cycle type question

Dear Ralf,

Post by Ralf Hemmecke
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.

Post by Ralf Hemmecke
Martin, perhaps you realise that I need this to compute an upper bound
for the functorial composition.
Take n=0 in Definition 2.2.4 of BLL. What happens?
Let \sigma\in S_0. 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.

Post by Ralf Hemmecke
Note that there is no restriction that G in the functorial composite
F\square G must have \card(G[0])=0.
So assume \card(G[0])=c>0. Is then G[\sigma] \in S_c? Second question.
Which of those c! many?

See above.

Ooops, that was my own question. So I would be happy if someone could
confirm my thinking.

Ralf

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Martin Rubey

2007-03-11 18:10:45 UTC

Dear Ralf,

Post by Ralf Hemmecke
Dear Ralf,

Hi Ralf!

Post by Ralf Hemmecke
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 :-)

Post by Ralf Hemmecke
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. Note
that - since it is a function from the empty set to the empty set, one cannot
really say that it takes an argument :-)

Post by Ralf Hemmecke
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...

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

Post by Ralf Hemmecke
So I would be happy if someone could confirm my thinking.

confirmed?

Martin

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Ralf Hemmecke

2007-03-11 21:20:49 UTC

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.

Post by Martin Rubey
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.

Post by Martin Rubey
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. :-)

Post by Ralf Hemmecke
So I would be happy if someone could confirm my thinking.

confirmed?

Thank you? ;-)

Ralf

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Martin Rubey

2007-03-11 21:51:31 UTC

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.

OK, that's a good argument for thinking in caegories...

Post by Ralf Hemmecke
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.

But that's exactly what I meant to say. (I didn't say that
G[\emptyset]=\emptyset, only that \sigma would be the identity on the empty
set)

Post by Martin Rubey
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. :-)

I'd vote for zero. You see, I think that lcm(0,n)=0 for any n is somewhat
problematic, since 0 is a multiple of any number, so why don't we put
lcm(n,m)=0 for any n and m?

Post by Ralf Hemmecke
So I would be happy if someone could confirm my thinking. confirmed?

Thank you? ;-)

? ;-)

Martin

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Ralf Hemmecke

2007-03-11 22:19:44 UTC

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

I'd vote for zero. You see, I think that lcm(0,n)=0 for any n is somewhat
problematic, since 0 is a multiple of any number, so why don't we put
lcm(n,m)=0 for any n and m?

Axiom and LibAlgebra give something else.

(3) -> lcm []
(3) ->
(3) 1

aldor -gloop
#include "algebra"
#include "aldorinterp"
import from Integer;
lcm(empty$List(Integer))

gives

%6 >>
1 @ AldorInteger

Mathematica doesn't allow to compute LCM[].

Maple states in its help: ilcm() = 1.

GAP4 doesn't allow Lcm([]).

Magma says