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Have there been any attempts to extend the "F_un" analogy to the representation theory of finite groups?

If I might be allowed some speculation:

If combinatorics can be regarded as analagous to linear algebra over the "field with one element", then are characters of finite groups over the "field with one element" simply permutation characters? Perhaps the analogous notion of irreducibility in this case would be transitivity?

In the complex character theory of finite groups, two irreducible characters are identical if and only if representations that afford them are equivalent. Perhaps by developing a representation theory of finite groups over the field with one element, we would find some analogous way of comparing "F_un characters" to check if two permutation representations of a given group are equivalent.

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    $\begingroup$ There already is an analog of character tables for permutation representation called the "table of marks," going back > 100 years to work of Burnside: en.wikipedia.org/wiki/Burnside_ring#Marks $\endgroup$
    – Sam Hopkins
    Commented Apr 8, 2023 at 15:53
  • $\begingroup$ Thanks! Has the study of the Burnside ring been important in answering more down-to-earth questions about finite group actions, just as ordinary character theory has been useful in answering "purely" group-theoretic questions? $\endgroup$
    – semisimpleton
    Commented Apr 8, 2023 at 16:17
  • $\begingroup$ I'll let somebody more knowledgeable answer that question about applications of the Burnside ring. Certainly there has been significant work on it since Burnside. My impression is that it is less "tractable" than the representation ring; as is usual in mathematics, linearizing makes life easier. $\endgroup$
    – Sam Hopkins
    Commented Apr 8, 2023 at 16:57
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    $\begingroup$ Weyl groups, including the symmetric groups, are often viewed as the corresponding algebraic group over a field with one element. See, for example, mathoverflow.net/questions/272498/… and en.wikipedia.org/wiki/Weyl_group. $\endgroup$
    – Andrew
    Commented Apr 9, 2023 at 12:20

1 Answer 1

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The table of marks, as defined by Burnside, has rows indexed by the transitive permutations $X$ and columns indexed by the conjugacy classes of subgroups $H\leqslant G$. The entry corresponding to $X$ and $H$ is $|X^H|$, the number of fixed points.

The Burnside ring $A(G)$ is the Grothendieck ring of permutation representations, with disjoint union as the addition and direct product as the multiplication. The ring homomorphisms from $A(G)$ to $\mathbb Z$ are exactly the fixed point functions in the table of marks, and they separate permutation representations. This is very analogous to character theory.

The ring $A(G)$ was extensively studied by Dress, who computed the index of the image of $A(G)$ in the sum of copies of $\mathbb Z$, one for each conjugacy class of subgroups, under the sum of these homomorphisms. There are many applications of $A(G)$ in representation theory, especially induction theorems and related topics.

Dress also discovered that there are no non-trivial idempotents in $A(G)$ if and only if $G$ is soluble. There was some speculation that one might be able to approach the odd order theorem this way, but nobody really got very far with it.

Burnside rings play a role in a lot of applications of group theory. In stable homotopy theory, for example, the stable cohomotopy of the classifying space $BG$ is compared with the completion of $A(G)$. This is the Segal conjecture, which was proved back in the eighties. This is analogous with the Atiyah completion theorem, which compares the unitary $K$-theory of $BG$ with the completion of the character ring.

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    $\begingroup$ Though for an interesting way in which marks & character theory are dis-analogous, see: mathoverflow.net/questions/358037/… $\endgroup$
    – Sam Hopkins
    Commented Apr 8, 2023 at 19:34
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    $\begingroup$ That's because the analogue in representation theory of the transitive permutation representations (induced from subgroups) isn't the basis of irreducible characters, but rather the induced characters from cyclic subgroups, as in Artin's induction theorem. $\endgroup$
    – Dave Benson
    Commented Apr 8, 2023 at 19:42
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    $\begingroup$ @SamHopkins Another important difference $\endgroup$
    – Gro-Tsen
    Commented Apr 8, 2023 at 20:59
  • $\begingroup$ Is there some heuristic argument for why the table of marks "should" be thought of as the representation theory of finite groups over the field with one element? $\endgroup$
    – Timothy Chow
    Commented Apr 9, 2023 at 2:25
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    $\begingroup$ One can always argue about what is a "correct" analogy, but I'm really referring to the relationship between detection and induction. For details, please see Chapter 5 of "Representations and Cohomology, I". $\endgroup$
    – Dave Benson
    Commented Apr 9, 2023 at 6:50

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