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Topic:
Queries about Species
Replies:
4
Last Post:
Jan 13, 2003 10:41 AM




Queries about Species
Posted:
Jan 5, 2003 3:45 AM


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)((2n2r)!!^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* this way.
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 7tuples of binary trees. It arises from the relation T = T^2 + 1. If this holds 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:
A001006: 1,1,2,4,9,21,51,127,323,835,.
Now, if a species is a categorification of an element of N[[x]], why are, Fiore and Leinster interested in categorifying N[x]/ (some polynomial in x), where 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 an integral.
David Corfield Faculty of Philosophy Oxford



