How to construct such a real algebraic curve

Question

Suppose $L$ is a real line in $\mathbb{RP}^2$, my question is: for a given postive integer $n$, is it possible to find a real projective algebraic curve $C$ of degree $n$ with maximum connected components such that $L$ does not intersect with any ovals of $C$?

It is known that, by Harnack's Theorem, the maximum number of connected components of a real projective algebraic curve of degree $n$ is equal to $H(n)=\frac{(n-2)(n-1)}{2}+1$. Furthermore, the connected components of M-curves (the curves with maximum connected components), is composed of one pseudo-line and $H(n)-1$ ovals if $n$ is odd, and are all ovels if $n$ is even.

An equivalent statement of my question is, can you construct a M-curve $C$ of degree $n$ such that all ovals of $C$ lie in $\mathbb{R}^2$, since you can move the line $L$ to the infinity $z=0$. There are some methods to construct a M-curve of degree $n$ (for example, the method given by Harnack), but it seems not easy to show there exists a line which does not meet any ovals of this M-curve.

Answer 1

You can always construct an M-curve with the property you asked for.

Let $L$ be a real line in the real projective plane and $d$ an even, respectively odd, integer. Then, there exists a real M-curve of degree $d$ intersecting $L$ in only pairs of complex conjugated points, respectively in $\frac {d-1} 2$ pairs of complex conjugated points and exactly one real point.

Proof: You can just modify a little bit the Harnack method. By Harnack’s construction, there exists a real M-curve $X$ of degree $d-1$ intersecting $L(\mathbb R)$ in $d-1$ real points (the line $L$ intersects only one real connected component of $X(\mathbb R)$). We can now perturb the union $X\cup L$ to a non-singular real curve $A$ of degree $d$ with the properties that we want. In fact, in order to construct $A$, it is enough to apply a small perturbation to the union $X\cup L$ with respect to $\frac d 2$ pairs of complex conjugated lines, respectively $\frac {d-1} 2$ pairs of complex conjugated lines and one real line (which must be chosen passing through a point of an appropriate connected component of $L(\mathbb R) \setminus X(\mathbb R)$).

To be more precise, we construct $A$ as the zero set of the following real homogeneous non-singular polynomial of degree $d$:

• let $f, g, h \in \mathbb R[x,y,z]$ be the homogeneous polynomials associated respectively to $X$, $L$ and the union of the lines chosen for the small perturbation;
• then, there exists a small enough real non-zero number $b$ such that $$A = \{(x,y,z) \in \mathbb R \mid f(x,y,z) g(x,y,z) + b h(x,y,z) = 0 \} \ .$$

Hope it was helpful.