# The first part of the Hilbert sixteenth problem for elliptic polynomials

A polynomial $$P(x,y)\in \mathbb{R}[x,y]$$ is called an elliptic polynomial if its highest homogeneous part does not vanish on $$\mathbb{R}^2\setminus\{0\}$$.

Inspired by the first part of the Hilbert 16th problem we ask that:

Is there an elliptic polynomial $$P(x,y)\in \mathbb{R}[x,y]$$ of degree $$n$$ which has a level set $$P^{-1}(c)$$ with more than $$n$$ connected components? Can elliptic polynomials produce $$M$$-curves?

• what do you mean by "last homogeneous part"? – user347489 May 20 at 20:53
• @user347489 for example $x^2+y^2$ is the last homogenous part of $x^2+y^2+ax+by+c$. – Ali Taghavi May 21 at 10:51
• Every polynomial is a sum if homogenous polynomials.the last homogenous part is the homogenous part with highest degree. – Ali Taghavi May 21 at 10:54
• @user347489 I revised the word "last" to "highest" – Ali Taghavi May 26 at 10:15
• What have you tried? Is it true for $n=2$? $n=4$? Also, possibly a more intrinsic way to describe your condition is in terms of projective space. Let $\overline P(x,y,z)\in\mathbb R[x,y,z]$ be the homogenization of your $P$. Then your condition is that $\overline P=0$ has no points on the line $z=0$ "at infinity" in $\mathbb P^2(\mathbb R)$. – Joe Silverman May 26 at 10:33

An elliptic polynomial of degree $$n=2m$$ can have at least $$m^2=n^2/4$$ real components. In particular the number of components can exceed $$n$$ once $$n \geq 6$$. For example, here is a Sage plot of the nine-component sextic curves $$(x^3-x)^2 + (y^3-y)^2 = \epsilon$$ for $$\epsilon = .02, .07, .14$$ in the square $$|x|,|y| < 1.25$$:

In general if $$P,Q$$ are polynomials of degree $$m=n/2$$, each with $$m$$ distinct real roots, then the degree-$$n$$ curve $$P(x)^2 + Q(y)^2 = \epsilon$$ has $$m^2$$ components for $$\epsilon>0$$ sufficiently small; as $$\epsilon \to 0$$ the components approximate ellipses (or circles) centered at the $$m^2$$ points $$(x,y)$$ with $$P(x)=Q(y)=0$$.

• Thank you very much for your very interesting answer. – Ali Taghavi Jun 1 at 9:53

You can get all the way up to $$\binom{n-1}{2}+1 = g+1$$, where $$g$$ is the genus. This is the maximal number of connected components a real curve of genus $$g$$ can have, so this is optimal.

To do this, I will use Viro's patchworking method. Itenberg and Viro already give an example of how to use patchworking to build a plane curve with $$g+1$$ connected components, so I will just show how to tweak it to use an elliptic polynomial. I can't write a better explanation of patchworking than the one I just linked, so I'll assume you read it.

Take the triangulation from Figure 7. .

Along each edge on the outside of the figure, there are $$2n$$ small triangles. Pair them into $$n$$ pairs, and merge each pair with its common neighbor to make $$n$$ trapezoids. Leave the rest of the figure as before. It is easy to check that this subdivision is still coherent. The resulting plane curve has the same topology as the original figure, but the big loop which crossed the line at infinity $$2n$$ times before is now disjoint from it.

Poking around Viro's website, I came across slides from a talk on Hilbert's 16th problem. On page 48, he says that Hilbert, in 1891, found a construction of a curve with $$g+1$$ real components by perturbing a union of two conics. .

The slides don't give a detailed explanation of Hilbert's construction, but it looks like it would produce an elliptic polynomial.