Is Jacobi Theta function same as Heat Kernel ? How to derive Jacobi Theta from Heat Kernel?

Question

My understanding is that the Jacobi Theta function is fundamental solution of heat equation:

$\displaystyle \vartheta (x,it)=1+2\sum _{n=1}^{\infty }\exp \left(-\pi n^{2}t\right)\cos(2\pi nx)$

The following heat kernel is also fundamental solution of heat equation:

$\Phi (x,t)={\frac {1}{\sqrt {4\pi kt}}}\exp \left(-{\frac {x^{2}}{4kt}}\right)$

But I do not see how to expand above heat kernel to get Jacobi Theta function.

Can anyone provide some hints on how to expand above heat kernel to get Jacobi Theta function ?

Thank you.

Answer 1

I recommend the book A Brief Introduction to Theta Functions by Richard Bellman reprinted by Dover Publications. The first expansion you wrote is the Fourier series of the theta function. Using the heat kernel you sum over a lattice of periods to match the periodicity of the theta function. Something like $\sum_{k\in \mathbb{Z}} \Phi (x+2\pi k,t)$. Notice the Fourier transform connection between the two exponentials in your two equations. The summation over a period lattice is equivalent to convolution by a periodic sum of shifted Dirac delta functions (a Dirac comb), and the Fourier transform of this is the point-wise product of the Fourier transform of the heat kernel with another Dirac comb.

Answer 2

HINT :

To derive Jacobi Theta from Heat Kernel just compare the respective PDE.

$$\vartheta_3 (z,q)=1+2\sum _{n=1}^{\infty }q^{n^{2}}\cos(2 nz)$$ This Jacobi Theta function is a solution of : $$4\,q\,\frac{\partial\,\vartheta_3}{\partial q}+\frac{\partial^2\,\vartheta_3}{\partial z\;^2}=0$$

It looks like the heat equation, isn't it ?

http://functions.wolfram.com/EllipticFunctions/EllipticTheta3/13/01/

With $z=\pi x$ and $q=e^{-\pi t}$ , I suppose that you can take it from here.