ABSTRACT
We will use the method exposed in Section 6.2 to construct the nodal pencils. Let T be the triangle (12), (67), (17) containing 8. The condition that 1, . . . , 8 realizes the list max(1ˆ = 8−) splits into eight disjoint subconditions. There is an ordering 14567 > (8ˆ, 1) > (8ˆ, 7) > (8ˆ, 6) > · · · > (8ˆ, 2) > 34567 in T : the point 8 lies between two consecutive of these curves. When 8 lies on a cubic (8ˆ, N) (otherwise stated, the eight chosen base points are on a cubic (1−, N)n), the pencil is singular, with 9 = N . By Bezout’s theorem with the cubics (8ˆ, N), the degeneration 9 = 8 may occur only if 8 lies between (8ˆ, 1) and (8ˆ, 7). Start with 8 between 14567 and (8ˆ, 1); the elementary change 8 goes to the outside of 14567, 2, 3 < 14567 writes: 2ˆ : 1− → 4+, 3ˆ : 1− → 4+, it is realizable. The essential elementary pair (2ˆ = 1−, 3ˆ = 1−) gives thus two cubics: (1−, 3), (12, L). The ninth base point 9 must lie on the odd component of (12, L). By Bezout’s theorem, 9 lies then on the arc 1X of (1−, 3). Move 3 toward 4 along the loop of (12, L), leaving the other seven chosen base points fixed. The ninth point 9 moves along the cubic but cannot cross X (otherwise 9 should come together with 1 or 2 at X). So 9 stays on the odd component of (12, L). So the first degeneration of the pencil occurs when 3 reaches 4 (as 3 → 4 = 0), and the close pair (3ˆ = 1−, 4ˆ = 1−) gives two admissible pairs of distinguished cubics: (3, 1−), (1−, 4) or (1−, 3), (4, 1−). We know already that (1−, 3) belongs to the pencil, so the correct pair is the first one. Using the other pairs of consecutive points A,B with A → B = 0, we get other cubics: the pair 4, 5 gives again (4, 1−) and a new cubic (1−, 5), the pairs 5, 6 and 6, 7 give (6, 1−) and (1−, 7). Note that the points 2, 3 verify 2 → 3 = 0 but the corresponding close pair (2ˆ = 1−, 3ˆ = 1−) is inessential. The position of 9 on each cubic is again obtained with Bezout’s theorem. Until now we have found six cubics. Note that all of them have an arc 81, so the missing cubics correspond to the openings of this arc. Starting from (12, L), move into
Curves in the Real
the portion formed of cubics with ovals. The next distinguished cubic will be (81, L), then it will be (81, C). The complete pencil is shown in the first row of Table 8.1. Let now 8 follow a path p crossing successively (8ˆ, 1) and (8ˆ, 7). We will prove that at some moment, 8 must indeed come together with 9. Letting 8 cross (8ˆ, 1) changes the pencil swapping the positions of 1 and 9 on all combinatorial cubics, see the second row of Table 8.1. The point 8 lies now in the zone between the arcs 17 of (8ˆ, 1) and (8ˆ, 7), and this zone is divided in an upper and a lower part by the path p. Applying Bezout’s theorem with (8ˆ, 1) and (1−, 81), 2X on one hand, and with (8ˆ, 7) and (1−, 7), 18 on the other hand, we get that 9 lies also in this zone. When 8 follows the path p from (8ˆ, 1) and (8ˆ, 7), the point 9 moves from 1 to 7. Let 9 cross p at some point P and assume that at this moment, 8 has the position Q. Assume first that Q lies ahead of P , see Figure 8.1. When 8 previously had the position P , the point 9 was still in the upper zone. But on the other hand, 9 must have been placed at Q (recall that the positions of eight base points determine the position of the ninth one). This is a contradiction. One gets a similar contradiction taking Q behind P on the path. So, at some moment, one must have 9 = 8. Letting 8 successively cross (8ˆ, 1), . . . (8ˆ, 2), we obtain in total the first nine pencils of Table 8.1, they are drawn in Figure 8.2 where
(G,H, I,A,B,C,D,E, F ) = (9, 2, 3, 4, 5, 6, 7, 8, 1), (1, 2, 3, 4, 5, 6, 7, 8, 9),
(1, 2, 3, 4, 5, 6, 7, 9, 8), (1, 2, 3, 4, 5, 6, 9, 7, 8), (1, 2, 3, 4, 5, 9, 6, 7, 8),
(1, 2, 3, 4, 9, 5, 6, 7, 8), (1, 2, 3, 9, 4, 5, 6, 7, 8), (1, 2, 9, 3, 4, 5, 6, 7, 8),
(1, 9, 2, 3, 4, 5, 6, 7, 8).
