1.4. THE WRONSKI MAP AND THE SHAPIRO CONJECTURE 7 We display this geometric configuration in Figure 1.1. There, (i) is the line tangent to γ at the point γ(i). The hyperboloid has two rulings. One ruling contains our (−1) (0) (1) γ (−1) γ Figure 1.1. hyperboloid containing three lines tangent to γ. three tangent lines and the other ruling (which is drawn on hyperboloid) consists of the lines which meet our three tangent lines. Now consider the fourth line (s) which is tangent to γ at the point γ(s). This has the parameterization (s) = ( 6s2 1 , 7 2 s3 + 3 2 s , 3 2 s 1 2 s3 ) + t ( 12s , 21 2 s2 + 3 2 , 3 2 3 2 s2 ) . We compute the intersection of the fourth line with the hyperboloid. Substituting its parameterization into (1.7) and dividing by −12 gives the equation (s3 s)(s3 s + t(6s2 2) + 9st2) = 0 . The first (nonconstant) factor s3 s vanishes when (s) is equal to one of (−1), (−0), or (1)–for these values of s every point of (s) lies on the hyperboloid. The second factor has solutions t = 3s2 1 ± 3s2 + 1 9s . Since 3s2 + 1 0 for all s, both solutions will be real. In fact, for s = −1/3, this will have exactly two solutions. We may also see this geometrically. Consider the fourth line (s) for 0 s 1. In Figure 1.2, we look down the throat of the hyperboloid at the interesting part of this configuration. This picture demonstrates that (s) must meet the hyperboloid in two real points. Through each point, there is a real line in the second ruling which meets all four tangent lines, and this proves the Shapiro Conjecture for m = p = 2. Example 1.11 (Rational functions with real critical points). When m = 2, the Shapiro Conjecture may be interpreted in terms of rational functions. A rational function ρ(t) = f(t)/g(t) is a quotient of two univariate polynomials, f and g. This
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