Sum over the branches of a composition of an entire function with the branches of an algebraic function is entire.

Question

Let $l(t)$ be the solution of the polynomial equation $g(l,t)=\det(lE-(A-tB))=0$, where E is the identity and A, and B are $n\times n$ matrices. The natural domain of definition is a Riemann surface the choice of which possesses a degree of arbitrariness. In any case it has exactly $n$ sheets and $l$ has $n$ branches $l_1,\dots,l_n$.

How can we see that if $f$ is an entire function then $\sum_{i=1}^n f(l_i)$ is an entire function? Is this true for more general functions $l$?

You may assume the branches are analytic in a punctured neighborhood of infinity with a simple pole at infinity and that $l_j$ behaves as $a_{jj}-tb_{jj}+O(\frac 1 t)$ as $t\to\infty$. Here $a_{ij}$ and ${b_ij}$ are the entries of $A$ and $B$ where $B$ is diagonal and $A$ projects diagonally onto the degenerate eigenspaces of $B$.

Context: The problem comes from Herbert Stahl’s proof of the BMV conjecture. Stahl claimed the result is an immediate consequence of the monodromy theorem but this is not obvious. How does it follow? Is there another argument?

Answer 1

First of all, we note that

1. $G(t)=\sum_{i=1}^n f(l_i(t))$ is defined on $\mathbb{C}$, so as soon as we prove that it is holomorphic around every point, it is an entire function
2. $f(z)=\sum_{m=0}^\infty a_m z^m$

Now, denote by $\sigma_k(t)=\sum_{i=1}^n l_i(t)^k$; we will prove that \begin{equation}\sum_{m=0}^\infty a_m\sigma_m(t)\end{equation} converges uniformly on compact set. If it is the case, then obviously it converges to $G(t)$.

A. First of all, $\sigma_k$ is a holomorphic entire function for every $k$

Just write $$g(l,t)=p_t(l)=l^n + b_{n-1}(t)l^{n-1}+\ldots+b_1(t)l+b_0(t)$$ (by your definition $g$ is a monic polynomial in $l$). Then $\sigma_k(t)$ the sum of the $k$-th powers of the roots of $p_t(l)$, which is a polynomial in the symmetric functions of such roots, i.e. it is a polynomial function of $\sigma_h(t)$ and $b_j(t)$ for $h<k$ and $j=1,\ldots, n-1$ (Newton relations). So it is holomorphic.

B. The series $\sum_m a_m\sigma_m$ converges uniformly on compact sets

Take $R>0$; there is a constant $C=C(R)$ such that for every $t_0$ with $|t_0|<R$, the roots of $p_{t_0}(l)$ are contained in the disc $\{|l|<C\}$ (roots of a polynomial are continuous functions, hence sending compacts to compacts).

Therefore, if $|t_0|<R$, $$|\sigma_k(t_0)|\leq nC^n$$ so $$\left|\sum_m a_m\sigma_m(t_0)\right|\leq \sum |a_m|C^m<+\infty$$ as $f$ is entire.

Now, by the uniform convergence, $G(t)$ is holomorphic.