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

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David Corfield

Posts: 10
Registered: 12/3/04
Re: Queries about Species
Posted: Jan 6, 2003 11:37 AM
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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" <baez@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?

> I've been meaning to think about this ever since I read Connes'
> comments on Bernoulli numbers in this book:
>

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

David Corfield




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