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

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    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.
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    (Original post by xfootiecrazeesarax)
    ,,
    Way out of my comfort zone, but some thoughts:

    Do you have conditions on A? (I think you must, as I think A has to be positive definite or the integral won't converge). If you do, can you justify that A has a square root (e.g. if A is diagonalisable)? If so, then if B^2 = A, I think you'll get a lot of what you want by making the transformation y = Bx.

    Also, on the latex side,[latex]{\bf x}[/latex] will give you a bold face  {\bf x} . (IIRC, the braces are necessary to limit the region put into bold). It's a massive pain to do this whereever needed, so TBH I wouldn't bother if it's reasonably clear that x is supposed to be a vector (as is the case here).

    Edit: also, did you mean \frac{x^tAx}{t} \to x^t x? In particular the t on the bottom? It doesn't look right. You probably also want to be careful to distinguish transpose from the 't' variable. (I would write x^Tx not x^tx for clarity).
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    (Original post by DFranklin)

    Edit: also, did you mean \frac{x^tAx}{t} \to x^t x? In particular the t on the bottom? It doesn't look right. You probably also want to be careful to distinguish transpose from the 't' variable. (I would write x^Tx not x^tx for clarity).
    Apologies that should be \frac{x^tAx}{b} \to x^t x
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    (Original post by DFranklin)
    Way out of my comfort zone, but some thoughts:

    Do you have conditions on A? (I think you must, as I think A has to be positive definite or the integral won't converge). If you do, can you justify that A has a square root (e.g. if A is diagonalisable)? If so, then if B^2 = A, I think you'll get a lot of what you want by making the transformation y = Bx.
    .
    Okay, thanks.

    Apologies yes A is positive definite.

    Okay so because then I'd have  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
    =   y^{T}y/b if  (B^{-1})^T=B^{-1}

    I am not up to scratch with matrix properties, diagonalising etc, do we a condition that  (B^{-1})^T=B^{-1} is true?

    (I shall now go and have a look into why it can be diagnolized and why this then implies it can be square-rooted)

    thanks
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    (Original post by xfootiecrazeesarax)
    Okay, thanks.

    Apologies yes A is positive definite.

    Okay so because then I'd have  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
    =   y^{T}y/b if  (B^{-1})^T=B^{-1}

    I am not up to scratch with matrix properties, diagonalising etc, do we a condition that  (B^{-1})^T=B^{-1} is true?

    (I shall now go and have a look into why it can be diagnolized and why this then implies it can be square-rooted)
    I'm out of practice enough that you should check all of this, but if A is symmetric, then you should be able to diagonalise A in the form P\Lambda P^T. If A is positive definite, all the coefficients of lambda are non-negative, and so you can square root them to form the square root of lambda. Then B = P \sqrt{\Lambda} P^T will do exactly what you want. I think, (in particular B and its inverse will turn out to be symmetric).
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    (Original post by DFranklin)
    I'm out of practice enough that you should check all of this, but if A is symmetric, then you should be able to diagonalise A in the form P\Lambda P^T. If A is positive definite, all the coefficients of lambda are non-negative, and so you can square root them to form the square root of lambda. Then B = P \sqrt{\Lambda} P^T will do exactly what you want. I think, (in particular B and its inverse will turn out to be symmetric).
    okay thank you, so  P the matrix of eigenvectors normalized is orthonormal so it's inverse is equal to it's transpose.

    And the symmetry of B gives  B^2 = BB^T from which it easily be seen that  B^2 = A

    However I'm unsure how to show that  P \sqrt{\Lambda} P^{T} is orthonormal using the fact that  P is..?
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    (Original post by xfootiecrazeesarax)
    However I'm unsure how to show that  P \sqrt{\Lambda} P^{T} is orthonormal using the fact that  P is..?
    It isn't. Why do you need it to be?

    (It might be instructive for you to consider what happens in some very simple cases).

    E.g. in 1 dimension (so everything is just numbers), we have P = P^T = 1, A is just a real number > 0. For concreteness, suppose A is 3.

    Then B ends up being sqrt(3). So B isn't orthonormal (it would have to be +/-1).

    But you know the 1D case works, so i don't see this can be an issue.
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    (Original post by DFranklin)
    It isn't. Why do you need it to be?

    .

    As post 4
    I thought we wanted B^{-1}=B^{T} to allow the substitution to bring it to the simplified form y^{T}y/b .

    edit: apologies I meant orthogonal !
    So I.e how to show the statement in the bottom of your post 5 that B and it's inverse are symmetric.
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    (Original post by xfootiecrazeesarax)
    As post 4
    I thought we wanted B^{-1}=B^{T} to allow the substitution to bring it to the simplified form y^{T}y/b .

    edit: apologies I meant orthogonal !
    So I.e how to show the statement in the bottom of your post 5 that B and it's inverse are symmetric.
    To show B is symmetric, start from B = P \sqrt{\Lambda} P^T and take the transpose. You end up with B^T = = P \sqrt{\Lambda} P^T.

    You can do something similar with inv(B).
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    (Original post by DFranklin)
    To show B is symmetric, start from B = P \sqrt{\Lambda} P^T and take the transpose. You end up with B^T = = P \sqrt{\Lambda} P^T.

    You can do something similar with inv(B).
    Ahh yes, thanks.

    So now I need the substituion rule 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?

    Do you have a starting point or a link to some good notes to look at or anything?

    Many thanks
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    (Original post by xfootiecrazeesarax)
    Okay, thanks.

    Apologies yes A is positive definite.

    Okay so because then I'd have  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
    =   y^{T}y/b if  (B^{-1})^T=B^{-1}
    Oh I just realised there is still the  b divided by, which makes me look like I need another substituion to get the required result looking at the factor of b in the final result
 
 
 
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