theta function like boundary conditions

Question

I have the boundary conditions: $$f_\alpha(z+1)=\mathrm{e}^{-\pi N(2z-1)}f_\alpha(z)$$ and $$f_\alpha(z+i)=f_\alpha(z),$$ now, I know a theta function of the form $$f_\alpha(z|\tau)=\sum_{m\in\mathbb{Z}}\mathrm{e}^{i\pi(m+\alpha/N)^2N\tau+2\pi i(m+\alpha/N)Nz}$$ satisfies boundary conditions of the form $$f_\alpha(z+1)=f_\alpha(z)$$ and $$f_\alpha(z+\tau)=\mathrm{e}^{-i\pi N(2z-\tau)}f_\alpha(z),$$ where $\tau=\tau_1+i\tau_2,$ is there a straight forward way to get from one to the other?

My guess was: $$g_\alpha(z|\tau)=\sum_{m\in\mathbb{Z}}\mathrm{e}^{i\pi(m+\alpha/N)^2Nz+2\pi i(m+\alpha/N)N\tau},$$ where $\tau=i$. I'm having trouble with the completing the square, it doesn't seem to reduce to the boundary conditions. Maybe there is an error in my calculation or my $g_\alpha(z|\tau)$ was incorrect. However it could be of the form $g_\alpha(iz|i\tau)$ but this doesn't seem to work either. Some help would be greatly appreciated.

Answer 1

Thank you reuns for your post! With your help I believe I have found an answer. With $$f(z+i)=f(z),\hspace{3cm}(1)$$ $$f(z+1)=\mathrm{e}^{-\pi N(2z-1)}f(z),\hspace{3cm}(2)$$ From $(1)$ I postulate $f(z)$ has the form $$f(z)=\sum_{n\in\mathbb{Z}}c_n\mathrm{e}^{2\pi nz}.\hspace{3cm}(3)$$ From $(2)$ we have: $$\sum_{n\in\mathbb{Z}}c_n\mathrm{e}^{2\pi nz+2\pi n}=\sum_{n\in\mathbb{Z}}c_n\mathrm{e}^{2\pi nz-2\pi Nz+\pi N}.\hspace{3cm}(4)$$ Looking at the right hand side of $(4)$ we see that this is equal to $$\sum_{n\in\mathbb{Z}}c_n\mathrm{e}^{2\pi nz-2\pi Nz+\pi N}=\sum_{n\in\mathbb{Z}}c_n\mathrm{e}^{2\pi z(n-N)+\pi N}=\sum_{n\in\mathbb{Z}}c_{n+N}\mathrm{e}^{2\pi nz+\pi N}.$$ From $(4)$ these are only equal if $$c_n\mathrm{e}^{2\pi n}=c_{n+N}\mathrm{e}^{\pi N},$$ or $$c_n=c_{n+N}\mathrm{e}^{\pi N-2\pi n}.\hspace{3cm}(5)$$ I solved this by letting $c_n=b_n\mathrm{e}^{f(n)}$ where $b_{n+N}=b_n$ this is solved by $$f(n)=-2\pi n+\frac{n^2\pi}{N}.\hspace{3cm}(6)$$ Since $b_n$ is periodic with period $N$ and the sum of the $c_n$ is over the integers, we can write the index as $n=\alpha+mN$, where $\alpha=0,1,...,N-1$ and $m\in\mathbb{Z}$. Therefore $$f(z)=\sum_{m\in\mathbb{Z}}\mathrm{e}^{-2\pi (\alpha+mN)+\frac{(\alpha+mN)^2\pi}{N}}\mathrm{e}^{2\pi (\alpha+mN)z}.\hspace{3cm}(7)$$ Taking out a factor of $N$ we find $$f_\alpha(z)=\sum_{m\in\mathbb{Z}}\mathrm{e}^{-2\pi N (\frac{\alpha}{N}+m)+(\frac{\alpha}{N}+m)^2N\pi}\mathrm{e}^{2\pi N (\frac{\alpha}{N}+m)z}.\hspace{3cm}(8)$$

We now double check the periodicity: Since $\alpha,N$, and $m$ are integers, $(1)$ is clearly satisfied. As for $(2)$ we have $$f_\alpha(z+1)=\sum_{m\in\mathbb{Z}}\mathrm{e}^{-2\pi N (\frac{\alpha}{N}+m)+(\frac{\alpha}{N}+m)^2N\pi}\mathrm{e}^{2\pi N (\frac{\alpha}{N}+m)z+2\pi N (\frac{\alpha}{N}+m)},$$ Looking at the squared term and the extra term coming from the boundary condition we have $$(\frac{\alpha}{N}+m)^2N\pi+2\pi N (\frac{\alpha}{N}+m)=\pi N(m^2+m(\frac{2\alpha}{N}+2)+\frac{\alpha^2}{N^2}+\frac{2\alpha}{N})$$ completing the square we have $$\pi N(m^2+m(\frac{2\alpha}{N}+2)+\frac{\alpha^2}{N^2}+\frac{2\alpha}{N})=\pi N(m-1+\frac{\alpha}{N})^2-\pi N$$ thus, $$f_\alpha(z+1)=\sum_{m\in\mathbb{Z}}\mathrm{e}^{-2\pi N (\frac{\alpha}{N}+m)+\pi N(m-1+\frac{\alpha}{N})^2-\pi N}\mathrm{e}^{2\pi N (\frac{\alpha}{N}+m)z}=\sum_{m\in\mathbb{Z}}\mathrm{e}^{-\pi N-2\pi zN+2\pi N}\mathrm{e}^{-2\pi N (\frac{\alpha}{N}+m)+(\frac{\alpha}{N}+m)^2N\pi}\mathrm{e}^{2\pi N (\frac{\alpha}{N}+m)z}=\mathrm{e}^{-\pi N(2z-1)}f_\alpha(z).$$