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.