Weierstrass function

The Weierstrass function is:

$\wp(z) = \frac{1}{z^2} + \sum_{w \in L, w \neq 0}(\frac{1}{(z-w)^2} - \frac{1}{w^2})$

The Weierstrass function is a non-constant, doubly periodic function. Its significance appears to be immense, and it has the following properties:

• $\wp(z)$ converges absolutely and uniformly on compact sets not containing elements of $L$
• $\wp(z)$ is meromorphic in $\mathbb{C}$ and has double pole at each $w \in L$
• $\wp(z) = \wp(-z)$ for all $z \in \mathbb{C}$
• $\wp(z + w) = \wp(z)$ for all $w \in L$
• Every doubly periodic function is a rational function of $\wp$ and its derivative $\wp'.$

I didn’t delve too deeply into the first two points. I understood that the terms $\frac{1}{w^2}$ are included to ensure the sum converges. These terms make $|\frac{1}{(z-w)^2} - \frac{1}{w^2}|$ comparable to $|\frac{1}{w^3}|.$ And it holds:

$\sum_{w \in L, w \neq 0}\frac{1}{|w^k|}$

converges for $k > 2.$

The third point is easy to see. For the last point, let’s examine the derivative:

$\wp'(z) = \frac{-2}{z^3} + \sum_{w \in L, w \neq 0}\frac{-2}{(z-w)^3}$

$= \sum_{w \in L}\frac{-2}{(z-w)^3}$

This sum converges based on the result mentioned above.

We can see:

$\wp'(z + w_0) = \sum_{w \in L}\frac{-2}{(z + w_0-w)^3} = \sum_{w \in L}\frac{-2}{(z -w)^3} =\wp'(z)$

This implies $\wp(z+w) - \wp(z)$ is constant. By setting $z = -w/2,$ it quickly becomes evident that this constant is 0.

The fifth point surprises me. Let’s explore what’s behind it.

Every doubly periodic function is a rational function of Weierstrass function and its derivative

First, let’s see that every function can be expressed as the sum of an even function and an odd function:

$f(z) = \frac{f(z) + f(-z)}{2} + \frac{f(z) - f(-z)}{2}$

We can quickly observe that $\wp'(z)$ is an odd function. If a function $f(z)$ is odd, then $\frac{f(z)}{\wp'(z)}$ is even. Therefore, it suffices to prove the statement for even functions.

For even functions, we can consider zeros and poles grouped in pairs $F,$ each pair being $(w, w_3 - w).$ That’s useful because (being even and doubly periodic): $f(z) = f(-z) = f(w_3 - z).$

Remember, $w_3 = w_1 + w_2$, and $F$ represents the fundamental parallelogram.

Note that if zero or pole occurs at $w_3/2,$ we take some rational function of $f$ where this doesn’t happen.

So, let’s have zeros and poles of $f$ in $F:$

$(z_1, w_3 - z_1), ..., (z_n, w_3 - z_n)$

Let’s observe the function:

$\wp(z) - \wp(z_i)$

It has a double pole at 0 and zeros at $z_i$ and $w_3 - z_i.$

That means

$\pi_{i=1}^{n}(\wp(z) - \wp(z_i))$

has the same zeros and poles as $f.$

Let’s make them of the same order too:

$h(z) = \pi_{i=1}^{n}(\wp(z) - \wp(z_i))^{ord_{z_i}f}$

Now, $f(z) / h(z)$ has no zeros and no poles. That means, it is a constant.

We have proven that any even doubly periodic function is a rational function of $\wp(z).$ For odd doubly periodic functions, we also need $\wp'(z).$ Therefore, any doubly periodic function is a rational function of $\wp(z)$ and $\wp'(z).$

Weierstrass sigma function

$\sigma(z) = z \Pi_{w \in L, w \neq 0} (1 - \frac{z}{w}) e^{\frac{z}{w} + \frac{z^2}{2w^2}}$

Weierstrass sigma function has the following properties:

• $\sigma(z)$ is holomorphic for all $z \in \mathbb{C}$
• $\sigma(z)$ has simple zeros at each $w \in L$ and no other zeros
• $\frac{d^2}{dz^2}log(\sigma(z)) = -\wp(z)$
• Given $w \in L,$ there exists $a = a_w$ and $b = b_w$ such that $\frac{\sigma(z+w)}{\sigma(z)} = e^{az + b}.$

The first two points seem intuitive, I was lazy and didn’t follow the proofs. The third point can be quickly checked.

To prove the fourth point:

$\frac{d^2}{dz^2}log(\frac{\sigma(z+w)}{\sigma(z)}) = -\wp(z+w) + \wp(z) = 0$

So there exists $a = a_w$ and $b = b_w$ such that:

$log(\frac{\sigma(z+w)}{\sigma(z)}) = az + b$

Thus:

$\frac{\sigma(z+w)}{\sigma(z)} = e^{az + b}$

Special case of Abel-Jacobi theorem

I have no idea what the general form of the Abel-Jacobi theorem is. I would very much like to understand it, but currently, my algebraic geometry knowledge is too weak.

But it appears that the special case of the Abel-Jacobi theorem is:

Let $D = \sum n_i [w_i]$ be a divisor. Then $D$ is the divisor of a function if and only if $deg(D) = \sum n_i = 0$ and $\sum n_i w_i \in L.$

The definition of the divisor of a function $f$ is:

$div(f) = \sum_{w \in F} ord_w(f) [w]$

The notation $[]$ should just be regarded as a symbol.

That for the divisor of a function $f$, it holds $deg(div(f)) = 0$ and $\sum_{w \in F} w \cdot ord_w(f) \in L$ was discussed last time.

What about the other way around? Can $f$ be constructed such that it will have zeros and poles at $w_i,$ and $ord_{w_i}f = n_i?$

Let’s denote $l = \sum n_i w_i.$ So, we have $l \in L.$

Let’s define

$f(z) = \frac{\sigma(z)}{\sigma(z-l)} \Pi_i \sigma(z - w_i)^{n_i}$

From the fourth point, it follows that $\frac{\sigma(z)}{\sigma(z-l)}$ has no zeros and no poles. In $F,$ $\sigma(z-w_i)$ has simple zeros at $w_i.$ That means $div(f) = D.$

But is $f$ doubly periodic?

$\frac{f(z+w)}{f(z)} = \frac{\sigma(z+w)\sigma(z-l)}{\sigma(z) \sigma(z+w-l)} \Pi_i (\frac{\sigma(z - w_i + w)}{\sigma(z-w_i)})^{n_i}$

$= e^{az + b - a(z-l) - b} \Pi_i (e^{a(z-w_i)+b})^{n_i}$

$= e^{al} e^{\sum_i n_i(a(z-w_i)+b)}$

$= e^{al} e^{\sum_i n_i a z- n_i a w_i+ n_i b}$

$= e^{\sum_i n_i a z + n_i b}$

$= e^{(az + b) \sum_i n_i}$

$= 1$

since $\sum_i n_i = 0.$