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nLab > Latest Changes: table of marks

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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeDec 29th 2018

    started some minimum

    v1, current

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeDec 30th 2018
    • (edited Jan 19th 2019)

    expanded and added a remark (here) on how the table of marks is the table of cardinalities

    Mij≔|(G/Hj)Hi|=|HomGSet(G/Hi,G/Hj)|.

    diff, v5, current

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeJan 19th 2019

    fixed the statement about marks via homs of linear reps (removed it). That was just nonsense – my bad.

    diff, v7, current

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeJan 19th 2019
    • (edited Jan 19th 2019)

    added pointer to

    • Klaus Lux, Herbert Pahlings, section 3.5 of Representations of groups – A computational approach, Cambridge University Press 2010 (author page, publisher page)

    diff, v8, current

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeJan 27th 2019
    • (edited Jan 27th 2019)

    added that with suitable ordering of conjugacy classes of subgroups, the table of marks becomes a triangular invertible matrix (here)

    diff, v10, current

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeJan 27th 2019

    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

    diff, v10, current

  7. Nice! I spelled out the conclusion of the argument expressing the structure constants in terms of the table of marks more carefully/in more detail.

    diff, v14, current

  8. Added as a corollary that the table of marks determines the Burnside ring.

    diff, v14, current

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeJan 27th 2019
    • (edited Jan 27th 2019)

    Thanks for the edits.

    Next I’ll look into writing out more of the algorithms for Chern characters of finite group representations. This should be very interesting to compute, both purely mathematically (here) as well as for physics (here)

    • CommentRowNumber10.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 27th 2019

    Added an 𝔽1 remark.

    diff, v16, current

    • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeJan 27th 2019

    Oh, but is there more than “one might explain”?

  12. Tweaked proof that the table of marks is lower triangular to be a little more explicit/elementary, as, for me at least, the argument is then clearer.

    diff, v17, current

    • CommentRowNumber13.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 27th 2019

    It would be nice to say more. I wonder what there is. There must be comparisons of character tables over other fields.

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeJan 28th 2019

    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 𝔽1, 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 𝔽1 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.

    • CommentRowNumber15.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 28th 2019

    Ok, will get rid of it.

    diff, v18, current

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeJan 28th 2019

    But did you have some references that talked about this?

    • CommentRowNumber17.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 28th 2019
    • (edited Jan 28th 2019)

    Given the discussion we having at the time about cohomotopy as the 𝔽1 version of K-theory (around here), I was looking for a 𝔽1-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)

    • CommentRowNumber18.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 28th 2019

    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 Sn. Then T=KC.

    • CommentRowNumber19.
    • CommentAuthorUrs
    • CommentTimeJan 28th 2019

    Interesting, thanks. Will need to look into that.

    • CommentRowNumber20.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2019
    • (edited Jan 31st 2019)

    added this basic statement (which is maybe what you (David C.) were after):

    For S∈GSetfin a finite G-set, for k any field and k[S]∈Repk(G) the corresponding permutation representation, the character χk[S] of the permutation representation at any g∈G equals the Burnside marks of S under the cyclic group ⟨g⟩⊂G generated by g:

    χk[S](g)=|X⟨g⟩|∈ℤ↪k.

    Hence the mark homomorphism (Def. \ref{BurnsideCharacter}) of G-sets restricted to cyclic subgroups coincides with the characters of their permutation representations.

    diff, v19, current

    • CommentRowNumber21.
    • CommentAuthorDavid_Corfield
    • CommentTimeJan 31st 2019

    Yes, that’s the sort of thing. Do you mean “any field” for k, or characteristic 0? Otherwise, what do you mean ℤ↪k for a finite field?

    • CommentRowNumber22.
    • CommentAuthorUrs
    • CommentTimeJan 31st 2019

    Right, thanks. It applies to any field whatsoever, but then I should write ℤ⟶k not ℤ↪k. Fixed now.

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