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Let $S$ be a Riemann surface. What can be said of the greatest integer $n$ such that the group of biholomorphisms of $S$, $\mathrm{Aut}(S)$, acts $n$-transitively on $S$ ?

(for the Riemann sphere, it is 3 for instance)

In particular, is there any easy way to see it is always greater than one ?

Edit : by Riemann surface, I mean connected complex holomorphic 1-dimensional manifold

Albert
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  • You'd better at least assume $S$ is connected. – Chris Eagle Jul 09 '13 at 15:54
  • According to the definition of Riemann surface I use, it is connected. I'll clarify it in the question – Albert Jul 09 '13 at 15:58
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    And, if you remove $k > 3$ points from the sphere, you get a Riemann surface that in general has trivial automorphism group. So it's not always greater than $1$. – Daniel Fischer Jul 09 '13 at 15:59
  • Interesting ! I guess the reasonning is that if you have an automorphism, it somehow extends to a Möbius transformation, which must permute the removed points, which is not always possible ? Does that mean that this number depends on the conformal structure more than the topology ? – Albert Jul 09 '13 at 16:03
  • Right. You have isolated singularities, and by the injectivity of $f \in \operatorname{Aut}(S)$, these must be removable singularities (viewing a pole as a removable singularity, since we're dealing with maps to the sphere). So it's not a topological invariant, but a biholomorphic invariant. – Daniel Fischer Jul 09 '13 at 16:06
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    Also, in $\operatorname{Aut}(\mathbb{D})$, you can only map two pairs of points to each other if the hyperbolic distance of the points in the pairs is the same, so $\operatorname{Aut}(\mathbb{D}\setminus {0})$ consists just of the rotations, and you have a surface with nontrivial automorphism group which doesn't act transitively. $\operatorname{Aut}(\mathbb{C}\setminus {0})$ on the other hand acts transitively. – Daniel Fischer Jul 09 '13 at 16:11
  • OK, so it can be $0,1,2,3$. I don't think it can be more than 3. So is there an easy way to compute it ? – Albert Jul 09 '13 at 16:47

2 Answers2

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If the automorphism group of a Riemann surface is transitive, the it must be of uncountable size since for any point p there must exist a distinct group element g(p,q) to carry it to each point q.

For a compact Riemann surface M_g of genus g >= 2, the size of the automorphism group Aut(M_g) is bounded above by a theorem of Hurwitz:

|Aut(M_g)| <= 84(g-1).

This shows Aut(M_g) cannot be transitive for g >= 2.

Also, even though the automorphism group of the disk is 1-transitive, as mentioned above ... but if Aut(D) is extended to act on S^1 = ∂D, then the action restricted to S^1 is in fact 3-transitive.

Dan Asimov
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I just remarked this : if $M$ is a hyperbolic Riemann surface, then $\mathrm{Aut}(M)$ cannot act 2-transitively on $M$.

Indeed, if $M$ is hyperbolic then its automorphisms are isometries for the hyperbolic metric, and so map couple of points to couple of points at the same distance. So $n=0$ or $1$ for all hyperbolic Riemann surfaces.

Besides, the non-hyperbolic Riemann surfaces are the plane, the disk, the sphere, the annuli and the torii.

  • In the case of the sphere, $n=3$

  • In the case of the plane, $n=2$

  • In the case of the disk, $n=0$

  • In the case of the annulus, $n=0$ (there are only rotations)

  • In the case of the torus, the automorphisms are the translations and the multiplications by units of the ring $\mathbb{Z}+ \tau \mathbb{Z}$. So $n \geq 1$ and I'm not sure yet if there is equality.

Dan Asimov
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Albert
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  • ℤ + ℤ is very rarely a ring, since it is very rarely closed under multiplication. For example if = 1 + i, then its square is not in ℤ + ℤ. – Dan Asimov Nov 09 '23 at 04:11