Date: Jan 5, 2003 3:45 AM
Author: David Corfield
Subject: Queries about Species
I have a few questions about combinatorics and species, and would be
grateful for any comments.
The species of permutations on even sets corresponds to the series
(1 - x^2)^ -1 = 1 + x^2 + x^4 + ...
Does it have a square root?
We have (1 - x^2) ^-1/2 . (1 - x^2) ^-1/2 = (1 - x^2) ^ -1
So, equating coefficients of x^2n:
(2n)! = sum over r from 0 to n ( (2n choose 2r) ((2r)!!^2)((2n-2r)!!^2)
Eg. 24 = 9 + 6 + 9, 720 = 225 + 135 + 135 + 225,
but I can't see how the permutations on 4 or 6 elements divides *nicely*
X/(1 - e^X) looks like a simple composition of species - pick out a one
element set and arrange a set of sets whose union is the remainder - yet
it can't be that simple to get at the Bernouilli numbers. I guess lots of
unwanted empty sets appear in the union. This would no doubt require
some of Cartier's mathemagics to make sense of it.
Why is it that you get the series expansions for species if they don't
blow up for X= 0, yet you're most interested in X= 1?
Fiore and Leinster have written a paper RA/0211454 on Gaussian
integers, for which one can give a species interpretation.
Lawvere noted that there's a simple bijection between binary trees and
7-tuples of binary trees. It arises from the relation T = T^2 + 1. If this
in a distributive category, then T^7 is isomorphic to T.
In Fiore and Leinster's case, we are dealing with P^5 isomorphic to P,
arising from the relation P = 1 + P + P^2. You can interpret this P as non-
empty rooted trees with at most two descendant nodes. The single node tree
corresponds to 1, then either you have one descendant and so a copy of P,
or two descendants and so P^2.
In the T case, to translate it into a species, you say
T = 1 + X.T^2, a binary tree being either empty or a node and then two
trees. You solve for T as a quadratic, giving the Catalan numbers, which
count the number of binary rooted trees (see Baez and Dolan's 'From finite
sets to Feynman diagrams' for more on this).
In the P case, the species equation is
P = 1 + X.P + X^2.P^2, where you treat the X as picking out a line,
rather than node.
Solving for P gives you the Motzkin numbers:
Now, if a species is a categorification of an element of N[[x]], why are,
and Leinster interested in categorifying N[x]/ (some polynomial in x),
the polynomial is a relation for the species?
Should one think of N[x,y]/(polynomial in x and y), then evaluate at x = 1
for a rig?
E.g., for binary trees Y = 1 + X.Y^2,
so at X = 1, we have the rig N[Y]/(Y - 1 - Y^2), in which Y^7 = Y.
Behind what I've been talking about there seems to be deep connections
between Catalan numbers, Motzkin numbers, and what Elkies shows
in math.CA/0101168 about Calabi's clever trick of showing that the zeta
function at even integers is a rational multiple of pi^2n by substitution in
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