Manuscript Em02-D@4 · Em02-D-v4.tex · section · fcf2fac7

Fixed inverse-cubic witness

Independent separated obstruction and necessary support › Fixed inverse-cubic witness · lines 1954–1983

Em02-D@4#eq:witness

In $O_{A}$ , let

d_{t} = t^{0} - ∣ t ∣, d_{y} = (y^{0} - a^{0}) - n \cdot (y - a) .

Take a nonnegative nonzero compact tangential bump $ψ_{A}$ on the positive seed and aligned crossed segment, with fixed normal cutoffs. Set

Φ_{A, 2} = \frac{d _{t} ^{2}}{2} d_{y} ψ_{A}, Φ_{A, 1} = 0.

Choose its support where $ζ = 1$ .

At $r = - t$ , the two $r^{0}$ derivatives in $C_{5 a}$ must hit $d_{t}^{2} /2$ . On the crossed segment, the output normal derivative must hit $d_{y}$ . Since $k_{n} \cdot \nabla_{x} d_{y} = 2$ ,

c_{A} = C_{5 a} (Φ_{A}) = \frac{i κ}{8 π} \int d^{4} a d R (s) d R (t) f (a + s) \int_{0}^{1} ψ_{A} (a, s, t, θ) d θ \neq = 0.

Displayed formulas in this passage

Em02-D@4#eq:witness · line 1974
$c_{A} = C_{5 a} (Φ_{A}) = \frac{i κ}{8 π} \int d^{4} a d R (s) d R (t) f (a + s) \int_{0}^{1} ψ_{A} (a, s, t, θ) d θ \neq = 0.$
unlabeled · line 1957
$d_{t} = t^{0} - ∣ t ∣, d_{y} = (y^{0} - a^{0}) - n \cdot (y - a) .$
unlabeled · line 1964
$Φ_{A, 2} = \frac{d _{t} ^{2}}{2} d_{y} ψ_{A}, Φ_{A, 1} = 0.$

Nearby in this section

theorem: Theorem: Separated raw failure