1
$\begingroup$

I am reading a proof of the fact that every even elliptic function $f$ with periods $1$ and $\tau$ is a rational function of the Weierstrass $\wp$ function.

The proof seems to use this fact often, but I don't understand where it comes from:

If $f$ is an even elliptic function with a zero (or a pole) at $0$, $1/2$, $\tau/2$, or $(1+\tau)/2$, the order of the zero (or pole) is even.

What I do know is that for elliptic functions [not necessarily even], (a) the number of poles (up to congruency) is $\ge 2$, and (b) the number of zeros (up to congruency) is the same as the number of poles. I don't understand why evenness of $f$ forces the order of the lattice points and half periods to be even. Why can't for example, one have a simple pole at $0$ and a simple pole at $1/2$? Apologies if this is very obvious...

So, if I have stated the above fact correctly, does this imply the following?

If $f$ is an even elliptic function, it has even order.

Thanks in advance!

$\endgroup$

1 Answer 1

4
$\begingroup$

If $\omega$ is a period of $f$ and $f$ is even, then we have $f(\omega/2+z) = f(\omega/2+z-\omega) = f(-\omega/2+z) = f(\omega/2-z)$, which shows that the order of vanishing of $f$ at $\omega/2$ is even.

If $f$ is even then $f'$ is odd, $f''$ is even, $f'''$ is odd, and so on. In particular, $f'(0) = f'''(0) = \ldots = 0$. So if $f$ is nonzero then the vanishing order of $f$ at $0$ cannot be odd, so it must be even.

Similarly, if $z \mapsto f(\omega/2+z)$ is even, its order of vanishing at $0$, which is the order of vanishing of $f$ at $\omega/2$, is even.

$\endgroup$
2
  • $\begingroup$ Sorry, I understand the equality that you wrote, but I don't understand how you arrived at the conclusion about the order of vanishing. $\endgroup$
    – angryavian
    Commented Dec 3, 2013 at 14:49
  • $\begingroup$ Thanks for the clarification! $\endgroup$
    – angryavian
    Commented Dec 3, 2013 at 15:33

You must log in to answer this question.

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