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Setting. Let $M$ be a Riemann surface and $\Gamma$ a discrete group that acts properly discontinuously on $M$ by holomorphic maps. It is well known that each $x \in M$ has a finite stabilizer, that the points with nontrivial stabilizer form a discrete set $R \subset M$ (ramification points), and that the projection $M \overset \pi \to M/ \Gamma$ is a (holomorphic) covering map outside of $R$.

Moreover, it is a consequence of the uniformization theorem that $M / \Gamma$ has a unique holomorphic structure for which $\pi$ is holomorphic. In particular, $M / \Gamma$ is a smooth manifold. (Note that this is not true in the case of, say the proper group action $\{\pm 1 \} \curvearrowright \mathbb R$ by multiplication: the quotient $\mathbb R / \pm 1 \cong [0, + \infty)$ is not a smooth manifold.)

We have that if $f : M \to \mathbb C$ is $\Gamma$-invariant and holomorphic, it descends to $M / \Gamma$ as a holomorphic function.

Question. What other smoothness properties are preserved under $\pi$? Ex:

  1. If $f : M \to \mathbb C$ is smooth and $\Gamma$-invariant, does it descend to $M / \Gamma$ as a smooth function? (Compare with this MO post)
  2. If $g$ is a Riemannian metric on $M$ and $\Gamma$ acts by isometries, does $g$ descends to $M / \Gamma$ as a smooth metric?

Is there a good reference on such questions? Perhaps in the particular case where $M$ is the complex upper half plane?

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    $\begingroup$ Part 2 fails even for the flat metric on ${\mathbb C}$ when $\Gamma$ is generated by the antipodal involution $z\mapsto -z$. $\endgroup$ – Moishe Kohan Feb 25 at 4:33
  • $\begingroup$ @MoisheCohen You're right, that's interesting. $\endgroup$ – rabota Feb 25 at 8:23
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Partial answer:

Part 1 is true, and follows from the answer to part 3 of this MO post: https://mathoverflow.net/questions/65264/integral-representation-of-higher-order-derivatives. For smoothness of $f$ at a branch point $\pi(p)$, the uniformization theorem allows us to reduce to the case of $\pi : B(0, 1) \to B(0, 1)$ given by $z \mapsto z^p$ and $\Gamma = \mathbb Z/p$ with the action given by multiplication by $p$th roots of unity. A special case of a theorem of G. Schwartz (Smooth functions invariant under the action of a compact lie group, Topology 14, 1975) implies that the $\mathbb Z/p$-invariant smooth functions $B(0, 1) \to \mathbb C$ are given by smooth functions in the invariant polynomials for this action. In this case, they are $Re(z^p)$ and $Im(z^p)$. Hence $f$ is a smooth function of $z^p$.

For part 2, we can reduce in the same way to a Riemannian metric on $B(0, 1)$ and $\pi(z) = z^n$. We may look at the metric as a smooth map $g : B(0, 1) \to SPD_2(\mathbb R)$ that takes values in (wlog) positive definite symmetric matrices. The invariance under $\Gamma$ means that for any $k \in \mathbb Z$ and $z \in B(0, 1)$, $$\zeta_p^{-k} g(\zeta_p z) \zeta_p^{k} = g(z) \tag{1}$$ where we identify a complex number $x+iy$ with its differential, $$\begin{pmatrix} x & -y \\ y & x \end{pmatrix}$$ The question is now equivalent to whether equation $(1)$ implies that $g(z)$ is a smooth function of $z^p$. The same result of Schwartz tells us that the trace and determinant of $g(z)$ are smooth functions of $z^p$, but this is not enough to conclude.

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