Algebraic function with extra condition, what can it be? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T11:05:57Z http://mathoverflow.net/feeds/question/87143 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87143/algebraic-function-with-extra-condition-what-can-it-be Algebraic function with extra condition, what can it be? Per Alexandersson 2012-01-31T15:40:52Z 2012-02-04T11:38:21Z <p>I have an unknown function $\psi(\xi_1,\dots,\xi_n)$, such that $\psi$ satisfy an (unknown) polynomial equation with coefficients polynomials in the $\xi_i$.</p> <p>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 $A$.</p> <p>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 $\xi_i$?</p> http://mathoverflow.net/questions/87143/algebraic-function-with-extra-condition-what-can-it-be/87352#87352 Answer by Will Sawin for Algebraic function with extra condition, what can it be? Will Sawin 2012-02-02T17:59:03Z 2012-02-02T17:59:03Z <p>Is your function entire?</p> <p>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.)</p> <p>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.</p> <p>So, up to multiplication by a root of unity, it must be the option you gave.</p> http://mathoverflow.net/questions/87143/algebraic-function-with-extra-condition-what-can-it-be/87519#87519 Answer by Per Alexandersson for Algebraic function with extra condition, what can it be? Per Alexandersson 2012-02-04T11:38:21Z 2012-02-04T11:38:21Z <p>So, here is my stab at a proof, which actually do not require algebraicness of $\psi$</p> <p>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.</p> <p>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</p> <p>$$\psi - n \xi_1 \psi'_1 = \psi i A'(1)$$ 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.</p> <p>We do need that $\psi$ is differentiable, but this should be true almost everywhere for algebraic functions.</p>