If two elliptic functions share the same poles and zeros (including multiciplity) then they are proportional

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I’m trying to understand the following statement found on my lecture notes:

If two elliptic functions share the same poles and zeros (including multiciplity) then they are proportional

I’m trying to reason the following way: if $f_1$ and $f_2$ are not constant, they must be meromorphic (since holomorphic elliptic functions are constant). Then they can be written as a rational function of two polynomials. I could now write the denominator and numerator of each of them as a product of linear factors (right?).

So $f_i=g_i/h_i$ where the zeros of $g_i$ are the zeros of $f_i$ and the zeros of $h_i$ are the poles of $f_i$. Finally, if $f_1$ and $f_2$ have the same zeros and poles then their numerators and denominators can only differ by a multiplicative constant respectively. Is that correct?

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A meromorphic function on $\Bbb{C}$ can't be globally written as a ratio of two polynomials in general. This is true if the meromorphic function has at worst a pole at infinity, but otherwise false. Also, since elliptic functions are biperiodic with respect to a rank $2$ lattice action, they have infinitely many poles (by the result you cite) so that if $e(z) = p_1(z)/p_2(z)$ were the expression of some elliptic function as a ratio of polynomials, it would follow that $e(z)$ has finitely many poles corresponding to the vanishing locus of $p_2(z)$.

Anyway, I'll suggest a few ways to see this:

As I mentioned in the comments, if you believe that an entire elliptic function is constant, then just note that a ratio of meromorphic functions $f_1,f_2$ with identical poles and zeros gives a global holomorphic function $f_1/f_2$ (after extending over removable discontinuities). Then, any condition of the form $f_i(z+\omega) = f_i(z)$ for $i=1,2$ is still satisfied by $f_1(z)/f_2(z)$.

Alternatively, here is a proof without assuming the fact about entire elliptic functions: $g = f_1/f_2$ extends (over some removable discontinuities) to a holomorphic function on all of $\Bbb{C}$. On the other hand, because $f_1$ and $f_2$ are periodic with respect to a rank $2$ lattice $\Lambda$ (by definition of being elliptic), $\lvert g(z)\rvert \le C$, where $C = \max\{g(z):z\in \overline{\Omega}\}$, where $\Omega$ is a fundamental domain for the $\Lambda$-action. Therefore, $g$ is an entire bounded function and hence constant by Liouville's theorem. So, $f_1(z) = \lambda f_2(z)$ for all $z$ and for some $\lambda \in \Bbb{C}$.

Alternatively, as elliptic functions correspond to meromorphic functions on a complex torus $\Bbb{C}/\Lambda$, it follows that they must have poles somewhere or else be constant, since a holomorphic function defined everywhere on a compact Riemann surface is constant.