Originally published at 狗和留美者不得入内. You can comment here or there.

I am writing this as a way to go through in detail the section on elliptic functions in Schlag’s book.

**Proposition 4.14.** *Let and set . For any integer , the series*

*defines a function with . Furthermore, the Weierstrass function*

*is an even elliptic function of degree two with as its group of periods. The poles of are precisely the points in and they are all of order .*

**Proof.** It suffices to prove that converges absolutely and uniformly on every compact set . Periodicity allows us to restrict to the closure of any fundamental region. There exists such that for all ,

.

Hence, when , then

provided is sufficiently large. In

,

there are occurrences of , which means the above converges when , and this, with the above bound, means . Periodicity implies . Moreover, the degree of (4.16) is determined by nothing that inside a fundamental region the series has a unique pole of order .

For the second part, we note that when ,

,

which means the series defining , which is clearly even, converges absolutely and uniformly on compact subsets of . For the periodicity of , note that is periodic relative the same lattice . Thus, for every ,

with some constant , which has to be zero by

.

▢

Another way to go about it to define such that

,

so that , from which by periodicity, we have

.

From this we can solve that

where the s are constants for .

**Lemma 4.15.** *With as before, one has*

*where , and are pairwise distinct. Furthermore, one has so that *(4.21)* can be written in the form*

*with constants and .*

View the torus as

.

is odd and has a pole of order at but no other poles in , which means has degree .

Oddness with periodicity applied at and yields that

are the three zeros of , each simple, and thus also the unique points where has valency apart from . The are distinct, because if not would assume such a value four times, impossible when the degree is .

Denoting the RHS of (4.21) by , we have that

with the zeros cancelled out, and thus equal to a constant.

At , the highest pole of is one of order with coefficient . In , we have essentially a cubic in with leading coefficient , and has pole of order with coefficient . In taking the limit towards zero, we only need to consider the term, which has the highest order pole, which is also of order with coefficient . That means our constant function is .

The final statement follows by observing from the Laurent series around zero, which is, from the geometric series of

.

Because is even, the odd coefficients must vanish. So we have

.

For now, let

.

What we want is to find the such that becomes analytic and thus constant, and to do that we must vanish out all the poles at . The coefficients tells us to multiply by . After that, we have from the coefficient that , which means

.

▢

**Proposition 4.16.** *Every is a rational function of and . If is even, then it is a rational function of alone.*

**Proof****.** Suppose that is non-constant and even. Then for all but finitely many values of , the equation has only simple zeros (since there are only finitely many zeros of ). Pick two such and denote them by . Moreover, we can ensure that the zeros of and are distinct from the branch points of . Thus, since is even and with , one has:

The elliptic functions

and

have the same zeros and poles which are all simple. It follows that for some . Solving this relation for yields the desired conclusion.

If is odd, then is even so where is rational. Finally, if is any elliptic function, then

is a decomposition into even/odd elliptic functions whence

with rational as claimed. ▢

We conclude with the following question: given disjoint finite sets of distinct points and in as well as positive integers for and for , respectively, is there an elliptic function with precisely these zeros and poles and of the given orders? In the case of yes iff since the degree must be constant throughout.

For the tori, we first observe that by the residue theorem one has

where is the boundary of a fundamental region such that no zero or pole lies on the boundary. Second, comparing parallel sides of the fundamental region and using the periodicity shows that the left-hand side is (4.25) is of the form with and thus equals modulo . (This follows from that is the difference of logarithms of the same value, which regardless of branch must be an integer multiple of .)

Now consider the edges in given by and respectively. By -periodicity of we infer that

.

The branch of logarithm here is irrelevant, since the arbitrary constant is differentiated away. By periodicity applied to the difference in this integral,

.

The other edge pair gives an element of , whence (4.24).

**Theorem 4.17.** *Suppose *(4.23)* and *(4.24)* hold. Then there exists an elliptic function which has precisely these zeros and poles with the given orders. This function is unique up to a nonzero complex multiplicative constant.*

**Proof.** Listing the points and expanded out with their respective multiplicities, we obtain sequences and of the same length, say . Shifting the s and s by lattice elements if needed, one has

.

Take

using the in (4.20). Then

which shows periodicity. ▢

Finally, we observe how we can solve (4.22) by integrating

where we choose some branch of the root, which yields

.

In other words, the Weierstrass function is the inverse of an elliptic integral. The integration path in (4.30) needs to be chosen to avoid the zeros and poles of , and the branch of the root is determined by .

Analogously, is satisfied by with similar restriction on the path and the choice of branch. This case though is a periodic function with one period, whereas in (4.30), there are two periods.

**References**

- Schlag, W.,
*A Course in Complex Analysis and Riemann Surfaces*, American Mathematical Society, 2014, pp. 153-157.