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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 ?

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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

1 Answer 1

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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.

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