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Computing representation number quad forms

  1. Jan 7, 2017 #1
    1. The problem statement, all variables and given/known data

    ## r_{A} (n) = ## number of solutions of ## { \vec{x} \in Z^{m} ; A[\vec{x}] =n} ##
    where ##A[x]= x^t A x ##, is the associated quadratic from to the matrix ##A##, where here ##A## is positive definite, of rank ##m## and even. (and I think symmetric?)

    I am solving for the ##r_{A}(1) ## for the two quadratic forms:

    ##Q(x,y,uv)= 2(x^2+y^2+u^2+v^2)+2xu+xv+yu-2yv##
    ##R(x,y,uv)=x^2+4(y^2+u^2+v^2)+xu+4yu+3yv+7uv##


    2. Relevant equations

    see above

    3. The attempt at a solution

    diagonalized these read:

    ##Q=2(x+u/2+v/4)^2+2(y-v/2+u/4)^2+11u^2/8 + 11v^2/8 ##
    ##R=(x+1/2u)^2+4(y+1/2u+3/8v)^2+11/4(u+v)^2+11/16v^2 ##

    Solving ##Q=1## with all ##x,y,u,v## integer, it is clear that ##u,v=0## is needed, and then ##x,y=\pm 1 ## gives ##r_{Q}(1)=4##.

    Now looking at ## r_{R}(1) ## by the same reasoning as above I would have said that we require ##v=0## , and then I' m not sure what to do.

    However the solution is:

    Must have ##u+v=1 ## & ##|v|=0,\pm 1 ##, this gives ## \pm(1,0,0,0)##, ##\pm(1,0,-1,1) ## , ##\pm(0,0,-1,1)##

    (the symbol that I interpreted as '&' in the solutions is a bit smudged, so looking at the solutions I'm not sure that this is supposed to be a 'or'? )

    Either way, I'm really confused, unsure where these conditions come from, how to think about this in a logical way...

    Many thanks for your help in advance.
     
  2. jcsd
  3. Jan 7, 2017 #2

    fresh_42

    Staff: Mentor

    Are you sure there are no typos? I calculated ##R(0,0,-1,1) = \frac{61}{64}##. It could also help to multiply the equations by ##8##, resp. ##16##, which would make the comparisons easier.
     
  4. Jan 7, 2017 #3

    haruspex

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    True, but not quite trivial.
    How do you get that? Don't those give Q=4?
    Then you would be wrong. I said it wasn't quite trivial.
    As fresh_42 writes, it will help to multiply through the equations to eliminate the fractions.
     
  5. Jan 11, 2017 #4
    oh right, the reason is that each term needs to be ##\leq 1 ## ?
     
  6. Jan 11, 2017 #5

    haruspex

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    Each term must be no more than 1, but I cannot say whether that is the "reason" you were wrong to conclude v=0 since I do not know how you concluded it. All I can say is that there is a solution with v not 0.
     
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