The following theorem gives counter-examples to the Ragsdale Conjecture (or the conjecture of Petrovsky) (cf. [8]).
a) there exists a nonsingular real algebraic plane projective curve of degree with
b) there exists a nonsingular real algebraic plane projective curve of degree with
Remarks
The term is positive when (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 . The corresponding patchworks (for and ) are shown in Figures 2 and 3.
Recently B. Haas [4] improved the construction presented below and obtained T-curves of degree with
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.
Suppose that the hexagon shown in Figure 9 is placed inside of the triangle in such a way that the center of 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 , let us take the convex primitive triangulation shown in Figure 9 and extend it to .
To apply the Patchwork Theorem we need to choose signs at the vertices. Inside of 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 has exactly one even oval more than the Harnack curve constructed above (i. e. now ). One can verify that the curve obtained has the real scheme shown in Figure 10.
Consider the partition of the triangle shown in Figure 11. Let us take in each shadowed hexagon the triangulation and the signs of the hexagon . The triangulation of the union of the shadowed hexagons can be extended to the primitive convex triangulation of . 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.
Calculation shows that for the corresponding piecewise-linear curve
To prove part b) of the theorem, let us take again the partition of the triangle shown in Figure 11 with the triangulation and the signs of each shadowed hexagon coinciding with the triangulation and the signs of . Fix, in addition, a triangulation of a neighborhood of the axis and the signs at the vertices of the triangulation as shown in Figure 13 (the case ). The chosen triangulation of the union of the shadowed hexagons and the neighborhood of the -axis can be extended to a primitive convex triangulation of . Outside of the union of the shadowed hexagons and the neighborhood of the -axis let us again choose the signs at the vertices of the triangulation using the Harnack rule.
The corresponding piecewise-linear curve has the real scheme shown in Figure 14. In this case .