Let $\Sigma$ be a closed orientable surface of genus $k \geq 2$. Suppose $U \subseteq \Sigma $ is an open subset with a Riemannian metric $g$ on $U$ such that (1) the Gaussian curvature $K$ of $g$ is constantly equal to $-1$ and (2) the volume form for $g$ on $U$ extends to a smooth $2$-form $\omega \in \Omega^2(\Sigma)$ that vanishes on $\Sigma \setminus U$.

In this case, is the volume of $U$ bounded by the volume of a hyperbolic metric (i.e. one having constant Gaussian curvature $K = -1$) on $\Sigma$? More specifically, by the Gauss-Bonnet theorem, bounded by $-2\pi \chi(\Sigma)=2\pi(2k-2)$.

I am also curious about necessary topological properties of the compact subset $\Sigma \setminus U$ that follow from properties (1) and (2).

  • $\begingroup$ What you're asking for with the extension seems unlikely. You're trying to have the volume form decay smoothly to $0$ as you approach $\partial U$ and yet maintain $K=-1$? Are you sure you can do this? $\endgroup$ Commented Jan 8, 2023 at 23:10
  • $\begingroup$ @TedShifrin This predicament has come about quite naturally. Given these assumptions on $\Sigma$ and $U$ I have tried to draw conclusions about the topology of $U$ (and $\partial U$) but have so far been unsuccessful. I have seen some work by Goldman, Hitchin, etc. that imply $\partial U$ may be a finite set of "cone singularities" where the metric degenerates (as in section 4 of Goldman's "Geometric Structures and Varieties of Representations" survey) but am not yet certain this is the case. $\endgroup$
    – Math
    Commented Jan 9, 2023 at 18:35

1 Answer 1


In fact, there are even complete hyperbolic metrics satisfying (1) and (2) and having arbitrarily large area. Here is a construction. Let $P\subset \Sigma$ be a subset consisting of $n$ distinct points, and let $U:=\Sigma \setminus P$. Then $U$ admits a complete metric $g$ of constant curvature $-1$ and finite area equal $-2\pi \chi(U)=\pi(2k-2) + 2\pi n$. Of course, $g$ itself is unlikely to satisfy (2). Here is how to modify it:

Let $\omega$ denote the volume/area form of $g$. There exists a 2-form $\omega'\in \Omega^2(\Sigma)$ such that:

(a) $\int_\Sigma \omega'= -2\pi \chi(U)$.

(b) $\omega'$ vanishes on $P$.

(c) $\omega'$ is a volume form on $U$, meaning that it does not vanish at any point of $U$.

The construction of $\omega'$ is an elementary exercise in differential topology and I will skip it.

The form $\omega'$ at this moment has nothing to do with $\omega$ except they have the same integral over $U$. According to a generalization of Moser's theorem due to Greene and Shiohama,

R. E. Greene and K. Shiohama, Diffeomorphisms and volume-preserving embeddings of noncompact manifolds, Trans. Amer. Math. Soc. 255 (1979), 403-414.

there exists a diffeomorphism $f: U\to U$ such that $f_*(\omega)=\omega'$. Now, take the Riemannian metric $g':= f_*(g)$ on $U$. This metric has the area form equal to $\omega'$ and, hence, meeting all your requirements.

Remark. The push-forward of a covariant tensor $T$ by a diffeomorphism $f$ of a smooth manifold is the same as $(f^{-1})^* T$.

As for your last question about restrictions on the subsets $\Sigma\setminus U$, there is none (as long as you are not requiring connectedness of $U$). The proof is similar to the one I gave above, just you have to allow incomplete metrics on $U$.

  • $\begingroup$ Thanks so much for this. It has helped me to realize the situation I have is slightly different from what I originally asked. I will open a new question. $\endgroup$
    – Math
    Commented Jan 9, 2023 at 19:52

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