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I have an unknown function $\psi(\xi_1,\dots,\xi_n)$, such that $\psi$ satisfy an (unknown) polynomial equation with coefficients polynomials in the $\xi_i$.

The function is homogeneous, that is, $\psi(t \xi_1,\dots, t \xi_n) = t^c \psi(\xi_1,\dots,\xi_n)$. I also know that $|\psi(e^{i \theta_1},\dots,e^{i \theta_n})|=1$ when $\sum_i \theta_i =0$, which implies that $$\psi(e^{i \theta_1},\dots,e^{i \theta_n}) = e^{i A(\theta_1,\dots,\theta_n)}$$ for some real-valued function $A$.

Clearly, one option is that $\psi(\xi_1,\dots,\xi_n) = \xi_1^{p_1} \cdots \xi_n^{p_n}$ such that $p_1+\cdots+p_n = c$, but can one exclude any other form? How do one take advantage of the fact that $\psi$ is algebraic in the $\xi_i$?

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So, here is my stab at a proof, which actually do not require algebraicness of $\psi$

Notice that we have $|\psi(\xi_1, \dots \xi_n)| = 1$ whenever $\xi_1 \cdots \xi_n = 1.$ Thus, using the homogeneity property, we may see that $$\psi(t^{1-n}\xi_1, t \xi_2, t\xi_3, \dots ,t\xi_n) = \phi(\xi_1,\dots,\xi_n) e^{i A(t)}$$ for any $\xi_1,\dots,\xi_n,$ since we may normalize the $\xi_i$:s with $(\xi_1 \xi_2 \cdots \xi_n)^{1/n}$ and move this to the other side ($\phi$). Now, changing $t$ will not change the modulus of the product of the parameters to the function, so it may only affect the argument, and hence the form above.

Now, using homogeneity again yields $$t^c\psi(t^{-n}\xi_1, \xi_2, \xi_3, \dots ,\xi_n) = \phi(\xi_1,\dots,\xi_n) e^{i A(t)}$$ Differentiating both sides w.r.t. $t$ and then putting $t=1$ gives

$$\psi - n \xi_1 \psi'_1 = \psi i A'(1) $$ Now, this is a differential equation, which is easy to solve, one sees that $\psi = \xi_1^\beta F(\xi_2,\dots,\xi_n)$ for some constant $\beta$ and unknown function $F$. However, this argument can be made for any of the variables, yielding the desired result.

We do need that $\psi$ is differentiable, but this should be true almost everywhere for algebraic functions.

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Is your function entire?

An entire algebraic function is just a polynomial function. So you know that it's a sum of terms of that type. (Unless it might have poles, in which case it's a rational function.)

Fix the terms $\xi_1,...,\xi_{n-1}$ at roots of unity and let $xi_n$ vary. Then $\xi_n$ is a polynomial that takes roots of unity. Therefore it must be a constant power of $\xi_n$ times a root of unity. (by Schwartz reflection) By continuity, the power cannot depend on $\xi_1,...,\xi_{n-1}$, so the only terms must have $\xi_n$ that power. We can continue this argument for each $\xi_i$, thus proving that the expresion consists of only a single term.

So, up to multiplication by a root of unity, it must be the option you gave.

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  • $\begingroup$ If your function is allowed to have poles, this is false. Let $f$ be any rational linear transformation of the Riemann sphere that preserves the roots of unity, and let $g$ be a homogeneous polynomial of degree zero, then multiplying by $f\circ g$ will preserve all the properties given, and you can make your function arbitrarily complicated like this. $\endgroup$
    – Will Sawin
    Commented Feb 2, 2012 at 18:20
  • $\begingroup$ well maybe not arbitrarily. But extremely. $\endgroup$
    – Will Sawin
    Commented Feb 2, 2012 at 18:20
  • $\begingroup$ @Will Sawin: A polynomial of degree zero? That sounds very much like a constant... $\endgroup$
    – Per Alexandersson
    Commented Feb 2, 2012 at 22:09
  • $\begingroup$ oh sorry. i meant a rational function of total degree zero like $\zeta_1^2\zeta_2^{-1}\zeta_3^{-1}$ $\endgroup$
    – Will Sawin
    Commented Feb 3, 2012 at 7:22
  • $\begingroup$ @Will Sawin: But the rational linear transform must be homogeneous as well... $\endgroup$
    – Per Alexandersson
    Commented Feb 4, 2012 at 11:24

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