Using Fenchel-Nielsen coordinates, the Weil-Petersson metric can be written as

$\omega_{WP} = \sum_{i} d\ell_i \wedge d \tau_i,$

where $i$ is an index labelling the curves of a pants decomposition of the surface in question and the pairs $(\ell_i,\tau_i)$ form the associated Fenchel-Nielsen coordinates. Although it is not manifest from the above expression, $\omega_{WP}$ is actual independent on the choice of pants decomposition.

Now, in this question it was asked and answered how the Fenchel-Nielsen coordinates change upon performing the so-called A-move or S-move on the pants decomposition. The answer is provided in this paper (and references there) by explicit transformation laws.

Let us consider the case of a once punctured torus, then a pants decomposition consists of a single curve and therefore we have $\omega_{WP} = d\ell \wedge d\tau$. I tried to check that $\omega_{WP} = d\ell \wedge d\tau = d\ell' \wedge d\tau'$, using the expressions given by Proposition 3.1 of the above paper (and setting $\ell_0 = 1$), but the equality does not seems to hold. Indeed, by putting numerical values one easily finds that the determinant of the transformation is not 1.

Where is the problem?

(Let me also add that it seems to me that there is a typo in the expression for the twist parameter, the last factor should be $\{ \dots \}^{(-1/2)}$ in order to agree with the same expression given in terms of the quantities $A$ and $B$ at pag. 7).


1 Answer 1


Apparently both the original paper by Okai and Proposition 3.1 in the paper mentioned by the OP contain typos in the expressions. Corrected formulas can be found in equations (5.13) and (5.22) of this recent arXiv submission (see also footnote 9 on page 75):

Jørgen Ellegaard Andersen, Gaëtan Borot, Séverin Charbonnier, Alessandro Giacchetto, Danilo Lewański, Campbell Wheeler, On the Kontsevich geometry of the combinatorial Teichmüller space, arXiv:2010.11806

Starting with their expressions it is easily checked with Mathematica that the determinant of the transformation is equal to 1:

Det@D[{2 ArcCosh[ Cosh[t/2]/Sinh[l/2] Sqrt[(Cosh[l] + Cosh[l0/2])/2]], 
      -2 ArcCosh[ Cosh[l/2] Sqrt[ 
           ((Cosh[t/2]^2 (Cosh[l] + Cosh[l0/2]) - 2 Sinh[l/2]^2) / 
           (Cosh[t/2]^2 (Cosh[l] + Cosh[l0/2]) + Sinh[l/2]^2 (Cosh[l0/2] - 1)))]]},
      {{l, t}}] // FullSimplify[#, t > 0 && l > 0 && l0 > 0] &

  • $\begingroup$ Another doubt: if you set the boundary lengths to 0 (or just equal) in the formula for the sphere, the formula becomes completely symmetric in i=1,2,3,4. But then it says that the length of the curve separating 12|34 and 13|24 are equal which doesn't make sense (for instance you could send them both to 0 simultaneously although they intersect). How is it possible? $\endgroup$ Aug 13, 2021 at 12:55
  • $\begingroup$ Can you elaborate? What formulae are you referring to? $\ell'(\ell, \tau)$ should depend non-trivially on $\tau$. $\endgroup$ Aug 14, 2021 at 5:51
  • $\begingroup$ I was thinking of formula 5.12, which gives $\ell'(\ell,\tau)$, where $\ell'$ is the length of the curve 12|34 and $\ell$ that of the curve 14|23. What I meant is that the formula seems to give the same result for the curves 12|34 and 13|24, because it is symmetric in the four boundary components and I would imagine that the formula giving the length $\ell''$ of the curve 13|24 is given by the same formula upon permuting the boundary labels? $\endgroup$ Aug 14, 2021 at 20:05
  • $\begingroup$ You cannot simply permute the boundary components, because you would have to transform $\tau$ accordingly as it is tied to the boundary components. $\endgroup$ Aug 15, 2021 at 5:06
  • $\begingroup$ But then, is there a formula that gives both $\ell'$ and $\ell''$ using a single $\tau$? After all, if $\ell$ and $\tau$ are fixed one should be able to write down the lengths of all geodesics in the sphere $\endgroup$ Aug 15, 2021 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.