2.6 The Simplest Restrictions on a Complex Scheme

To begin with, recall the following obvious restriction, which was used in Section 2.2.

2.6.A   A curve with empty real point set is of type II.$\qedsymbol$

The next theorem is in a sense dual to 2.6.A.

2.6.B   An M-curve is of type I.

Proof. Let $A$ be an M-curve of degree $m$. Then $\mathbb{R}A$ is the union of

$$\frac{(m-1)(m-2)}2+1$$

disjoint circles lying on $\mathbb{C}A$, which is a sphere with $\frac{(m-1)(m-2)}2$ handles. That many disjoint circles necessarily divide a sphere with $\frac{(m-1)(m-2)}2$ handles. Indeed, cut $\mathbb{C}A$ along $\mathbb{R}A$. The Euler characteristic of a surface has not changed. It equals

$$2-2\left(\frac{(m-1)(m-2)}2\right)=2-(m-1)(m-2).$$

Then cap each boundary circle with a disk. Each component of $\mathbb{R}A$ gives rise to 2 boundary circles. Therefore the number of the boundary circles is $(m-1)(m-2)+2$. The surface which is obtained has Euler characteristic $2-(m-1)(m-2)+(m-1)(m-2)+2=4$. However, there is no connected closed surface with Euler characteristic 4. (A connected closed oriented surface is a sphere with $g$ handles for some $g\ge0$; it has Euler characteristic $2-2g\le2$.) $\qedsymbol$

2.6.C Klein's Congruence (see [Kle-22, page 172])   If $A$ is a curve of type I of degree $m$ with $l$ ovals, then $l\equiv [\frac m2]\mod2$.

Proof. Consider a half of $\mathbb{C}A$ bounded by $\mathbb{R}A$. Its Euler characteristic equals the half of the Euler characteristic of $\mathbb{C}A$, i.e. $1-\frac{(m-1)(m-2)}2$. Cap the boundary components of the half with disjoint disks. This increases the Euler characteristics by the number of components of $\mathbb{R}A$. In the case of even degree $m=2k$, the Euler characteristic of the result is $1-(2k-1)(k-1)+l\equiv k+l\mod2$. In the case of odd degree $m=2k+1$, it is $1-k(2k-1)+l\equiv k+l\mod2$. In both cases the Euler characteristic should be even, since the surface is closed orientable and connected (i.e. sphere with handles). Thus in both cases $k\equiv l\mod 2$, where $k=[m/2]$. $\qedsymbol$

2.6.D A Nest of the Maximal Depth (see [Rok-78, 3.6]).   A real scheme of degree $m$ containing a nest of depth $k=[m/2]$ is of type I.

Such a scheme exists and is unique for any $m$ (for even $m$ it is just the nest, for odd $m$ it consists of the nest and the one-sided component). To realize the scheme, perturb the union of $k$ concentric circles and, in the case of odd $m$, a line disjoint from the circles. The uniqueness was proved in 1.3, see 1.3.C.

I preface the proof of 2.6.D with a construction interesting for its own. It provides a kind of window through which one can take a look at the imaginary part of $\mathbb{C}P^2$.

As we know (see Section 2.2), the complex point set of a real line is divided by its real point set into two halves, which are in a natural one-to-one correspondence with the orientations of the real line. The set of all real lines on the projective plane is the real point set of the dual projective plane. The halves of lines comprise a two-dimensional sphere covering this projective plane. An especially clear picture of these identifications appears, if one identifies real lines on the projective plane with real planes in $\mathbb{R}^3$ containing 0. A half of a line is interpreted as the corresponding plane with orientation. An oriented plane corresponds to its positive unit normal vector, which is nothing but a point of $S^2$. The complex conjugation $conj$ maps a half of a real line to the other half of the same line. It corresponds to the reversing of the orientation, which, in turn, corresponds to the antipodal involution $S^2\to S^2:x\mapsto-x$.

There is a unique real line passing through any imaginary point of $\mathbb{C}P^2$. To construct such a line, connect the point with the conjugate one. The connecting line is unique since a pair of distinct points determines a line, and this line is real, since it coincides with its image under $conj$.

Consequently, there is a unique half of a real line containing an imaginary point of $\mathbb{C}P^2$. This construction determines a fibration $p:\mathbb{C}P^2\smallsetminus \mathbb{R}P^2\to S^2$. The fibres of $p$ are the halves of real lines. Note that conjugate points of $\mathbb{C}P^2\smallsetminus \mathbb{R}P^2$ are mapped to antipodal points of $S^2$.

Proof. [Proof of 2.6.D] Let $A$ be a real projective curve of degree $m$ with a nest of depth $[m/2]$. Choose a point $P\in\mathbb{R}P^2$ from the domain encircled by the interior oval of the nest. Consider the great circle of $S^2$ consisting of halves of real lines which pass through $P$. Since each line passing through $P$ intersects $\mathbb{R}A$ in $m$ points, it cannot intersect $\mathbb{C}A\smallsetminus \mathbb{R}A$. Therefore the great circle has no common point with the image of $\mathbb{C}A\smallsetminus \mathbb{R}A$ under $p:\mathbb{C}P^2\smallsetminus \mathbb{R}P^2\to S^2$. But the image contains, together with any of its points, the antipodal point. Therefore it cannot be connected, and $\mathbb{C}A\smallsetminus \mathbb{R}A$ cannot be connected, too. $\qedsymbol$