Wednesday, March 27, 2019

Entry 38 The Jacobi, Borwein, and derived Jacobi theta functions

For consistency, let \(q\) be the nome's square \(q=e^{2\pi i \tau}\,\)throughout. 

I. The null Jacobi theta functions (with \(z=0\)) are,
$$\qquad\qquad\quad \vartheta_3(q) = \sum_{m=-\infty}^{\infty}q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}+\frac{4\,\eta^2(16\tau)}{\eta(8\tau)}=\frac{\eta^5(2\tau)}{\eta^2(\tau)\eta^2(4\tau)}$$$$
\vartheta_4(q) = \sum_{m=-\infty}^{\infty}(-1)^n\,q^{n^2} = \frac{\eta^2(\tau)}{\eta(2\tau)}$$$$
\vartheta_2(q) = \sum_{m=-\infty}^{\infty}q^{(n+1/2)^2} = \frac{2\,\eta^2(4\tau)}{\eta(2\tau)}$$
II. The Borwein cubic theta functions are,
$$a(q)=\sum_{m,n=-\infty}^{\infty}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)} +\frac{9\,\eta^3(9\tau)}{\eta(3\tau)}$$$$
b(q)=\sum_{m,n=-\infty}^{\infty}\zeta^{m-n}q^{m^2+mn+n^2} = \frac{\eta^3(\tau)}{\eta(3\tau)}\quad\quad$$$$
\quad c(q)=\sum_{m,n=-\infty}^{\infty}q^{(m+1/3)^2+(m+1/3)(n+1/3)+(n+1/3)^2} = \frac{3\,\eta^3(3\tau)}{\eta(\tau)}$$
where \(\zeta =e^{2\pi i/3}\). 

III. The derived Jacobi theta functions are,
$$\qquad\qquad d(q)= \vartheta_4^4(q)+2\,\vartheta_2^4(q) = \frac{\eta^8(\tau)}{\eta^4(2\tau)} +\frac{32\,\eta^8(4\tau)}{\eta^4(2\tau)}$$$$
\quad e(q)= \vartheta_4^4(q)= \frac{\eta^8(\tau)}{\eta^4(2\tau)}$$$$
\quad f(q)= \tfrac12 \vartheta_2^4(q^{1/2})= \frac{8\,\eta^8(2\tau)}{\eta^4(\tau)}$$ These obey the beautiful relations, $$\vartheta_3^4(q)=\vartheta_4^4(q)+\vartheta_2^4(q)$$$$ a^3(q)=b^3(q)+c^3(q)$$$$d^2(q)=e^2(q)+f^2(q)$$ As well as, $$\vartheta_3(q)\vartheta_3(q^3)=\vartheta_4(q)\vartheta_4(q^3)+\vartheta_2(q)\vartheta_2(q^3)$$$$ a(q)a(q^2)=b(q)b(q^2)+c(q)c(q^2)$$$$d(q)d(q)=e(q)e(q)+f(q)f(q)$$ the last of which naturally leads to the third one above. Also, $$\vartheta_3(q^4)=\vartheta_4(q)+\vartheta_2(q^4)$$$$ a(q^3)=b(q)+c(q^3)$$$$d(q^2)=e(q)+f(q^2)$$
The Jacobi thetas can be expressed in terms of each other,
$$\vartheta_3(q) = \vartheta_4(q)+2\,\vartheta_2(q^4)$$ $$\vartheta_4(q) = 2\,\vartheta_3(q^4)-\vartheta_3(q)$$ $$\vartheta_2(q) = \vartheta_3(q^{1/4})-\vartheta_3(q)$$ similarly for the Borwein thetas, $$a(q) = b(q) + 3\,c(q^3)$$ $$2\,b(q) = 3a(q^3)-a(q)$$ $$2\,c(q) = a(q^{1/3})-a(q)$$ and also similarly for the derived Jacobi thetas, $$d(q) = e(q) + 4\,f(q^2)$$ $$3\,e(q) = 4d(q^2)-d(q)$$ $$3\,f(q) = d(q^{1/2})-d(q)$$Furthermore, we have the similar, $$\big(\vartheta_3(q)\big)^2 = 1+4\sum_{n=0}^\infty\left(\frac{q^{4n+1}}{1-q^{4n+1}}-\frac{q^{4n+3}}{1-q^{4n+3}}\right)$$ $$\quad\quad a(q) = 1+6\sum_{n=0}^\infty\left(\frac{q^{3n+1}}{1-q^{3n+1}}-\frac{q^{3n+2}}{1-q^{3n+2}}\right)$$