because there are only finitely many automorphisms of the associated algebraic curve. I should know more, but I don’t. – Ted Shifrin Jun 14 '24 at 20:53

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Not a complete answer. Let me restrict attention to Riemannian metrics; I don't know what the non-Riemannian case looks like. If a Riemannian surface is homogeneous then in particular it must have constant curvature. The classification of surfaces of constant curvature (via the uniformization theorem) implies that the $n$-punctured plane, $n \ge 2$, can only be given a Riemannian metric of constant negative curvature, making it a hyperbolic surface.

I think, but don't know how to prove, that the $n$-punctured plane $X_n, n \ge 2$, as a hyperbolic surface, is never homogeneous. I think Hurwitz's automorphism theorem implies that closed hyperbolic surfaces can only have finitely many isometries, but I don't know how to handle the open case.

Edit: From Moishe's comments, this argument by Ian Agol that the isometry group of a hyperbolic surface is discrete unless it is the disk or annulus solves it.

Qiaochu Yuan
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    It is not homogeneous. One can prove even more, if $S$ is a connected surface which admits a transitive Lie group action, then $\chi(S)\ge 0$. – Moishe Kohan Jun 14 '24 at 21:41
  • @Moishe I was wondering about that in general. It feels like something I knew decades ago. Is this discussed in some MSE post? – Ted Shifrin Jun 14 '24 at 22:23
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    @TedShifrin For general smooth Lie group action, see my answer here. Discreteness of isometry group of a hyperbolic surface of negative Euler characteristic is a very soft fact, see for instance Agol's explanation here. – Moishe Kohan Jun 15 '24 at 02:19
  • @Moishe Many thanks. – Ted Shifrin Jun 15 '24 at 03:12