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

[binbagsss]

Homework Statement

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)

Homework 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

The Attempt at a Solution

[/B]
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 $\frac{x^t A x}{t} \to x^t x$ , I am 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.

 

[Ray Vickson]

Homework Statement

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)

Homework 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

The Attempt at a Solution

[/B]
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 $\frac{x^t A x}{t} \to x^t x$ , I am 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.

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 $u_i(x)= \sum_{j=i}^n u_{ij} x_j$.

 

[binbagsss]

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 $u_i(x)= \sum_{j=i}^n u_{ij} x_j$.

okay thanks,
however I am still unsure what to do next with this...
possible more hint anyone?

 

[binbagsss]

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