4.2 Fiedler's Alternation of Orientations

Consider the pencil of real lines passing through the intersection point of real lines $L_0$, $L_1$. It is divided by $L_0$ and $L_1$ into two segments. Each of the segments can be described as $\{L_t\}_{t\in[0,1]}$, where $L_t$ is defined by equation $(1-t)\lambda _0(x)+t\lambda _1(x)=0\}$ under an appropriate choice of equations $\lambda _0(x)=0$ and $\lambda _1(x)=0$ defining $L_0$ and $L_1$, respectively. Such a family $\{L_t\}_{t\in[0,1]}$ is called a segment of the line pencil connecting $L_0$ with $L_1$.

A point of tangency of two oriented curves is said to be positive if the orientations of the curves define the same orientation of the common tangent line at the point, and negative otherwise.

The following theorem is a special case of the main theorem of Fiedler's paper [Fie-82].

4.2.A. Fiedler's Theorem.   Let $A$ be a nonsingular curve of type I. Let $L_0$, $L_1$ be real lines tangent to $\mathbb{R}A$ at real points $x_0$, $x_1$, which are not points of inflection of $A$. Let $\{L_t\}_{t\in I}$ is a segment of the line pencil, connecting $L_0$ with $L_1$. Orient the lines $\mathbb{R}L_0$, $\mathbb{R}L_1$ in such a way that the orientations turn to one another under the isotopy $\mathbb{R}L_t$. If there exists a path $s:I\to \mathbb{C}A$ connecting the points $x_0$, $x_1$ such that for $t\in(0,1)$ the point $s(t)$ belongs to $\mathbb{C}A\smallsetminus \mathbb{R}A$ and is a point of transversal intersection of $\mathbb{C}A$ with $\mathbb{C}L_t$, then the points $x_0$, $x_1$ are either both positive or both negative points of tangency of $\mathbb{R}A$ with $\mathbb{R}L_0$ and $\mathbb{R}L_1$ respectively.

Figure 26:

I give here a proof, which is less general than Fiedler's original one. I hope though that it is more visualizable.

Roughly speaking, the main idea of this proof is that, looking at a curve, it is useful to move slightly the viewpoint. When one looks at the intersection of the complexification of a real curve with complexification of real lines of some pencil, besides the real part of the curve only some arcs are observable. These arcs connect ovals of the curve, but they do not allow to realize behavior of the complexification around. However, when the veiwpoint (= the center of the pencil) is moving, the arcs are moving too sweeping ribbons in the complexification. The ribbons bear orientation inherited from the complexification and thereby they allow to trace relation between the induced orientation of the ovals connected by the arcs. See Figure 26

Proof. [Proof of 4.2.A] The whole situation described in the 4.2.A is stable under small moves of the point $P=L_0\cap L_1$. It means that there exists a neighbourhood $U$ of $P$ such that for each point $P'\in U$ there are real lines $L'_0$, $L'_1$ passing through $P'$ which are close to $L_0$, $L_1$, and tangent to $A$ at points $x'_0$, $x'_1$; the latter are close to $x_0$, $x_1$; there exists a segment $\{L'_t\}_{t\in I}$ of the line pencil connecting $L'_0$ with $L'_1$ which consists of lines close to $L_t$, and, finally, there exists a path $s':I\to \mathbb{C}A$ connecting the points $x'_0$ and $x'_1$, which is close to $s$, such that $s'(t)\in\mathbb{C}A\cap \mathbb{C}L'_t$.

Choose a point $P'\in U\smallsetminus \bigcup_{t\in I}\mathbb{R}L_t$. Since, obviously, $\mathbb{R}A$ is tangent to the boundary of the angle $\bigcup_{t\in I}L_t$ from outside at $x_0$, $x_1$, the new points $x'_0$, $x'_1$ of tangency are obtained from the old ones by moves, one of which is in the direction of the orientation of $\mathbb{R}L_t$, the other - in the opposite direction (see Figure 26). Since $P'\notin\bigcup_{t\in I}L_t$, it follows that no line of the family $\{L_t\}_{t\in I}$ belongs to the family $\{L'_t\}_{t\in I}$ and thus

$$s(\operatorname{Int}I)\cap s'(\operatorname{Int}I)\subset(\bigcup... ...hbb{R}L_t)\cap (\bigcup_{t\in I}(\mathbb{C}L']_t-\mathbb{R}L'_t))=\varnothing .$$

Thus the arcs $s(I)$ and $s'(I)$ are disjoint, and bound in $\mathbb{C}A$, together with the arcs $[x_0,x'_0]$ and $[x'_1,x_1]$ of $\mathbb{R}A$, a ribbon connecting arcs $[x_0,x'_0]$, $[x_1,,x'_1]$. This ribbon lies in one of the halves, into which $\mathbb{R}A$ divides $\mathbb{C}A$ (see Figure 26). Orientation, induced on the arcs $[x_0,x'_0]$, $[x_1,x'_1]$ by an orientation of this ribbon, coincides with a complex orientation. It proves, obviously, 4.2.A $\qedsymbol$

The next thing to do is to obtain prohibitions on complex schemes using Fiedler's theorem. It takes some efforts because we want to deduce topological results from a geometric theorem. In the theorem it is crucial how the curve is positioned with respect to lines, while in any theorem on topology of a real algebraic curve, the hypothesis can imply some particular position with respect to lines only implicitely.

Let $A$ be a nonsingular curve of type I and $P\in\mathbb{R}P^2\smallsetminus \mathbb{R}A$. Let $Z=\{L_t\}_{t\in I}$ be a segment of the pencil of lines passing through $P$, which contains neither a line tangent to $\mathbb{R}A$ at a point of inflection of $\mathbb{R}A$ nor a line, whose complexifications is tangent to $\mathbb{C}A$ at an imaginary point. Denote $\bigcup_{t\in I}\mathbb{R}L_t$ by $C$.

Fix a complex orientations of $A$ and orientations of the lines $\mathbb{R}L_t$, $t\in I$, which turn to one another under the natural isotopy. Orient the part $C$ of the projective plane in such a way that this orientation induces on $\mathbb{R}L_0$, as on a part of its boundary, the orientation selected above. An oval of $A$, lying in $C$ is said to be positive with respect to $Z$ if its complex orientation and orientation of $C$ induce the same orientation of its interior; otherwise the oval is said to be negative with respect to $Z$.

A point of tangency of $A$ and a line from $Z$ is a nondegenerate critical point of the function $A\cap C\to I$ which assigns to $x$ the real number $t\in I$ such that $x\in L_t$. By index of the point of tangency we shall call the Morse index of this function at that point (zero, if it is minimum, one, if it is maximum). A pair of points of tangency of $\mathbb{R}A$ with lines from $Z$ is said to be switching, if the points of the pair has distinct indices and one of the points is positive while the other one is negative; otherwise the pair is said to be inessential. See Figure 27.

Figure 27:

If $A$ is a nonsingular conic with $\mathbb{R}A\ne\varnothing$ and $\mathbb{R}A\subset C$ then the tangency points make a switching pair. The same is true for any convex oval. When an oval is deforming and loses its convexity, new points of tangency may appear. If the deformation is generic, the points of tangency appear and disappear pairwise. Each time appearing pair is an inessential pair of points with distinct indices. Any oval can be deformed (topologically) into a convex one. Tracing the births and deaths of points of tangency it is not difficult to prove the following lemma.

4.2.B [Fie-82, Lemma 2].   Let $\Gamma$ be a component of $\mathbb{R}A\cap C$ and $M$ be the set of its points of tangency with lines from $Z$. If $\Gamma \cap\partial C=\varnothing$, then $M$ can be divided into pairs, one of which is switching, and all others are inessential. If $\Gamma \cap\partial C\ne\varnothing$, and $\Gamma$ connects distinct boundary lines of $C$ then $M$ can be decomposed into inessential pairs. If the end points of $\Gamma$ are on the same boundary lines of $C$ then $M$ with one point deleted can be decomposed into inessential pairs.$\qedsymbol$ARRAY(0x8a8481c)

Denote the closure of $(\mathbb{C}A\smallsetminus \mathbb{R}A)\cap(\bigcup_{t\in I}\mathbb{C}L_t)$ by $S$. Fix one of the decomposions into pairs of the set of points of tangency of lines from $Z$ with each component of $\mathbb{R}A\cap C$ existing by 4.2.B. By a chain of points of tangency call a sequence of points of tangency, in which any two consecutive points either belong to one of selected pairs or lie in the same component of $S$. A sequence consisting of ovals, on which the selected switching pairs of points of tangency from the chain lie, is called a chain of ovals. Thus the set of ovals of $A$ lying in $C$ appeared to be decomposed to chains of ovals. The next theorem follows obviously from 4.2.A.

4.2.C   The signs of ovals with respect to $Z$ in a chain alternate (i.e. an oval positive with respect to $Z$ follows by an oval negative with respect to $Z$, the latter oval follows by an oval positive with respect to $Z$). $\qedsymbol$

The next theorem follows in an obvious way from 4.2.C. Contrary to the previous one, it deals with the signs of ovals with respect to the one-sided component in the case of odd degree and outer ovals in the case of even degree.

4.2.D [Fie-82, Theorem 3].   If the degree of a curve $A$ is odd and ovals of a chain are placed in the same component of the set

$$C\smallsetminus ($$one-sided component of $$\mathbb{R}A)$$

then the signs of these ovals alternate. If degree of $A$ is even and ovals of a chain are placed in the same component of intersection of $\operatorname{Int}C$ with the interior of the outer oval enveloping these ovals, then the signs of ovals of this chain alternate.$\qedsymbol$