I asked this question almost a month ago on Math SE. After waiting three weeks for an answer or a comment, I opened a bounty on the question in hope that it might get an answer this way. The bounty expires in a day, and sadly I didn't get any answer or hint on it until now. However, someone said in a comment that this question might be better suited for MathOverflow, and this is why I decided to post it now on MO. I apologize if this question isn't appropriate for MathOverflow.

For better readability, I decided not to fully copy-paste my question and only write the important parts. For the research and attempt I've done on the question and for further references, I kindly refer to the original question found here:

The original question on Math SE

Setting: It is well known that the Teichmüller space $T_{g,b}$ of an oriented Riemann surface $S_{g,b}$ of genus $g \geq 2$ with $b \geq 1$ geodesic boundary components (satisfying $2g + b \geq 3$) can be globally parametrized by $6g - 6 + 3b$ geodesic length functions, a number which equals its dimension (see Schmutz [1]).

However, in the case of a closed surface $S_{g,0}$ (i.e. without boundary components), the minimal number of geodesic length functions needed for a parametrization of the corresponding Teichmüller space $T_{g,0}$ is $6g-5$ (first proven by Schmutz [1]), which is greater than its dimension (which is $6g-6$). The fact that $6g-5$ is minimal is not proven in Schmutz' paper. Note that Schmutz' given parametrization is to be understood as embedding of $T_{g,0}$ into $\mathbb{R}^{6g-5}$ via the length functions of $6g-5$ geodesics, and not as homeomorphism between $T_{g,0}$ and $\mathbb{R}^{6g-5}$.

Question: The question is why a global parametrization of the Teichmüller space $T_{g,0}$ (of a closed surface $S_{g,0}$) by $6g-6$ geodesic length functions is not possible, i.e. why the number $6g-5$ is minimal.

What I am looking for: A proof to the question or a reference where I can find a proof. Also hints that may lead to a proof would be kindly appreciated. I know from several references (see original question on Math SE) that it has something to do with Scott Wolpert's studies on the convexity of the geodesic length functions and that the result might be stated in terms of trace functions for marked Fuchsian groups (these references also state that Wolpert first found the result). Overall, I'd also be interested in the reason why the parametrization can be done by a number of geodesic length functions that equals the dimension of the Teichmüller space in the case of surfaces with boundary components, but not in the case of closed surfaces.

Edit: Thanks to a comment of Mr. Agol I noticed a mistake in my initial statement of Hamenstädt's result. What is shown in [2] is that for surfaces with $n \geq 1$ punctures, $6g - 5 + 2n$ geodesic length functions provide coordinates on $T_{g,n}$, and not $6g - 6 + 2n$ as I stated. So in case of surfaces with punctures, we have the same situation as for closed surfaces.


[1] P. Schmutz, Die Parametrisierung des Teichmüllerraumes durch geodätische Längenfunktionen, Comment. Math. Helv. 68, 1993, no. 2, 278-288 (found here in german or here in french, sadly not available in english)

[2] U. Hamenstädt, Length functions and parametrizations of Teichmüller space for surfaces with cusps, Ann. Acad. Sci. Fenn. Math. Vol. 28, 2003, 75 - 88 (found here)

  • $\begingroup$ It is $2b$, not $3b$, right? It is complex, so has even dimension. $\endgroup$ Jul 4, 2016 at 15:57
  • $\begingroup$ As far as I know it's $3b$ if we are considering surfaces with geodesic boundary and $2b$ if we are considering surfaces with punctures instead (see e.g. here). $\endgroup$
    – user331406
    Jul 4, 2016 at 16:08
  • $\begingroup$ Since you were using Hamenstädt as a principal reference, I assumed you meant punctures. $\endgroup$ Jul 4, 2016 at 16:20
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    $\begingroup$ Yes you're right, maybe I should add a remark on this. I'm mostly interested in surfaces with boundary components and closed surfaces, as treated in Schmutz' paper $\endgroup$
    – user331406
    Jul 4, 2016 at 16:53

1 Answer 1


I think Scott's argument is that the lengths of $6g-6$ curves can't form coordinates for Teichmuller space. If one has $6g-6$ geodesics which parameterize, then they must be filling (they meet every simple closed curve). But the length of a filling (immersed) curve is proper in Teichmuller space, and hence the infimum of the length is achieved (in fact, at a unique point). Hence the lengths of $6g-6$ geodesics can't form local coordinates near the global (and hence local) infimum of the sums of the lengths.

Now we apply invariance of domain to conclude that we cannot even have a parameterization.

The reason that the geodesics must be filling is that otherwise, there is a disjoint simple closed curve. For any choice of lengths of the curves, we may twist about the disjoint simple closed geodesic, showing that the length of the curve does not uniquely determine the metric.

The reason that the length of the (geodesic representative of) a filling curve is proper is that if we consider the part of Teichmuller space in which the curve has bounded length, then any given (simple closed) curve will have bounded length as well, since it has bounded intersection with the filling curve, and hence can be cut and pasted out of boundedly many pieces of the filling curve. Moreover, there is also a lower bound on the length of simple closed curves by the collar lemma (if a simple closed curve were too short, then it would have a large collar, and force a filling curve to have length greater than the width of the collar). If we take a union of two pants decompositions of the surface which fill, then it is not hard to show that the space of metrics in which the lengths of the cuffs of the pants remain bounded (above and below) lies in a compact part of Teichmuller space (one may use Fenchel-Nielsen coordinates with respect to one pants decomposition, and use the boundedness of the lengths of the other pants decomposition to bound the twist parameters).

  • $\begingroup$ @Agol By this argument, any number of simple closed curves cannot parametrize the Teichmuller space which is clearly false. What am I missing? $\endgroup$
    – Cusp
    Jul 4, 2016 at 18:51
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    $\begingroup$ That's correct. Your phrasing of Schmutz' theorem is unfortunate (although I see that this unfortunate phrasing is present in the Math Review of his paper). He did not prove that $6g-5$ lengths parameterize Teichmuller space in the strong sense of providing a homeomorphism to $\mathbb{R}^{6g-5}$. That's a good thing, because Teichmuller space is homeomorphic to $\mathbb{R}^{6g-6}$ and we would not want to contradict invariance of domain. Instead what he proved is that there exist $6g-5$ length functions which taken together yield an embedding of Teichmuller space into $\mathbb{R}^{6g-5}$. $\endgroup$
    – Lee Mosher
    Jul 4, 2016 at 19:16
  • $\begingroup$ @Agol Thank you very much for your answer. I read about these properties of the length functions before, but I couldn't form them into an argument. I will accept your answer. $\endgroup$
    – user331406
    Jul 5, 2016 at 7:56
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    $\begingroup$ @user331406: For the Teichmuller space of surfaces with boundary, the space is naturally a manifold with corners (the boundary is obtained by setting boundary components to have length 0, and lowest-dimensional corner strata when all boundary components are 0). The infimum of a collection of curves will occur when boundary components have length 0 (I think), hence no contradiction to local coordinatization occurs. As for the case of punctures, I couldn't find it stated in the literature that you referenced that one could parameterize with 6g-6+2b curves (Hamenstadt uses 6g-5+2b). $\endgroup$
    – Ian Agol
    Jul 8, 2016 at 21:10
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    $\begingroup$ I looked at the Seppala-Sorvali and Schmutz references, but I didn't see the statement. They do prove that a non-orient able surface has a coordinatization by 3g-3 curves. Wolpert's argument doesn't give a contradiction, since the curves don't need to fill: if the disjoint curve is 1-sided, you can't twist about it. $\endgroup$
    – Ian Agol
    Jul 10, 2016 at 2:52

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