>> On 03/11/2007 03:25 PM, Ralf Hemmecke wrote:
>>> Could someone tell me what the cycle type of \sigma \in S_0 is. I
>>> somehow think that (), i.e. the empty tuple is a good candidate.
>> What else.
> 
> Hmmm, I wanted to say the sequence which consists of zeros only.
> Well, that's about the same thing, I guess :-)

Also quite OK. At one point I thought that one should rather take the
point of view that all permutations are automorphisms of N (natural
numbers). But that doesn't quite fit to the fact that a species is a
endo-functor of the category B of *finite* sets and bijections.

>>> Take n=0 in Definition 2.2.4 of BLL. What happens?  Let \sigma\in
>>> S_0.

> So, \sigma is the only permutation of the empty set. Something rather
>  imaginary... There is exactly one such permutation, which explains
> 0!=1.

Of course there must be one. \emptyset is an object in B and in a
category there is always the 1 arrow from an object to itself.

> Note that - since it is a function from the empty set to the empty
> set, one cannot really say that it takes an argument :-)

I don't say that.

>>> What is G[\sigma]? Ah, sure, it is a permutation in some S_m.
>>> First question: what is m?
>> The only good choice is m=\card G[0]. And to be functorial
>> G[\sigma] must be the identity permuation of S_m.
> 
> I just wanted to look that up. OK, I did: one can say (although I
> don't really have a good argument for that) that the sigma above is
> the identity on the empty set...

No. G[\emptyset] is not necessarily empty. So the G transports the arrow

1: \emptyset -> \emptyset to G[1]: G[\emptyset] -> G[\emptyset]. And the 
latter is most certainly the identity on G[\emptyset], because G is a 
functor.

> Note that axiom says lcm(0,n)=0 for any n.

I don't think that is important since you suggested to take

   n := lcm([i for i in 1..#p | p.i ~= 0]);

and I would have to compute the lcm of the empty list. That is more 
interesting. :-)

>> So I would be happy if someone could confirm my thinking.
> 
> confirmed?

Thank you? ;-)

Ralf

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