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added pointer to
added statement and proof of the expression of the Burnside ring structure constants in terms of the table of marks (here)
Will copy this also into the entry Burnside ring
Oh, but is there more than “one might explain”?
It would be nice to say more. I wonder what there is. There must be comparisons of character tables over other fields.
Okay, but then maybe we should wait with suggesting that there is such a relation, until we know that there is.
A priori it looks implausible to me that the the marks are literally the characters over $\mathbb{F}_1$, as they are parameterized over different combinatorial data: Taken at face value, no change of base field changes the fact that a character is a function on conjugacy classes of group elements, hence, if you wish, on conjugacy classes of cyclic subgroups. But the marks are functions on the set of conjugacy classes of all subgroups.
Now I don’t know if maybe some $\mathbb{F}_1$ magic makes away with that difference, somehow. But it seems rather implausible at face value.
On the other hand, it is manifest that the marks are a decategorification of the incarnation of a $G$-set as a presheaf on the orbit category, as in Elmendorf’s theorem.
But did you have some references that talked about this?
Given the discussion we having at the time about cohomotopy as the $\mathbb{F}_1$ version of K-theory (around here), I was looking for a $\mathbb{F}_1$-version of the usual character table, and turned up the table of marks. I can’t remember anything more explicit than:
Much like character theory simplifies working with group representations, marks simplify working with permutation representations and the Burnside ring. (Wikipedia)
But incidentally, there does appear to be a relationship in some cases between the table of marks and the character table :
Let $T$ be the table of marks with respect to the Young subgroups, and $K$ be the Kostka matrix, and $C$ be the character table of the symmetric group $S_n$. Then $T=K C$.
Interesting, thanks. Will need to look into that.
added this basic statement (which is maybe what you (David C.) were after):
For $S \in G Set_{fin}$ a finite G-set, for $k$ any field and $k[S] \in Rep_k(G)$ the corresponding permutation representation, the character $\chi_{k[S]}$ of the permutation representation at any $g \in G$ equals the Burnside marks of $S$ under the cyclic group $\langle g\rangle \subset G$ generated by $g$:
$\chi_{k[S]}\big( g \big) \;=\; \left\vert X^{\langle g\rangle} \right\vert \;\in\; \mathbb{Z} \hookrightarrow k \,.$Hence the mark homomorphism (Def. \ref{BurnsideCharacter}) of $G$-sets restricted to cyclic subgroups coincides with the characters of their permutation representations.
Yes, that’s the sort of thing. Do you mean “any field” for $k$, or characteristic $0$? Otherwise, what do you mean $\mathbb{Z} \hookrightarrow k$ for a finite field?
Right, thanks. It applies to any field whatsoever, but then I should write $\mathbb{Z} \longrightarrow k$ not $\mathbb{Z} \hookrightarrow k$. Fixed now.
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