# Real intersections of plane cubic curves

Let $C$ and $D$ be plane cubic curves over the real numbers. Suppose that the real loci in the projective plane of $C$ and $D$ consist of two connected components. Denote $C_0$ and $D_0$ the bounded component (which is often referred as "oval") of $C$ and $D$, and denote $C_1$ and $D_1$ the unbounded component of $C$ and $D$ respectively.

I have following question: What is the maximal number of real intersections of $C_1$ and $D_1$ in the projective plane?

Of course a trivial bound is 9 given by Bezout's theorem, but I conjecture that the bound is rather 5. I think this should be well known, but I couldn't find any reference.

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The answer is $9$. Choose a curve $C$ with the required topology. Choose $8$ points $x_1$, $x_2$, ..., $x_8$ on $C_1$. The conditions of passing through the $x_i$ impose $8$ linear conditions on the $10$ dimensional space of cubics, so we can find a second cubic $E$ passing through the $x_i$ and not proportional to $C$. Let $x_9$ be the ninth intersection of $C$ and $E$.

I claim that $x_9$ is also on $C_1$. Proof: Consider $C_1$ as a loop in $\mathbb{RP}^2$, and let $\tilde{C}_1$ be the preimage in the universal cover $S^2$. The loop $C_1$ is not contractible so, if we travel all the way around $C_1$, we will go from $(x,y,z)$ to $(-x,-y,-z)$ in $S^2$. Since $E$ is an odd degree polynomial, $E(x,y,z) = - E(-x,-y,-z)$. So $E$ must change sign an odd number of times along $C_1$. We already know that it changes sign at $x_1$, ..., $x_8$; the only place to get an additional sign change is at $x_9$. (Note that this argument shows that $C_1 \cap D$ is always odd, counting with multiplicity. I think that a careful count will show that $C_1 \cap D_1$ is odd, and the other three possibilities are always even.)

Now, $E$ may not have the right topology. But, for $\epsilon$ sufficiently small, the curve $D= C+\epsilon E$ also passes through $x_1$, ..., $x_9$, and it does have the right topology.

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I think it's easier to show that $C_0 \cap D$ is even - it's just the Jordan curve theorem. Each time $D$ goes inside it must go back outside. – Will Sawin Aug 28 '12 at 14:46

[Revised because I read too quickly and thought the problem was to find two connected cubics meeting in nine points]

Here's a version of the construction with two sets of three lines that may be easier to parse visually:

Each triplet of lines forms an equilateral triangle; these are plots of $C=40c$ and $C=240c$ where $$C = (y-3) (y+6-3^{1/2}x) (y+6+3^{1/2}x),$$ $$c = (x+1) (x-2-3^{1/2}y) (x-2+3^{1/2}y).$$

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Are there disconnected curves in that pencil? – David Speyer Aug 28 '12 at 18:36
Apparently not, though there are two other reducible curves, $y^3 - y = \pm (x^3-x)$, each of which contains a line $y = \pm x$. – Noam D. Elkies Aug 28 '12 at 18:46
[except, of course, for the curves $x^3-x=0$ and $y^3-y=0$ themselves, though in the projective plane each of these reducible curves becomes connected through its point at infinity.] – Noam D. Elkies Aug 28 '12 at 19:09
@W. Sawin: Yes they do, projectively. Any smooth cubic $C$ has at most two connected components in ${\bf RP}^2$. There are lines $l$ such that $l \cap C$ consists of a single point; putting such a line at infinity by a projctive change of coordinates yields an equivalent model for which the one or two components remain connected also in the finite plane ${\bf R}^2$. Here such a line $l$ is $x=2$ (a vertical line separating the rightmost branches of the red and blue curves from the rest). The resulting picture is equivalent though without threefold symmetry. – Noam D. Elkies Aug 29 '12 at 3:33
As it happens such a change of coordinates is not required for the statement of the original question, which specified two components in the projective plane. For each curve the complement of the middle oval is projectively connected: it goes North to infinity, comes back up north from below, then Southwest to infinity and back, then Southeast to infinity and back where we began. – Noam D. Elkies Aug 29 '12 at 4:09

Another approach, building on Noam Elkies's answer:

$$(y - 5) (x - 2 y + 10.5) (x + 2 y - 10.5) = -0.001$$ $$(x - 5) (y - 2 x + 10.5) (y + 2 x - 10.5) = -0.001$$

The images below show the same graph at different resolutions. Unfortunately, getting far enough out to see all 9 points makes it hard to see the ovals cleanly. But the basic idea is to choose two triples of lines which form triangles (not which pass through a common point as in Noam's answer) and deform a little to make the topology work.

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