How to determine bounds on the extremal length around annuli?

I wish to determine bounds for the sum of moduli of a family of topological annuli in the complex plane. Towards that end I would like to ask a question about the closely related concept of extremal length.

Let a quadrilateral $Q$ be a simply connected region of the complex plane, with Jordan curve boundary, and with two disjoint Jordan arcs $\gamma$ and $\gamma'$ making up part of the boundary. The two curves are each parameterised injectively by the closed unit interval, with the image $\gamma(t)$ (or $\gamma'(t)$) moving anticlockwise as $t$ increases. It is well known that for every quadrilateral there exists a unique $k \in \mathbb{R^+}$ and unique conformal homeomorphism $\phi$ mapping $Q$ onto the interior of a geometrical rectangle with vertices $0$, $1$, $1+ki$, and $ki$, and with $\gamma$ and $\gamma'$ mapped to the top and bottom edges respectively. In such a case the extremal length of curves joining $\gamma$ and $\gamma'$ in $Q$ is $k$. If the quadrilateral $Q$ satisfies the further conditions that $Re(\phi(\gamma(t))) = 1-t$ and $Re(\phi(\gamma'(t))) = t$ we say that $Q$ is a rectangle.

We construct a family of annuli by gluing together a collection of rectangles and generic quadrilaterals. For each $n$ let $(R^n_1, R^n_3, \ldots, R^n_{2m-1})$ be an m-tuple of rectangles, with extremal lengths $k_1n, k_3n, \ldots, k_{2m-1}n$ respectively, where each $k_i$ is a positive scalar fixed for all $n$. Likewise, let $S_2, S_4, \ldots, S_{2m}$ be a sequence of generic quadrilaterals. For each $n$, we form an annulus by taking the first element $R^n_1$ in our sequence of rectangles and identifying the edge $\gamma'_1$ of $R^n_1$ with the edge $\gamma_2$ of $S_2$, identifying $\gamma'_1(t)$ to $\gamma_2(t-1)$, for each $t$. We then identify the edge $\gamma'_2$ of $S_2$ with the edge $\gamma_3$ of $R_3$, again identifying $\gamma'_2(t)$ to $\gamma_3(t-1)$, and so on, finally identifying the edge $\gamma'_{2m}$ of $S_{2m}$ with $\gamma_1$ of $R^n_1$.

Intuitively, it seems to me, as $n$ increases one should expect the sections of the annuli made up of rectangles to predominate over those sections made up of generic quadrilaterals. See the figure below, where the red shaded regions represent images of the rectangles. Thus for $n$ large one would expect the extremal length around the annuli $A_n$ to closely approximate the sum $(k_1 + k_3 + \ldots + k_{2m-1})n$. Could someone confirm that this is indeed the case, and if it is the case suggest a reference or some suggestions as to how one might go about proving it? (source)

• You do not want $\gamma,\gamma'$ to be closed curves, you want them to be embedded arcs. – Lee Mosher Aug 3 '12 at 21:22
• And when you speak about "generic" quadrilaterals $S_2,...$, do you perhaps mean, as your picture suggests, that the similarity class of $S_2$ is fixed, etc.? – Lee Mosher Aug 3 '12 at 21:28
• Hi Lee. You're quite right on the first point, I'll edit that straight away. On the second point, I only mean to imply by the use of "generic" that the parameterisation of the arc does not necessarily induce the linear parameterisation needed to call it a rectangle, as defined above. For a given $k$ the parameterisation of $\gamma$ and $\gamma'$ induces an equivalence relation between quadrilaterals. I'll think about how to express this more clearly in the question. – uncooltoby Aug 3 '12 at 22:42
• About the parameterization, in the definition of a "quadrilateral" that I usually see, $\gamma$ and $\gamma'$ do not come equipped with any special parameterization. Of course, a parameterization emerges under the conformal map to the standard rectangle, but otherwise it is not really relevant to your problem. – Lee Mosher Aug 3 '12 at 23:04
• You're right that my definition is a bit non-standard. However, we need this extra condition so that the extremal length of two glued together quadrilaterals is well determined. – uncooltoby Aug 3 '12 at 23:58

This is a comment, but it is a bit too long, so I will post it as an answer. At best, this is the restatement of the obvious proof strategy.

Let $A$ be an annulus, and let $\Gamma_1$ be the family of curves homotopic to the generator of the fundamental group of the annulus.
From the definition,

$$Mod(A)^{-1} = {\rm Ext}(\Gamma_1) = \sup_{\rho} \inf_{\gamma \in \Gamma_1} \frac{\ell_\rho(\gamma)^2}{{\rm Area}_\rho(A)}$$,

where the supremum is over the metrics in the conformal class of $A$. This means that if you do the calculation for a specific metric $\rho$, you get a lower bound for ${\rm Ext}(\Gamma_1)$ which gives you an upper bound for ${\rm Mod(A)}$. The picture you draw is suggestive for the possible choices of $\rho$.

To get lower bounds, one way is to use the identity

$$Mod(A) = {\rm Ext}(\Gamma_2) = \sup_{\rho} \inf_{\gamma \in \Gamma_2} \frac{\ell_\rho(\gamma)^2}{{\rm Area}_\rho(A)}$$,

where $\Gamma_2$ is the family of curves connecting the boundaries of the annulus. Now doing the calculation for a specific $\rho$ gives you a lower bound for ${\rm Mod(A)}$.

Perhaps by choosing the metrics to some interpolated versions of the flat metrics on the rectangles you can get reasonable upper and lower bounds.

• Cheers Alex, this might have just been a 'restatement of the obvious proof strategy', but being new to this sort of thing I hadn't in fact considered determining bounds on the extremal length across as opposed to around the annuli. Doing this with the obvious flat metric on the rectangular regions , as you say, gives the required lower bounds. The upper bounds follow from the standard additive property of extremal length - namely that the extremal length around the annuli is always greater than or equal to the sum of the extremal length of its constituent parts. – uncooltoby Aug 13 '12 at 1:23
• One extra comment though, your definition of extremal length above is wrong. It should include 'sup inf' instead of 'sup sup', perhaps you or someone else with the required privileges would like to change it? – uncooltoby Aug 13 '12 at 1:27
• Typos fixed, thanks. I am glad you got it to work! – Alex Eskin Aug 13 '12 at 7:40