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Let me define a degree $n$ colored arrangement of circles on $S^2$ to be a collection $\mathcal{C}$ of $n$ disjoint, smoothly embedded circles $C_1,\dotsc, C_n\subset S^2$ together with a continuous map (temperature)

$$ T: S^2 \setminus (C_1\cup\cdots \cup C_n)\to \lbrace 1,-1\rbrace $$

such that for any circle $C_k$, the temperatures on opposite sides of $C_k$ are different. In other words, if $\gamma:(-\varepsilon,\varepsilon)\to S^2$, $t\mapsto \gamma(t)$, is a short smooth path that intersects exactly one circle $C_k$, and it does so transversally at $t=0$, then

$$\lim_{t\searrow 0} \; T\bigl(\; \gamma(t)\;\bigr)= - \lim_{t\nearrow 0}\; T\bigl(\;\gamma(t)\;\bigr). $$

Think of the components of $T^{-1}(-1)$ as icy regions and of the components of $T^{-1}(1)$ as liquid regions. We will refer to $T$ as a coloring of the arrangement of circles $\mathcal{C}$.

Two arrangements of circles $\mathcal{C}$ and $\mathcal{C}'$ are called equivalent if there exists an orientation preserving diffeomorphism of $S^2$ that maps one arrangement to the other. We denote by $\mathcal{A}_n$ the set of equivalence classes of arrangements on $n$ disjoint circles on $S^2$.

Two degree $n$ colored arrangements $(\mathcal{C}, T)$ and $(\mathcal{C}', T')$ are called equivalent if there exists an orientation preserving diffeomorphism $\Psi: S^2\to S^2$ mapping the circles in $\mathcal{C}$ to the circles in $\mathcal{C}'$ and such that $T'\circ \Psi= T$.

Two colorings of a given arrangement $\mathcal{C}$ of $n$ disjoint circles on $S^2$ are called equivalent if the colored arrangements $(\mathcal{C}, T)$ and $(\mathcal{C}, T')$ are equivalent. We get in this fashion a map

$$ K_n : \mathcal{A}_n\to\mathbb{Z}$$

that associates to an arrangement $\mathcal{C}$ the nummber of inequivalent colorings of $\mathcal{C}$. Here are some questions I find interesting.

$\mathbf{Q_0}$ Find the cardinality of $\mathcal{A}_n$ ans its large $n$ asymptotics.

$\mathbf{Q_1}$ Find the cardinality of $\mathcal{S}_n$ and its large $n$ asymptotics.

$\mathbf{Q_1^*}$ Investigate the map $K_n$.

$\mathbf{Q_2}$ We declare an colored arrangement $(\mathcal{C}, T)$ to be selfdual if it is equivalent to $(\mathcal{C}, -T)$. Find the number of selfdual colored arrangements of degree $n$ and its large $n$ asymptotics.

$\mathbf{Q_3}$ Equip $\mathcal{S}_n$ with the uniform probability density and denote by $X_n:\mathcal{S}_n\to \mathbb{R}$ the random variable o that associates to a colored arrangement $(\mathcal{C}, T)$ the Euler characteristic of the liquid region $T^{-1}(1)$. Denote by $\mu_n$ the probability distribution of $X_n$. Investigate the large $n$ behavior of $\mu_n$.

$\mathbf{Q_4}$ Can you think of a good notion of a random arrangement of $n$ disjoint circles on $S^2$?

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up vote 5 down vote accepted

Consider the graph obtained by assigning to each region a vertex, and two vertices being joined by an edge iff the two regions are neighbors. It is easy to see that this is a tree, with the empty circles as terminal vertices: once you crossed a circle, the only way to return to the same region is by the same edge.

The tree isomorphism corresponds to the equivalence of arrangements of circles.

So the cardinality of $\mathcal A_n$ is the same as for the unlabeled trees. In the same wiki reference is mentioned he asymptotic estimate given by Otter (1948). For the cardinality of $\mathcal A_n$ is no known formula.

For the $\mathcal S_n$, there are only two ways to color a tree so that each edge connects two vertices of opposite colors. They may be not distinct (i.e. may be selfdual), since it is possible to have a tree automorphism which switches the colors. The asymptotics is similar to the uncolored case, because there are only one or two distinct colorings of the same tree.

The notion of randomness for trees depends on the purpose.

The answers are partial, mainly because the answers to the corresponding questions for the trees are partial known, but I hope that the connection with trees may help.

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Thank you! Very useful answer. –  Liviu Nicolaescu Mar 16 '12 at 19:12
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