Hello,
On 03/28/2007 10:51 PM, Nicolas M. Thiery wrote:
>> Are you sure?
> This was explained to me by François Bergeron, so pretty much yes :-)
Since I am a notorious non-believer, I have still have my doubts. Let me
explain.
1) Section 1.4 equation (14) states the
"..., we have the combinatorial equation S = E \circ C".
2) Section 1.2 "Combinatorial Equality" states 3 types of equality.
a) identity (written =)
b) equipotency (written \equiv)
c) combinatorial equality (written =)
Clearly, under 1) it is not written \equiv. And since we don't have
identity, it can only mean 2c).
> (up to the point that we are speaking about the same thing). Actually,
> I think he told me this example was in the Book.
> The point is that relabeling a permutation in cycle notation
> corresponds to the action of S_n on itself by *conjugation*, and not
> *on the right* as when you relabel a permutation in array notation.
> Example, relabeling by the transposition (1,2):
>
> (1,2) (3) = 213
>
> || ||
> \/ \/
>
> (1,2) (3) <> 123
In fact, I started to implement permutations as being in an array
representation. When I wanted to implement the generation of isomorphism
types, I realized that I had not enough bits to represent a
representative of an isomorphism type of a permutation. So the
representation became
Array List L
for Permutation and
List L
for Cycle and LinearOrder. That is the fun if one implements labelled
and unlabelled structures at the same time. One realises quickly that
something doesn't work.
I am pretty sure that S = E \circ C stands for an equality of functors
(i.e there is a "natural" isomorphism between the two functors).
Of course LinearOrder \ne Permutation = Compose(SetSpecies, Cycle)
(where the latter is only an isomorphism in AC).
Ralf
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