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Generalising theta transformation formula from 1-d to m-d

  1. Jan 9, 2017 #1
    1. The problem statement, all variables and given/known data

    For the theta series given by : ##\theta(t) = \sum\limits_{n\in Z} e^{2\pi i n^2 t} ## in 1-d there is the transformation formula:

    ##\theta(-1/4t)=\sqrt{-2it\theta(t)}##

    To prove this one uses the fact that this theta function is holomorphic and so by Riemann theorem it is sufficient to prove this on one line only- let this line chosen be an imaginary line ##t=ib##, so need to show:

    ##\sum\limits_{n\in Z} e^{\frac{- \pi n^2} {2b}} = \sqrt{2b}\sum\limits_{n\in Z} e^{- 2 \pi b n ^{2}} ## [1]

    The proof then follows from the Poisson summation formula and that the Fourier transform of a Gaussian returns itself when the fourier transform is defined as, taking the function gaussian :

    ##F(e^{- \pi x^{2}})= \int\limits^{\infty}_{-\infty} e^{- \pi x^{2}} e^{-2 i \pi xy } dx = e^{-\pi y ^2}## [2]

    (And where the Poisson summation formula is ##\sum\limits_{n \in Z} F(f(n))=\sum\limits_{n \in Z} f(n) ##

    and then take the fourier transform of the LHS of [1] , and make the change of variables ##x'=x/(2b)^{1/2}## to make use of [2] shows

    ##F(e^{- \pi x^{2}/2b}) = \sqrt{2b} e^{- 2 \pi y^2 } ##

    and then the result follows from the Poisson summation formula.

    QUESTION

    I want to generalise this to the m-d case where here the theta transformation formula is

    ##\theta(t,A)=\sum\limits_{x \in Z^m} e^{\pi i A[x] t} ##

    where ##A[x] = x^t A x ##, ##x## a vector (sorry i don't know how to use vector notation)

    2. Relevant equations

    The Poisson summation formula and Fourier transform generalised from 1-d , as above, to m-d are obtained by sending ##\sum_{n \in Z} \to \sum_{n \in Z^m} ## for the Poisson summation and sending ##e^{-2 \pi i xy} \to e^{ -2 \pi i x^t y } ## in the fourier transform

    3. The attempt at a solution

    So the proof is to follow the above really. Again proving on an imaginary line ##t=ib## so we then want to show :

    ##\sum_{x \in Z^m} e^{- \pi A[x] / b } = \frac{\sqrt{b}^{m}}{\sqrt{det A}} \sum_{x \in Z^{m}} e^{- \pi b A^{-1} [x] }##

    So again the idea is to show that ##F(e^{- \pi A[x] / b }) = \frac{\sqrt{b}^{m}}{\sqrt{det A}} e^{- \pi b A^{-1} [x]} ###

    and then result follows from the poisson.

    So we have that [2] holds in m-d and again need a transform of variable.

    MY QUESTION

    This is where I am stuck , in order to make use of the Gaussian result I need to transform ## \frac{x^t A x}{t} \to x^t x ## , Im unsure how to do this since the ##x^t x ## compared to simply just ##x## is throwing me, and then secondly I am unsure of the general rules for substitution when a matrix ##A## is involved, that will give arise to the required ##det A## that is needed.
     
  2. jcsd
  3. Jan 9, 2017 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your matrix ##A## had better be positive-definite, since otherwise your multi-dimensional version will not be a straight generalization of the one-d version.

    So, if ##A## is positive-definite it can be written as ##A = U^T U##, where ##U## is an upper-triangular matrix with positive numbers along the diagonal; in fact for large ##n##, using the algorithm to compute ##U## is the best way to actually test for positive-definiteness of ##A## (avoiding all the usual determinant tests, etc.) See, eg, Choleski Factorization or Choleski's Algorithm; this is so straightforward that I once programmed it on an HP calculator to handle matrices up to ##8 \times 8##.

    There are other ways to represent a positive-definite ##A## as a "square" of other another matrix, but the Choleski method is most convenient when you want to write ##x^T A x## as a sum of squares: ##x^T A x = (Ux)^T Ux = \sum_{i=1}^n u_i(x)^2## where ##u_i(x)= \sum_{j=i}^n u_{ij} x_j##.
     
    Last edited: Jan 9, 2017
  4. Jan 9, 2017 #3
    okay thanks,
    however I am still unsure what to do next with this...
    possible more hint anyone?
     
  5. Jan 12, 2017 #4
    Okay , so because ## A## is symmetric it can be diaganolised in the form ## P \Lambda P^{T}##, ##P## the eigenvectors, chosen orthonormally, and ##\Lambda## the eigenvalue matrix , and because it is positive definite all eigenvalues are ## \geq 0 ## .

    Now if I let ##B=P \sqrt{\Lambda } P^{T}##, then ##B^{2}=A##, I can make the substitution ##y=Bx## and then

    ## x=B^{-1}y ##

    So ##x^{T}Ax/b = (B^{-1}y)^{T}B^2(B^{-1}y)/b ##

    ##= y^{T}(B^{-1})^{T}BBB^{-1}y/b ##




    MY QUESTION

    This gives me the substitution I need, but I am unsure of the 'substitution rule'- change of variables in the integral,

    of what ##dx ## will go to in terms of ##B ## and ##dy ##on making the substituion##y=Bx ##


    I have never came across a subsitution like this with matrices, and am struggling to find anything online?

    Can anyone help out, or a link to some good notes to look at or anything?

    Many thanks
     
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