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Suppose $\Gamma$ is an analytic Jordan curve in the complex plane $\mathbb{C}$. What are ways to test whether $\Gamma$ is (contained in) an algebraic curve, i.e., whether there exists a real polynomial $P=P(x,y)$ such that $P(x,y)=0$ for all $x,y \in \mathbb{R}$ with $x+iy \in \Gamma$?

I know that Bézout's theorem can be used to show that certain curves are not algebraic, for instance curves that intersect a line in infinitely many points. But analytic curves in general also do not intersect lines in infinitely many points, so that's not helpful here.

Apologies if the question is too vague, but any comments or references would be appreciated.

Thank you!

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    $\begingroup$ Given $n$ points you can test for a polynomial curve of degree $d$ containing those $n$ points by linear algebra. This is a necessary condition for a curve of degree $d$ to contain the whole curve and for $n$ large one can heuristically guess if is close enough to sufficient in practice $\endgroup$
    – Will Sawin
    Commented Nov 14 at 22:12
  • $\begingroup$ @WillSawin By the way, is it true that for any curve which is not a union of intervals in an algebraic one, there always be some algebraic curve intersecting it in infinitely many points? $\endgroup$
    – Denis T
    Commented Nov 15 at 0:58
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    $\begingroup$ @DenisT No. A compact non-algebraic analytic curve will intersect any algebraic curve in finitely many points. $\endgroup$
    – Will Sawin
    Commented Nov 15 at 1:08
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    $\begingroup$ If your analytic curve satisfies a linear differential equation with polynomial coefficients, then the book D-Finite Functions by Manuel Kauers contains much information. See link.springer.com/book/10.1007/978-3-031-34652-1. $\endgroup$ Commented Nov 15 at 2:41
  • $\begingroup$ @RichardStanley Thank you for the reference! It seems to contain a lot of information indeed. I will take a closer look. $\endgroup$ Commented Nov 15 at 5:44

1 Answer 1

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Let $\Gamma$ be a non-algebraic analytic curve. Let $z_1,\dots, z_n$ be points of $\Gamma$. Let $X_{n,d}$ be the set of algebraic curves of degree $d$ that contain $z_1,\dots, z_n$. Now consider a random point $z_{n+1}$ on $\Gamma$ and the analogous space $X_{n+1,d}$. I claim $\dim X_{n+1,d} <\dim X_{n,d}$ almost surely.

Why is this? Let $Y_{n,d}$ be the set consisting of pairs of a curve in $X_{n,d}$ and a point $z \in \Gamma$ contained in that curve. Then $Y_{n,d}$ maps to $X_{n,d}$ and the fibers, being the intersection of two analytic curves, are discrete, and, lying on a Jordan curve, are bounded, hence are finite. So $\dim Y_{n,d} \leq \dim X_{n,d}$. Now $Y_{n,d}$ also maps to $\Gamma$, and $X_{n+1,d}$ is the fiber of $Y_{n,d}$ over $z_{n+1}$. If an analytic space maps to a curve then the fiber over a random point of the curve has dimension one less with high probability, so $\dim Z_{n+1,d} = \dim Y_{n,d}-1 \leq \dim X_{n,d}-1$.

By induction, it follows that $\dim X_{n,d} \leq \frac{d+2}{2}-1 - n$ since $\frac{d+2}{2}-1 $ is the dimension of the space of algebraic curves of degree $d$.

Thus if $n \geq \frac{d+2}{2}$ then with high probability if $z_1,\dots, z_n$ are random points then $X_{n,d}$ is empty.

On the other hand, if $\Gamma$ is a non-algebraic curve of degree $d$ then for any points $z_1,\dots, z_n$, $X_{n,d}$ is nonempty.

We can calculate $X_{n,d}$ using linear algebra and the coordinates of $z_1,\dots,z_n$ so this gives an empirical method to determine with high confidence whether $\Gamma$ is an algebraic curve of degree at most $d$. Of course the calculations are not completely precise in practice so we will more accurately get a method to determine if $\Gamma$ is close to an algebraic curve of degree at most $d$.

Then of course we can heuristically assume that if $\Gamma$ is not an algebraic curve of degree at most $d$ for some fixed $d$ then it probably isn't an algebraic curve at all, and if $\Gamma$ is close to an algebraic curve within the limits of the precision of our calculations then it probably is an algebraic curve.

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  • $\begingroup$ Ah I understand the heuristic test you had in mind now, thank you for taking the time to add the details! $\endgroup$ Commented Nov 16 at 0:21

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