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Most mathematicians are aware that our species consists of two genders, denoted for simplicity by the multisets $\lbrace X,X\rbrace$ and $\lbrace X,Y\rbrace$, with offspring given by $\lbrace A,B\rbrace$ for $A\in\lbrace X,X\rbrace$ and $B\in \lbrace X,Y\rbrace$.

I am asking for the existence of combinatorial structures generalizing this construction.

More precisely, given a finite set $\mathcal C$ of $n$ distinct elements ($n=2$ and $\mathcal C=\lbrace X,Y\rbrace$ in the above example) and an integer $k$, we denote by $\mathcal M_k$ the set of all multisets containing exactly $k$ not necessarily distinct elements of $\mathcal C$.

A gender partition of a subset $\mathcal S\subset \mathcal M_k$ is a partition of $\mathcal S$ into $k$ non-empty parts $\mathcal G_1,\dots,\mathcal G_k$ called genders such that the following two conditions hold:

(i) Given $(g_1,\dots,g_k)\in\mathcal G_1\times \dots \times \mathcal G_k$, every element $(x_1,\dots,x_k)\in g_1\times \dots\times g_k$ gives rise to a multiset $\lbrace x_1,\dots,x_k\rbrace$ which is in $\mathcal S$.

(ii) Every multiset $g\in\mathcal S$ is of the form $\lbrace x_1,\dots,x_k\rbrace$ for $(x_1,\dots,x_k)\in g_1\times \dots\times g_k$ where $g_i\in \mathcal G_i$ are suitable elements.

Examples with $\mathcal C=\lbrace X,Y\rbrace$ are:

(a) $\mathcal S= \lbrace X,X\rbrace\cup \lbrace X,Y\rbrace$ with $\mathcal G_i$ given by singletons.

(b) $\mathcal S= \lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace \cup \lbrace X,Y\rbrace$ with $\mathcal G_1=\lbrace \lbrace X,X\rbrace,\lbrace Y,Y\rbrace\rbrace$ and $\mathcal G_2$ consisting of $\lbrace X,Y\rbrace$.

(c) An example with $k=3$ (easily generalizable to arbitrary values of $k$) is given by $\mathcal S=\lbrace \lbrace X,X,X\rbrace,\lbrace X,X,Y\rbrace, \lbrace X,Y,Y\rbrace\rbrace$ with $\mathcal G_i$ given by singletons.

More examples of gender partitions $\mathcal S=\mathcal G_1\cup \dots\cup \mathcal G_k$ are fairly easy to construct. (And there are fairly easy notions for "products", "quotients", one can split an element of $\mathcal C$ into several new elements, etc.)

The following additional condition is more difficult to satisfy:

Call a gender partition $\mathcal S=\mathcal G_1\cup \dots\cup\mathcal G_k$ balanced if $\mathcal S$ admits a stationary probability measure $\mu$ giving equal weight $\frac{1}{k}$ to all genders $\mathcal G_i$. A probability measure $\mu$ on $\mathcal S$ is stationary if the probabiliy $\mu(\lbrace x_i,\dots,x_k\rbrace)$ of every offspring of $(g_1,\dots,g_k)$ (with respect to uniform choices for $x_i\in g_i$) is proportional to $\prod_{i=1}^k\mu(g_i)$. (Stationary probability measures exist always and are unique if $\mathcal S$ is minimal in some sense.)

Example: The examples (a) and (b) above are balanced, (c) is not balanced.

Question: Produce other examples of balanced gender partitions. Is there for example a balanced gender partition for $k=3$?

Remark: One can also consider probabilities on offsprings which depend on the choice of $x_i\in g_i$. Example (c) is not balanced even in this more general framework.

Variation: Instead of working with multisets, one can also work with sequences of length $k$. An offspring of $k$ sequences $g_1=(g_1(1),\dots,g_1(k)),\dots,g_k=(g_k(1),\dots,g_k(k))$ with $g_i\in\mathcal G_i$ is then given by $g_1(\sigma(1)),\dots,g_k(\sigma(k))$ where $\sigma$ is a (not necessarily arbitrary) permutation of $\lbrace 1,\dots,k\rbrace$.

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Your first phrase is Jane Austen material! – Mariano Suárez-Alvarez May 11 '10 at 16:07
In other universes, they are not restricted to a finite set of genders, and I've heard that the mating rituals there are quite a party. – Joel David Hamkins May 11 '10 at 16:14
I have considered adding the case of infinitely many genders (and/or infinite sets $\mathcal C$) but I fear that the measure part gets less trivial. Concerning, the party, practical details for the construction of offsprings are off-topic. – Roland Bacher May 11 '10 at 16:27
Note that the XY system is not the only one observable on earth. While two sexes or only one sex are the most represented options, you have subvariants. ==== Extract from Wikipedia: "In Lepidoptera (moths and butterflies), examples of Z0, ZZW and ZZWW females can be found. This suggests that the W chromosome is essential in female determination in some species (ZZW), but not in others (Z0)." – ogerard May 11 '10 at 16:39
There is also a precedent in Isaac Asimov's novel "The Gods themselves" where an extra-universe species has three different genders. – ogerard May 11 '10 at 16:43
up vote 5 down vote accepted

So, I hope I understand the definitions correctly. Here's a way to construct an example with $k = 3$ genders (say A, B and C) using $n = 9$ sex chromosomes, which I will take to be the elements of $\mathbb{Z}/9\mathbb{Z}$: start with all 165 multisets of chromosomes unused. Choose any unused multiset $\{x, y, z\}$, and add it, together with the multisets $\{x + 3, y + 3, z + 3\}$ and $\{x + 6, y + 6, z + 6\}$, to gender A, add the multisets $\{x + 1, y + 1, z + 1\}, \{x + 4, y + 4, z + 4\}, \{x + 7, y + 7, z + 7\}$ to gender B and add the multisets $\{x + 2, y + 2, z + 2\}, \{x + 5, y + 5, z + 5\}, \{x + 8, y + 8, z + 8\}$ to gender C. Continue until every multiset of chromosomes is used up. The resulting genders (each consisting of 55 multisets of chromosomes) are symmetric and so (if I'm not mistaken) are balanced.

It's easy to see how to construct a wide variety of similar gender partitions for given $k$ if we may choose $n$ appropriately. These partitions have nice symmetry and use all possible multisets of chromosomes. This says nothing at all about, say, constructing gender partitions for $n = 2$ (though I believe that I've confirmed by case analysis that there are no balanced gender partitions for $k = 3$ and $n = 2$).

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I am not sure that your construction works. The 9 multisets in your construction are not necessarily all different. This is not very annoying if multiplicities do not cross genders, otherwise it will not work. Concerning the case $n=2$ (two chromosomes), it is not hard to show that there are no balanced solutions for any number $k\geq 3$ of genders. – Roland Bacher May 19 '10 at 9:36
It does work, I just omitted the special case of the triples {0, 3, 6}, {1, 4, 7} and {2, 5, 8}: these are the reason that 165 is not divisible by 9 and 55 is not divisible by 3. As stated, my algorithm includes three copies of each, but the three copies all land in the same gender, or as you put it, the multiplicities do not cross genders. The same construction works (with an analogous special case) for $k$ prime and $n = k^2$, but it may fail if $k$ is composite for the reason you mention. – JBL May 19 '10 at 13:17
Ah, I see, I was imagining quadratically-many pairs for $n = 2$ but it's only $k + 1$. Also, it may be helpful for visualizing my construction to view the chromosomes as $n$-th roots of unity. – JBL May 19 '10 at 13:33
I am convinced. Balancedness follows from the fact that $\mathbb Z/3\mathbb Z$ acts transitively on the genders (by adding a common constant to all elements). This action is compatible with the reproduction law and yields thus equal probability for the gender of the offspring. – Roland Bacher May 19 '10 at 16:45

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