Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi, I was thinking about the following question ; I will appreciate it if somebody can give me a full or partial answer or can at least cite any reference(s)/ papers etc :

By $ \bar{P} $ , we denote a topological pair of pants ( that is, a 2-sphere with three open disks removed ) with boundary, and by $P$, we would mean the same without any boundary. Let $\mathcal {Mod(\bar{P})}$ denote the space of all complete hyperbolic metric on $ \bar{P} $ which make the boundary components of $P$ geodesics ; call these geodesics $\gamma_1, \gamma_2, \gamma_3 $ . Since these metrics are completely determined by the hyperbolic lengths $ l_1, l_2, l_3 $ of the three boundary components, we can give $ \mathcal { Mod(\bar{P} ) } $ the co-ordinates $ l_1, l_2, l_3 $ and so think of $ \mathcal { Mod(\bar{P} ) } $ as $\mathbb{R_+}^3$. Consider the following map $ F : \mathcal {Mod(\bar{P})} \cong \mathbb{R_+}^3 \to \mathbb{R_+}^3 $ defined as follows :

$m_i$ is the module of the maximal ring domain in $ \bar{P} $ whose core curve is homotopic to $ \gamma_i $, i.e. for all ring domain $R_i$'s whose core curve is homotopic to $ \gamma_i $, we have $ m_i = sup mod (R_i) $ and this supremum is achieved by a ring domain, which is the the pair of pants $\bar{P}$ minus the unique geodesic joining $ \gamma_{i+1 modulo 3}, \gamma_ {i+2 modulo 3} . $. Now define the map : $ F : \mathcal {Mod(\bar{P})} \to \mathbb{R_+}^3 $ by $ F(l_1,l_2,l_3) = (m_1,m_2,m_3) $. I was wondering whether this map is a homeomoprhism of $ \mathbb{R_+}^3 $. I guess constructing $F$ explicitly would be really hard. Is there any literature along this line ?

share|improve this question
You mean, "$m_i$ is the modulus of the maximal ring domain . . . " (not "module"), right? –  macbeth Mar 10 '12 at 16:46
I think different authors use different terminologies, module is also in use , for example see Lehto and Virtanen : Quasiconformal Mappings in the plane and also Ahlfors : Lectures on Quasiconformal Mappings. However, there is a possibility that the term 'module' might be a bit old. –  Analysis Now Mar 10 '12 at 22:41
add comment

1 Answer

up vote 3 down vote accepted

I think the map $F$ is injective, but not surjective.

There is a unique conformal map of the interior of the pants to the complement of 3 slits in $\mathbb{RP}^1\subset \mathbb{CP}^1$, up to the action of $PGL_2(\mathbb{R})$. If we parameterize the slits as $[a_0,a_1],[a_2,a_3],[a_4,a_5]$, where $a_0 < a_1 < a_2 < a_3 < a_4 < a_5$ in $\mathbb{R}$, then the configuration is invariant under complex conjugation, which induces a hyperbolic isometry of the pants. So the pant seams are the intervals $[a_1,a_2], [a_3,a_4], [a_5,a_0]$ taken in cyclic order on the circle $\mathbb{RP}^1$. In fact, one may think of the map as "sewing" the cuffs of the pants along the two intervals connecting the seams. The rings $R_i$ are the complements of the slits $[a_i,a_{i+1}]\cup [a_{i+3},a_{i+4}]$, indices taken $(\mod 6)$ (where $R_i \cong R_{i+3}$). The points are uniquely determined by the modulus of $R_i$, up to the action of $PSL(2,\mathbb{R})$, or by the cross-ratio $[a_i,a_{i+1};a_{i+3},a_{i+4}]$.

Therefore the modulus of each ring determines uniquely another ring $U_i$ which is the complement of the slits $[a_{i+1},a_{i+3}]\cup [a_{i+4},a_i]$ (where the intervals are determined by the cyclic order and orientation of $\mathbb{RP}^1$). Then $R_{i+1} \supset U_i$, $R_{i+2}\supset U_i$, as homotopy equivalences. So the moduli of the rings $R_{i+1}$ and $R_{i+2}$ bound the modulus of the ring $U_i$, by the monotonicity of moduli of annuli, which uniquely determines the modulus of $R_i$. So this shows one cannot achieve all possible triples of moduli, so the map $F$ is not surjective.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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