Given two conjugate pairs of points in general position in $\mathbb{CP}^2$, there is a pencil of real conics containing these four points. Is there a real empty conic in this pencil?

  • $\begingroup$ What is perhaps more interesting (although a bit obvious in this case) is that the real manifold parameterizing smooth, real "empty" conics containing the four points is itself connected. I believe the analogue can fail, for instance, for empty quartic elliptic curves in $\mathbb{R}P^3$. $\endgroup$ – Jason Starr Aug 1 '14 at 13:39

Yes, there is a real empty conic containing the points. Denote the first conjugate set of points by $\{p,\overline{p}\}$, and denote the second conjugate set of points by $\{q,\overline{q}\}$. If the pairs are "general", then no three of these four points are collinear. In particular, the lines $L=\text{span}(p,q)$ and $\overline{L}=\text{span}(\overline{p},\overline{q})$ are distinct and intersect in a unique real point $r$. The union $L\cup \overline{L}$ is invariant under conjugation, hence it has a (homogeneous, degree 2) defining polynomial $F$ that with real coefficients. Choose a real chart $B$ about $r$ (or rather about a preimage point in $\mathbb{R}^3\setminus\{0\}$, or alternatively, dehomogenize $F$ on $B$). Since the only zero of $F$ on $B$ is at $r$, either $F$ is nonnegative with unique zero at $r$, or $F$ is nonpositive with unique zero at $r$. Up to replacing $F$ by $-F$, assume that $F$ is nonnegative.

Now consider the lines $P=\text{span}(p,\overline{p})$ and $Q=\text{span}(q,\overline{q})$. If $r$ were contained in $P$, then $P$ would equal $L$ so that $q$ is contained in $P$, i.e., $p$, $\overline{p}$ and $q$ are collinear, contrary to the hypothesis that the pairs are "general". Thus $r$ is not contained in $P$. By the same argument, also $r$ is not contained in $Q$. Thus $r$ is not contained in $P\cup Q$. Since $P\cup Q$ is invariant under conjugation, there is a defining equation $G$ with real coefficients. Since $r$ is not contained in $G$, either $G$ is positive or negative on $r$. Up to replacing by $-G$, assume that $G$ is positive. Up to shrinking $B$, we may assume that $G$ is positive on all of $B$.

Since $F$ has only one zero in $\mathbb{R}P^2$ at $r$, and since $G$ is positive on a neighborhood $B$ of $r$, for sufficiently small $\epsilon>0$, $F+\epsilon G$ has no zeroes in $\mathbb{R}P^2$ (I am using that $\mathbb{R}P^2$ is compact).

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