ABSTRACT
The motivation for this chapter comes from a simple observation. Consider the hyperbolic curve of degree m = 2k or 2k + 1, consisting of k nested ovals plus one pseudo-line if m is odd. The lines of a pencil based at a point in the innermost oval intersect all this curve at m real points. This fact may be used to prove that the hyperbolic curves are dividing. More generally, let us say that a real pencil of curves is totally real with respect to some real algebraic curve A if it has only real intersections with CA. Rokhlin proved in [70] that if an algebraic curve is swept out by a totally real pencil of lines, then this curve is dividing. The argument generalizes to pencils of curves of higher degrees. Can conversely any dividing curve be endowed with some totally real pencil? The answer to this question has been found recently to be affirmative, even when the base points of the pencil are not contained in the curve, see [53]. But the technique developed in that paper does not give an explicit description of the pencil. We study here the case of M − 2-sextics with real scheme 〈2q 1〈6〉〉 or 〈6q 1〈2〉〉. By a congruence due to Kharlamov [45], these two real schemes are of type I. Recall that the rigid isotopy class of a non-singular sextic is determined by its real scheme, plus its type I or II (see Section 11.1). Thus, any two M−2-sextics realizing the same real scheme among our two are rigidly isotopic. We shall call here the interior ovals inner ovals, and the exterior empty ovals outer ovals.
