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Counter-Examples to Ragsdale Conjecture

The following theorem gives counter-examples to the Ragsdale Conjecture (or the conjecture of Petrovsky) (cf. [8]).

Theorem 8   For each integer $ k \geq 1$

a) there exists a nonsingular real algebraic plane projective curve of degree $ 2k$ with

$\displaystyle p = \frac{3k(k-1)}{2} + 1+
\left[\frac{(k - 3)^2 + 4}{8} \right]$

b) there exists a nonsingular real algebraic plane projective curve of degree $ 2k$ with

$\displaystyle n = \frac{3k(k-1)}{2} +
\left[\frac{(k - 3)^2 + 4}{8} \right] $

Remarks

The term $ \left[\frac{(k - 3)^2 + 4}{8} \right]$ is positive when $ k \geq 5$ (i. e., the first counter-examples to the Ragsdale Conjecture appear in degree 10; the patchwork of them is shown in Figures 2 and 3). On the other hand, it is known that there is no counter-example to the Ragsdale Conjecture among curves of lower degree. These counter-examples of degree 10 have the real schemes shown in Figure 8. One of these counter-examples can be improved in order to obtain a curve of degree 10 with $ n = 32$. The corresponding patchworks (for $ p = 32$ and $ n = 32$) are shown in Figures 2 and 3.

Figure 8: Schemes of counter-examples to Ragsdale Conjecture of degree 10
\begin{figure}\centerline{\epsffile{sch-d10.eps}}\end{figure}

Recently B. Haas [4] improved the construction presented below and obtained T-curves of degree $ 2k$ with

$\displaystyle p=\frac{3k(k-1)}{2}+1+\left[\frac{k^2-7k+16}{6}\right].$

Neither the counter-examples provided by the above theorem, nor curves constructed by Haas are M-curves. Moreover, as we mentioned above, it is not known if the conjecture of Petrovsky holds for M-curves. The only known counter-examples to the Ragsdale Conjecture among M-curves are the curves constructed by Viro [16] (see also [18]). It is curious that we did not succeed in presenting those M-counter-examples as T-curves.

Proof. Proof of Theorem Let us show, first, how to construct a curve of degree $ m = 2k$ with $ p =\frac {3k(k-1)}{2} + 2$.

Suppose that the hexagon $ S$ shown in Figure 9 is placed inside of the triangle $ T$ in such a way that the center of $ S$ has both coordinates odd. Any convex primitive triangulation of a convex part of a convex polygon is extendable to a convex primitive triangulation of the polygon. Inside of the hexagon $ S$, let us take the convex primitive triangulation shown in Figure 9 and extend it to $ T$.

Figure 9: Hexagon $ S$ and the patchwork fragment produced by it.
\begin{figure}\centerline{\epsffile{hex.eps}}\end{figure}

To apply the Patchwork Theorem we need to choose signs at the vertices. Inside of $ S$ put signs according to Figure 9; outside use the Harnack rule of distribution of signs.

It is easy to calculate that the corresponding piecewise-linear curve $ \mathcal{L}$ has exactly one even oval more than the Harnack curve constructed above (i. e. now $ p =\frac {3k(k-1)}{2} + 2$). One can verify that the curve obtained has the real scheme shown in Figure 10.

Figure 10: Real scheme of a curve of degree $ 2k$ with $ p =\frac {3k(k-1)}{2} + 2$.
\begin{figure}\centerline{\epsffile{sch-p2.eps}}\end{figure}

Consider the partition of the triangle $ T$ shown in Figure 11. Let us take in each shadowed hexagon the triangulation and the signs of the hexagon $ S$. The triangulation of the union of the shadowed hexagons can be extended to the primitive convex triangulation of $ T$. Let us fix such an extension. Outside of the union of the shadowed hexagons choose the signs at the vertices of the triangulation using the Harnack rule.

Figure 11: Partition of $ T$ for part a.
\begin{figure}\centerline{\epsffile{partofta.eps}}\end{figure}

Calculation shows that for the corresponding piecewise-linear curve $ \mathcal{L}$

$\displaystyle p = \frac{3k(k-1)}{2} +1+ a$

where $ a$ is the number of shadowed hexagons, and

$\displaystyle a = \left[\frac{(k - 3)^2 + 4}{8} \right]$

This curve has the real scheme shown in Figure 12.

Figure 12: Real scheme of a curve of degree $ 2k$ with $ p = \frac {3k(k-1)}{2} +1+ a$.
\begin{figure}\centerline{\epsffile{sch-pa.eps}}\end{figure}

To prove part b) of the theorem, let us take again the partition of the triangle $ T$ shown in Figure 11 with the triangulation and the signs of each shadowed hexagon coinciding with the triangulation and the signs of $ S$. Fix, in addition, a triangulation of a neighborhood of the axis $ OY$ and the signs at the vertices of the triangulation as shown in Figure 13 (the case $ k \equiv 1(\mod4)$). The chosen triangulation of the union of the shadowed hexagons and the neighborhood of the $ y$-axis can be extended to a primitive convex triangulation of $ T$. Outside of the union of the shadowed hexagons and the neighborhood of the $ y$-axis let us again choose the signs at the vertices of the triangulation using the Harnack rule.

Figure 13: Partition of $ T$ for part b.
\begin{figure}\centerline{\epsffile{partoftb.eps}}\end{figure}

The corresponding piecewise-linear curve $ \mathcal{L}$ has the real scheme shown in Figure 14. In this case $ n=\frac {3k(k-1)}{2}+a$. $ \qedsymbol$

Figure 14: Real scheme of a curve of degree $ 2k$ with $ n=\frac {3k(k-1)}{2}+a$.
\begin{figure}\centerline{\epsffile{sch-pb.eps}}\end{figure}


next up previous
Next: Bibliography Up: Patchworking Algebraic Curves Disproves the Previous: Patchworking Harnack Curves
2002-11-16