The Algebraic Correspondence of the Isosceles Pedal Map

We associate to the isosceles reduction of the Jonah pedal map an algebraic correspondence C t is a subset of CP to the 1 times CP to the 1, defined by ‎F(s,s prime )=(s prime ) squared D t(s) minus Q t(s) squared s squared =0‎. We prove that D t has degree exactly 4, so C t has bidegree (6,2). The defect identity D t minus Q t squared =(1 minus s squared )(1 minus 2Q t) isolates the defect polynomial Delta t=1 minus 2Q t, whose roots are s=1/2 (the equilateral triangle, for every t) and s=(2t minus 1)/(2t) (the isosceles triangle with that apex, which is the Jonah triangle at t=7/10). By Bézout's theorem the diagonal meets C t in 8 points counted with multiplicity, and we give their exact multiplicities. At t=7/10, the branch s prime =+G t(s) passes through the Jonah point with slope 451/343. An exact computation of the second iterate (a resultant of degree 36 at five values of t) is reported as a computation, not as a general proof. The conjecture N n=6 to the n+2 to the n for the number of fixed points of the n-th iterate remains open. No consequence for the Riemann Hypothesis is established. Status: identities and multiplicities proved; second-iterate computation exact at five values of t; slope verified numerically at six values of t; the conjecture on N n and the Julia set statement are labelled [C].