identity involving symmetric group

Discussion in 'Math Research' started by Kevin Buzzard, Nov 27, 2003.

  1. Hi. Here's a puzzle.

    *****************************

    Let j>=1 be an integer. Let n_1,n_2,...,n_j be j positive integers with
    the following two properties:

    1) the average of the n_i is 2
    2) none of the n_i are 2

    [e.g. j=5 and the integers could be 1,1,1,1,6]

    Consider the product

    j(j-n_1)(j-n_1-n_2)(j-n_1-n_2-n_3)...(j-n_1-n_2-...-n_j)

    (note that the last term in the product is just -j)

    Well, this product is unfair because it relies on the order of the n_i.
    So now sum this product over all j! permutations of the n's.

    [i.e. form sum_{g in Symm(j)} j(j-n_{g1})(j-n_{g1}-n_{g2})...(-j) ]

    Is this sum always zero?

    ***********************************

    Remark 1): if one lets some of the n_i be 2 the sum sometimes isn't 0.

    Remark 2): this is actually a *highly disguised* question about mod p
    modular forms! [more precisely, about their expansions around supersingular
    points of the modular curve] It's not an idle question at all. A grad student
    of mine has run up against it and it's got to the stage where he can't
    do it, I can't do it, and we both just want it out of the way.

    Thanks for any help,

    Kevin Buzzard
     
    Kevin Buzzard, Nov 27, 2003
    #1
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  2. Kevin Buzzard

    Graham Jones Guest

    Your puzzle is reminiscent of the calculation of the character table of
    the symmetric group Symm(j). I'm looking at "Introduction to group
    characters", W Ledermann, 1977. My knowledge dates from that era too,
    and is very rusty by now.

    If you put m_i = n_i - 1, you can rewrite the conditions as

    (1) {m_i} is a partition of j
    (You might prefer to say that m_i >= 0 and the non-zero elements form a
    partition of j, but in the character table calculation, the zeroes are
    added to make things go smoothly.)

    (2) {m_i} corresponds to a conjugacy class in Symm(j) whose elements act
    fixed-point-free on {1,2,...j}

    I don't see anything resembling your product in the character table
    calculation, but there are lots of determinants of j by j matrices
    around.
     
    Graham Jones, Nov 28, 2003
    #2
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  3. Let j>=1 be an integer. Let n_1,n_2,...,n_j be j positive integers with
    I find it a bit unnerving that no-one has replied---questions like
    this usually get eaten for breakfast in this forum. I take some
    heart in the fact that a substantial percentage of our readers are
    currently eating turkey, but am a little scared that someone will
    ultimately find a counterexample.

    As a result of my posting here I have had several replies by email,
    all unfortunately telling me things I already knew (I have thought
    about the problem a fair bit myself); to stem the flow a bit I
    thought I'd post some of these things here.

    1) if j is even then it's trivial (the orderings m_1,m_2,...,m_j
    and m_j,m_{j-1},...,m_2,m_1 cancel each other out)

    2) if all but one of the n_i are 1 then the result boils down
    to "alternating sum of binomial coefficients is zero",
    or more precisely to the fact that (1-1)^(j-1)=0.

    3) I cannot do the case where all but two of the n_i are 1.
    Robin Chapman tells me that he has verified the case 1,1,1,1,...,1,a,b
    by computer for 3 <= a <= b < 50 (and the sum was always zero)
    but the combinatorics of the general case still elude me.
    I still cannot say that I "understand" why it works for 1,1,1,3,4,
    but it does.

    4) Robin also tells me that he has checked the case 1,1,1,...,1,a,b,c
    by computer for 3 <= a <= b <= c <= 20 (and the sum was always zero).
    This makes me more optimistic that the sum really always is zero.

    Kevin Buzzard
     
    Kevin Buzzard, Nov 28, 2003
    #3
  4. Let j>=1 be an integer. Let n_1,n_2,...,n_j be j positive integers with
    The answer is "yes" and Noam Elkies provided me with a proof. Here
    are hopefully sufficiently many of the details to enable people
    to reconstruct the full argument. We work in the field of fractions F
    of the ring C[t_1,t_3,t_4,...][[1/z]].

    Lemma: if T is a power series t_1/z+t_3/z^3+t_4/z^4+... (note no t_2 term)
    and A is the differential operator on F defined by
    A(f)=z*t*f'
    then the coefficient of z^{-j} in A^j(z^j) is always zero.

    Proof: one checks that it's possible to change variables
    to w=cz+(lower order terms) so that A becomes d/dw. Then
    the lemma becomes obvious (after some unravelling).

    Now the result we want follows from the lemma, by looking at
    the coefficient of z^{-j}.

    Thanks to all who responded,

    Kevin Buzzard
     
    Kevin Buzzard, Dec 1, 2003
    #4
  5. Kevin Buzzard

    Graham Jones Guest

    Another possible clue:

    If k of the n_i are bigger than one, then the above sum seems (from
    experiments) to have k! sub-sums which are zero.

    Since rearranging the ones has no effect, determining whether the sum is
    zero is a matter of summing over the k! orderings of the non-unity n_i,
    and for each of those, the (j choose k) ways that ones can be inserted
    between them. And when evaluated like that, you don't just get k!
    numbers which sum to zero, you get k! zeroes. According to my
    unsystematic experiments, that is.
     
    Graham Jones, Dec 1, 2003
    #5
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