4
$\begingroup$

The Jacobi theta function, $\theta(u;\tau)$ (in some convention which will be implicit below), has the following elliptic transformation behavior:

$$ \theta(u+ m + n \tau;\tau) = (-1)^{m+n} e^{2 \pi i (-n u - \frac{1}{2} n^2 \tau)} \theta(u,\tau) $$

In particular, this means that $\log \theta(u;\tau)$ picks up a linear shift under such a transformation. Then if we take two derivatives with respect to $u$, $(\log \theta(u;\tau))''$, this kills this linear piece and we find an elliptic function with a double pole at $u=0$, which is precisely the Weierstrass p-function. On the other hand, if we take a single derivative, $(\log \theta(u;\tau))' = \theta'(u;\tau)/\theta(u;\tau)$, we get a function with a single simple pole at $u=0$, but this can still pick up shifts by integer multiples of $2 \pi i$ under elliptic transformations. Then it is natural to consider the exponential:

$$ f(u) = \exp \bigg(\frac{\theta'(u;\tau)}{\theta(u;\tau)} \bigg) $$ Then $f(u)$ is an elliptic function with an essential singularity at $u=0$. I was wondering if this function (or its log) has any special name or significance in the study of elliptic functions.

$\endgroup$
1
  • 1
    $\begingroup$ the log of $f$ is, up to a linear function, the Weiestrass zeta function $\endgroup$
    – user8268
    Dec 5, 2018 at 18:02

1 Answer 1

0
$\begingroup$

It can be written as the sum $$\sum_{k\ge0}\frac1{k!}s_k(u;\tau),$$where$$ s_k(u;\tau):=\sum_{n=0}^k(-1)^n{k\choose n}\left(\frac{\theta'}\theta\right)^{k-n}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)$$and $s_k$ are the $a$-coefficients of$$\frac{\theta(u-a;\tau)}{\theta(u;\tau)}\exp\left({a\frac{\theta'}\theta}(u;\tau)\right).\tag1$$To show the latter function is elliptic, plug in your transformation identity for $\theta(u;\tau)$: \begin{align}\frac{\theta(u+m+n\tau-a;\tau)}{\theta(u+m+n\tau;\tau)}\exp\left({a\frac{\theta'}\theta}(u+m+n\tau;\tau)\right)&=\frac{(-1)^{m+n} e^{2\pi i (-n (u-a) -\frac{1}{2} n^2 \tau)} \theta(u-a,\tau)}{(-1)^{m+n} e^{2\pi i (-n u -\frac{1}{2} n^2\tau)} \theta(u,\tau)}\exp\left(a\left[-2in\pi+\frac{\theta'}\theta(u;\tau)\right]\right)\\ &=e^{2an\pi i }\frac{\theta(u-a,\tau)}{ \theta(u,\tau)}\exp\left(-2ain\pi+a\frac{\theta'}\theta(u;\tau)\right)\\ &=\frac{\theta(u-a;\tau)}{\theta(u;\tau)}\exp\left(a\frac{\theta'}\theta(u;\tau)\right) \end{align} Note that from expanding $(1)$ into powers of $a$, you'll get $s_0(u;\tau)=1$, $s_1(u;\tau)=0$, and $s_2(u;\tau)=(\theta''/\theta-\theta'^2/\theta^2)(u;\tau)=\wp(u\,|1/2;\tau/2)$. To find $s_{k\ge3}(u;\tau)$ more straightforwardly, first differentiate $s_k$'s expression with respect to $u$:\begin{align}s_k'(u;\tau)&=\sum_{n=0}^k(-1)^n{k\choose n}\left[(k-n)\left(\frac{\theta'}\theta\right)^{k-n-1}(u;\tau)\cdot s_2(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)-\left(\frac{\theta'}\theta\right)^{k-n}(u;\tau)\cdot\frac{\theta^{(n)}\theta'}{\theta^2}(u;\tau)+\left(\frac{\theta'}\theta\right)^{k-n}(u;\tau)\cdot\frac{\theta^{(n+1)}}\theta(u;\tau)\right]\\ &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+s_2(u;\tau)\sum_{n=0}^k(-1)^n\frac{k!}{(k-n-1)!n!}\left(\frac{\theta'}\theta\right)^{k-n-1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)+\sum_{n=0}^k(-1)^n{k\choose n}\left(\frac{\theta'}\theta\right)^{k-n}(u;\tau)\cdot\frac{\theta^{(n+1)}}\theta(u;\tau)\end{align}Then, with some reindexing and manipulations,\begin{align}s_k'(u;\tau)&=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)\sum_{n=0}^k(-1)^n\frac{(k-1)!}{(k-n-1)!n!}\left(\frac{\theta'}\theta\right)^{k-n-1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[1ex] &\quad+\sum_{n=1}^{k+1}(-1)^{n-1}{k\choose n-1}\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)\sum_{n=0}^k(-1)^n{k-1\choose n}\left(\frac{\theta'}\theta\right)^{k-1-n}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[1ex] &\quad+\sum_{n=1}^{k+1}(-1)^{n-1}\frac{k!}{(k-n+1)!(n-1)!}\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)\\[1ex] &\quad+ks_2(u;\tau)\left(\sum_{n=0}^{k-1}(-1)^n{k-1\choose n}\left(\frac{\theta'}\theta\right)^{k-1-n}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)+(-1)^k{k-1\choose k}\frac\theta{\theta'}(u;\tau)\frac{\theta^{(k)}}\theta(u;\tau)\right)+\sum_{n=1}^{k+1}(-1)^{n-1}\frac{k!n}{(k+1-n)!n!}\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)\\[1ex] &\quad+ks_2(u;\tau)s_{k-1}(u;\tau)+\sum_{n=1}^{k+1}(-1)^n\frac{k!(k+1-n)-(k+1)!}{(k+1-n)!n!}\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)s_{k-1}(u;\tau)\\[1ex] &\quad+\sum_{n=1}^{k+1}(-1)^n\left({k\choose n}-{k+1\choose n}\right)\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)s_{k-1}(u;\tau)-(-1)^0\left({k\choose 0}-{k+1\choose 0}\right)\left(\frac{\theta'}\theta\right)^{k+1}(u;\tau)\cdot\frac{\theta^{(0)}}\theta(u;\tau)\\[1ex] &\quad+\sum_{n=0}^{k+1}(-1)^n\left({k\choose n}-{k+1\choose n}\right)\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)s_{k-1}(u;\tau)-s_{k+1}(u;\tau)\\[1ex] &\quad+\sum_{n=0}^{k+1}(-1)^n{k\choose n}\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)s_{k-1}(u;\tau)-s_{k+1}(u;\tau)\\[1ex] &\quad+\sum_{n=0}^k(-1)^n{k\choose n}\left(\frac{\theta'}\theta\right)^{k-n+1}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)+(-1)^{k+1}{k\choose k+1}\left(\frac{\theta'}\theta\right)^0(u;\tau)\cdot\frac{\theta^{(k+1)}}\theta(u;\tau)\\[4ex] &=-\frac{\theta'}\theta(u;\tau)s_k(u;\tau)+ks_2(u;\tau)s_{k-1}(u;\tau)-s_{k+1}(u;\tau)\\[1ex] &\quad+\frac{\theta'}\theta(u;\tau)\sum_{n=0}^k(-1)^n{k\choose n}\left(\frac{\theta'}\theta\right)^{k-n}(u;\tau)\cdot\frac{\theta^{(n)}}\theta(u;\tau)\\[4ex] &=ks_2(u;\tau)s_{k-1}(u;\tau)-s_{k+1}(u;\tau) \end{align}Thus, $s_k(u;\tau)$ is elliptic, due to being expressible in terms of $s_2$ and its derivatives! My opening sum follows from substituting $a=1$ into $(1)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .