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# Generalising theta transformation formula proof from 1-d to m-d watch

1. For the theta series given by :

in 1-d there is the transformation formula:

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 , so need to show:

[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 :

[2] (And where the Poisson summation formula is

and then take the fourier transform of the LHS of [1] , and make the change of variables to make use of [2] shows 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 where , 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 for the Poisson summation and sending 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 so we then want to show :

So again the idea is to show that 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 , Im unsure how to do this since the compared to simply just is throwing me, and then secondly I am unsure of the general rules for substitution when a matrix is involved, that will give arise to the required that is needed.
2. (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 .

Also, on the latex side,${\bf x}$ will give you a bold face . (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 ? 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 not for clarity).
3. (Original post by DFranklin)

Edit: also, did you mean ? 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 not for clarity).
Apologies that should be
4. (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 .
.
Okay, thanks.

Apologies yes A is positive definite.

Okay so because then I'd have
So

if

I am not up to scratch with matrix properties, diagonalising etc, do we a condition that 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
5. (Original post by xfootiecrazeesarax)
Okay, thanks.

Apologies yes A is positive definite.

Okay so because then I'd have
So

if

I am not up to scratch with matrix properties, diagonalising etc, do we a condition that 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 . 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 will do exactly what you want. I think, (in particular B and its inverse will turn out to be symmetric).
6. (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 . 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 will do exactly what you want. I think, (in particular B and its inverse will turn out to be symmetric).
okay thank you, so the matrix of eigenvectors normalized is orthonormal so it's inverse is equal to it's transpose.

And the symmetry of B gives from which it easily be seen that

However I'm unsure how to show that is orthonormal using the fact that is..?
7. (Original post by xfootiecrazeesarax)
However I'm unsure how to show that is orthonormal using the fact that 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.
8. (Original post by DFranklin)
It isn't. Why do you need it to be?

.

As post 4
I thought we wanted to allow the substitution to bring it to the simplified form .

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.
9. (Original post by xfootiecrazeesarax)
As post 4
I thought we wanted to allow the substitution to bring it to the simplified form .

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 and take the transpose. You end up with .

You can do something similar with inv(B).
10. (Original post by DFranklin)
To show B is symmetric, start from and take the transpose. You end up with .

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

So now I need the substituion rule of what will go to in terms of and on making the substituion

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

Apologies yes A is positive definite.

Okay so because then I'd have
So

if
Oh I just realised there is still the 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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