# Adding segments to an annulus - a question regarding the conformal modulus.

Let $A \subset \mathbb{C}$ be an topological annulus, i.e. a region of $\mathbb{C}$ bounded by two disjoint Jordan curves.

Let $B \subset \mathbb{C}$ be a quadrilateral, i.e. a topological disc with four distinct marked points $(z_1,z_2,z_3,z_4)$ arranged anticlockwise on the boundary.

Both annuli and quadrilaterals, as defined above, have a conformal invariant, in both cases known as the modulus. Let $A_R$ be a geometrical annulus with inner boundary a circle of radius $1$ and outer boundary a circle of radius $R$. If $A$ can be mapped conformally and bijectively onto $A_R$ we say that $A$ has modulus $\ln{R}/2\pi$. Likewise, let $Q_m$ be a geometrical rectangle with vertices $(0, m, i+m, i)$, we say that $B$ has modulus $m$ if there exists a conformal bijection mapping $B$ onto $Q_m$ with $z_1$,$z_2$,$z_3$, and $z_4$ being mapped onto $0$, $m$, $i+m$, and $i$ respectively. The two concepts are linked by the fact that if we take a geometric annulus centred on the origin with modulus $m$, remove all the points on the positive real line from the annulus, and then take the preimage under the exponential map, we obtain (infinitely many copies of) a geometric rectangle also with modulus $m$.

My questions relate to the degree in which the choice of branch cut is arbitrary.

More particularly:

Let $C \subset \mathbb{C}$ be another annulus and represent $C$ as the union $R_1 \cup R_2$ of two simply connected regions with the intersection $R_1 \cap R_2$ consisting of two smooth simple curves $\gamma_1$ and $\gamma_2$, disjoint from each other and each of which has one end point, $x_1$ and $x_2$ respectively, on the inner boundary and one end point, $y_1$ and $y_2$ on the outer boundary. Assume we have a continuous function $f_1$ mapping $R_1$ onto $A$, and such that the restriction of $f_1$ to the interior of $R_1$ is a conformal bijection and such that the image of $\gamma_1$ and $\gamma_2$ is a single simple curve from the inner to the outer boundary of $A$, with $f_1(x_1) = f_1(x_2)$ and $f_1(y_1) = f_1(y_2)$. Similarly assume that $f_2$ is a bijection mapping $R_2$ onto $B$, conformal on the interior, with $\gamma_1$ being mapped onto the component of the boundary joining $z_1$ and $z_2$ and $\gamma_2$ being mapped to the component of the boundary joining $z_3$ and $z_4$. See the diagram below.

My first question is whether the modulus of $C$ is determined entirely by the modulus of $A$ and $B$? (This seems unlikely to me, but no harm in asking)

Secondly, if this is not the case can we give any bound for the modulus of $C$ given the modulus of $A$ and $B$.

Thirdly, are there any conditions we can impose to reduce these bounds? For example specifying that the curves $\gamma_1$ and $\gamma_2$ meet the boundary curves of $C$ orthogonally.

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A very similar, but somewhat easier question, is the following: take $R$ a rectangle, say $[0,m]\times[0,1]$ identifying $\mathbb C$ to $\mathbb R^2$ for nicer notation. This has modulus $m$. Now let $\gamma$ be a simple curve connecting the top to the bottom sides of $R$, while staying within the rectangle. It splits it into two simply connected domains $R_1$ and $R_2$ that are topological rectangles. So each one has a modulus, say the moduli are $m_1$ and $m_2$. Do $m_1$ and $m_2$ determine $m$?

The answer is no here, because of the following two examples.

First, looking at $[0,m_1+m_2]\times[0,1]$ split vertically by $\{m_1\} \times [0,1]$, if the answer was yes, then it would have to be $m=m_1+m_2$.

Second, one can split $[0,m]\times[0,1]$ using a curve that goes back and forth many times between the neighborhood of $\{0\}\times[0,1]$ and the neighborhood ot $\{m\}\times[0,1]$. This will make both $m_1$ and $m_2$ very small, and is incompatible with their sum being equal to $m$.

Something very similar can be done in your case: take $\gamma_1$ straight, and $\gamma_2$ going back and forth between the left and right sides of $\gamma_1$. This will make the moduli of both $A$ and $B$ very small.

One thing which on the other hand is true, in the simply connected case at least, is a sub-additivity relation: with the above notation, one always has $m \geqslant m_1 + m_2$. One way to prove this is via extremal length, as follows. (Not sure the terminology is completely standard.)

Let $\rho : R=[0,m]\times[0,1] \to \mathbb R_+$. The $\rho$-length of a rectifiable curve $\gamma$ is $L_\rho(\gamma) := \int_\gamma \rho ds$ (in terms of the curvilinear coordinate). The $\rho$-width $W_\rho(R)$ of $R$ is the shortest $\rho$-length of a curve connecting the two vertical sides of $R$. The $\rho$-area of $R$ is $A_\rho(R) = \int_R \rho^2 |dz|$ (with respect to the Lebesgue measure on $R$). Then, the extremal width of $R$ is defined as $$W(R) := \sup_\rho \frac {W_\rho(R)^2} {A_\rho(R)}.$$

It is easy to define this for any topological rectangle, and to check that $\rho$ is conformally invariant. In particular it has to be a function of the modulus. In the case of the rectangle, the supremum is reached when $\rho$ is constant, in which case $W_\rho(R)=m$ and $A_\rho(R)=m$, leading to $W(R)=m$. In other words: the extremal width is just the same as the modulus.

But now, given a rectangle $R$ split into two rectangles $R_1$ and $R_2$, one can put maximizing functions $\rho_1$ and $\rho_2$ for the previous variational problem on them. This defines $\rho$ on $R$, which does not do better than the optimal one for $R$: the last sentence is exactly the inequality $m \geqslant m_1 + m_2$.

Maybe things have to be tweaked a little bit in your case, because you are gluing objects of different topologies, but the same philosophy will apply. Given more geometrical information, one might also use extremal length to derive better bounds.

Reference for all that: Ahlfors, Lars V. (1973), Conformal invariants: topics in geometric function theory, New York: McGraw-Hill Book Co., MR 0357743.

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