Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Representation number via quad forms of theta quadratic form

  1. Dec 17, 2016 #1
    ##\theta(\tau, A) = \sum\limits_{\vec{x}\in Z^{m}} e^{\pi i A[x] \tau } ##

    ##=\sum\limits^{\infty}_{n=0} r_{A}(n)q^{n} ##,

    where ## r_{A} = No. [ \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?)

    So I thought that this meant to solve the quadratic ##A[x]= \vec{x^t} A \vec{x} = n ##, for each ##n##, and the representation number is then given by the number of solutions to this?, subject to ## \vec{x} \in Z^{m} ## ,

    What is ##Z^{m}## here please? ( z the integer symbol)

    Many thanks
     
  2. jcsd
  3. Dec 17, 2016 #2

    fresh_42

    Staff: Mentor

    ## \vec{x} \in Z^{m} ## is simply ##\vec{x} = (x_1, \ldots , x_n)^\tau \in \underbrace{\mathbb{Z} \times \ldots \times \mathbb{Z}}_{n-\ times}##.
    Was that your question?
     
  4. Dec 17, 2016 #3
    no not quite, i needed to check this before i can post my full question, and that my interpretation of what the representation number is correct? (otherwise the question im about to post may not make sense)
     
  5. Dec 18, 2016 #4
    Okay so on the attachment of extract from my book, I'm not understanding the comment '##Q_{1}(x,y) ## and ##Q_{2}(x,y) ## yeild the same series since they represent the same integers.'

    So as I said above my interpretation of how to compute the ##r(n)## was to :

    set ##2 Q(x,y) = A(x,y) = n ## , for each ##n## in turn and count the number of solutions to this for each ##n##.

    So looking at ##Q_{0}(x,y)##, should find ##2(x^{2}+xy+6y^2)=0## has one solution (i.e ##(x,y)=0##) , ##2(x^{2}+xy+6y^2)=1## should find 2 solutions and ##2(x^{2}+xy+6y^2)=2,3## has no solutions for ##x \in Z^m ##
    Is my understanding correct here?

    So then looking at ##Q_1 (x,y)## and ##Q_{2} (x,y) ## which differ only on the sign of the ##xy## term, I don't see how it is obvious that these will have the same number of solutions for ##Q(x,y) = n## for each ##n##?

    Many thanks in advance.
     

    Attached Files:

  6. Dec 18, 2016 #5

    fresh_42

    Staff: Mentor

    If you replace ##x## by ##-x## (or equivalently ##y## by ##-y\,##) you get the same number of pairs ##(x,y) \in \mathbb{Z}^2## with ##Q_i(x,y)=n##. Different pairs though, but equally many.
     
  7. Dec 29, 2016 #6
    oh right thanks,
    how is this obvious? e.g how do you know you won't end up getting complex solutions for the corresponding sign change, ta
     
  8. Dec 29, 2016 #7

    fresh_42

    Staff: Mentor

    $$\#\{(x,y)\in \mathbb{Z}^2\,\vert \,Q_1(x,y)=0\}=\#\{(-x,y)\in \mathbb{Z}^2\,\vert \,Q_1(-x,y)=0\}=\#\{(w,y)\in \mathbb{Z}^2\,\vert \,Q_1(-x,y)=0\, \wedge \, w=-x\,\}=\#\{(w,y)\in \mathbb{Z}^2\,\vert \,Q_2(x,y)=0\, \wedge \, w=-x\,\}=\#\{(w,y)\in \mathbb{Z}^2\,\vert \,Q_2(x,y)=0\, \wedge \, w=x\,\}=\#\{(x,y)\in \mathbb{Z}^2\,\vert \,Q_2(x,y)=0\}$$ because we consider all pairs in ##\mathbb{Z}^2##, so the sign doesn't make any difference in the total number of solutions, only in the way we write, resp. notate them: ##\#\{(x,y)\in \mathbb{Z}^2\,\vert \,x=1 \wedge y=2\}=\#\{(x,y)\in \mathbb{Z}^2\,\vert \,x=-1 \wedge y=2\}##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Representation number via quad forms of theta quadratic form
  1. Quadratic forms (Replies: 2)

  2. Quadratic forms (Replies: 1)

  3. Quadratic Forms (Replies: 5)

Loading...