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### David Corfield

Jan 5, 2003, 3:45:40 AM1/5/03
to
I have a few questions about combinatorics and species, and would be

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*
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
7-tuples 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

### John Baez

Jan 6, 2003, 2:25:30 AM1/6/03
to
In article <av8rch\$bki\$1...@news.ox.ac.uk>,
David Corfield <david.c...@philosophy.oxford.ac.uk> wrote:

>I have a few questions about combinatorics and species, and would be

I'll tackle the easiest one now and attempt the harder ones
later, but I sure hope other people try too.

>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 Bernoulli numbers. I guess lots of

>unwanted empty sets appear in the union.

I don't see what you're worrying about here, but I presume
it's related to the naive "0/0" you get when you evaluate
this expression at X= 0.

comments on Bernoulli numbers in this book:

Alain Connes, Andre Lichnerowicz and Marcel Paul Schutzenberger,
A Triangle of Thoughts, AMS, Providence, 2000.

He points out that if H is the Hamiltonian for some sort
of particle in a box and beta is the inverse temperature,

1/(1 - e^{-beta H}) = 1 + e^{-beta H} + e^{-2 beta H} + ...

is the operator you take the trace of to get the partition
function of a collection of an arbitrary number of particles of
this sort. And he claims that pondering this explains all the
appearances of X/(1 - e^X) and the Bernoulli numbers in topology!
See Milnor and Stasheff's book "Characteristic Classes" for an
introduction to *that* - but this book was written before
quantum theory invaded topology, so we're left to fit Connes'
clues together for ourselves.

>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?

Well, for now I'll just say that that species don't "blow up";
it's only when you decategorify them that you get divergent formal
power series. Then of course they usually diverge at X = 1,
because we are usually interested in structures that can be
put on finite sets in infinitely many different ways.

You probably knew all this and wanted a deeper answer;
I don't think there is one. I forget if you know what it
means to evaluate a species at an arbitrary groupoid X;
if not, maybe this would make you happier.

### David Corfield

Jan 6, 2003, 11:37:01 AM1/6/03
to
Chris Hillman kindly pointed out to me that
(1 - x^2) ^-1/2 = cosh(arctanh x)

To turn this into a species, note that cosh corresponds to being a
set of even cardinality, and arctanh corresponds to being a set of odd
cardinality up to cyclic order. Then from what Baez tells us in
Week 190, to compose species we look for an even numbered
set of odd numbered sets, the latter taken up to cyclic order.

This spits out 9 answers when a 4 element set is fed to it as it should.

After integrating this species to arcsin, do we find a case of the elusive
categorification of pi?

"John Baez" <ba...@galaxy.ucr.edu> wrote in message
news:avbb1a\$gmg\$1...@glue.ucr.edu...

> I'll tackle the easiest one now and attempt the harder ones
> later, but I sure hope other people try too.
>
> >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 Bernoulli numbers. I guess lots of
> >unwanted empty sets appear in the union.
>
> I don't see what you're worrying about here, but I presume
> it's related to the naive "0/0" you get when you evaluate
> this expression at X= 0.

What I was driving at is as follows:

X/(1 - e^X) as a species is constructed by multiplying X with the composite
of Permuations (i.e., 1/(1 - X)) acting on Set (i.e., e^X). Following your
rules in Week 190, let's give it a 2-element set {a, b}. First split this
set
into a one-element set and its complement. For the complement, find an
ordered set of unordered sets whose union is that complement.

So, if the X picks up {a}, the 1/(1 - e^X) will start listing:
<{b}>, <phi, {b}>, <{b}, phi>, <phi, phi, {b}>, ...
[phi the empty set]

Clearly an infinite number of such things, but all is not lost.
Decategorified the number concerned gives us:
1 + 2 + 3 + 4 +..., or zeta(-1), which we know to be -1/12.

So the coefficient of (X^2/2!) in X/(1 - e^X) is twice this, i.e., -1/6,
minus the 2nd Bernouilli number, as one would hope.

Cartier does this kind of thing (without species) in his fascinating
Mathemagics (A Tribute to L. Euler and R. Feynman)
www.mat.univie.ac.at/~slc/wpapers/s44cartier1.pdf

He presents this style as complementary to Hilbert-Bourbaki.
Will species bring them together?

> comments on Bernoulli numbers in this book:
>

As for how this relates to Connes' comments, I'll leave that to others.

David Corfield

### John Baez

Jan 8, 2003, 3:53:38 AM1/8/03
to
In article <av8rch\$bki\$1...@news.ox.ac.uk>,
David Corfield <david.c...@philosophy.oxford.ac.uk> wrote:

>I have a few questions about combinatorics and species, and would be

I'll take another crack at these now. By the way, you might have gotten
more replies if you'd phrased your question in terms of *generating
functions*, rather than *species*. Category theory is great, but most
people don't know this, so when trying to extract information from
them it's often best to play down this aspect of things, emphasizing
other buzzwords.

>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 Bernoulli 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.

Okay, I understand this remark now that you've expanded it...
and resummed to get a finite answer.

I'm getting more and more confused and tantalized by the relation
between Bernoulli numbers, species, statistical mechanics and quantum
theory. The reason I mentioned Connes' remarks is that he was
emphasizing the relation between Bernoulli numbers and *partition
functions*. Partition functions in statistical mechanics and
quantum theory are often just another way of looking at generating
perhaps shed a little light on what's going on - though not enough
to fully penetrate all the murk, yet.

If you have a system with allowed energy levels E_n, the probability
that it's in the nth one is proportional to

exp(-b E_n)

where b, usually written "beta", is the inverse temperature -
or really 1/kT where k is Boltzmann's constant and T is the
temperature. To get probabilities we need to normalize these
numbers by dividing by the "partition function"

Z(b) = sum_n exp(-b E_n)

You can do lots of great stuff in statistical mechanics if you
know the partition function of a system, mainly by differentiating
it with respect to b and playing various games.

Now, if we don't count the "zero-point energy", the harmonic
oscillator in quantum mechanics has allowed energy levels
0,1,2,3,4,... , so its partition function is

Z(b) = sum_n exp(-nb)

1
= -------------
1 - exp(-b)

This is where Connes' remark about Bernoulli numbers
comes in. The above partition function blows up as b -> 0
(the infinite temperature limit, where all energy levels
become equally probable). In fact it has a first-order pole
at b = 0. To deal with this, we can work with

b
b Z(b) = -------------
1 - exp(-b)

This is analytic at b = 0, and its Taylor series defines the
Bernoulli numbers, up to some signs:

b Z(b) = sum B_n (-x)^n / n!

What does all this have to do with species? I'm not sure.
But it's tantalizing, for two reasons! In my work with
Jim Dolan, the harmonic oscillator itself is categorified:
its Hilbert space becomes the category of species, and its
Hamiltonian (the "number operator") becomes a functor from
this category to itself. In your post, you try to interpret
B_n as the cardinality of a set of structures on the n-element
set. You get get a divergent series, basically the Riemann
zeta function evaluated at a negative integer, and you resum
this to get the right answer. This trick is common in quantum
field theory under the name of "zeta function regularization".

So, there seem to be two possible connections between
Bernoulli numbers and species via physics, but they don't
fit together neatly, and neither one is fully worked out -
though yours comes a lot closer.

http://www.maths.ex.ac.uk/~mwatkins/zeta/physics2.htm

On a different note:

>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?

Yikes... another twisted loop of ideas?

Here's all I can say:

We can indeed define "multivariable species" that serve as a
categorification of rigs like N[[x_1,...,x_n]]. To do this, just define an
"n-variable species" to be a type of structure that we can put
on an ordered n-tuples of finite sets. In other words, it's a functor

F: FinSet_0 x ... FinSet_0 -> Set

where FinSet_0 is the groupoid of finite sets and bijections,
and we take an n-fold Cartesian product of this category.

So, yes, we could define "complex species" whose generating
functions have complex coefficients by starting with two-variable
species and then imposing Fiore and Leinster's equation on one
variable to make that variable act like "i".