# Which smoothness properties are preserved under ramified covering maps?

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?

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

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.