Manuscript Em01-C@3 · Em01-C-v3.tex · corollary · f0953702

Corollary: Rank and definite quadratic defect

Exact characteristic normal coordinates · lines 1011–1043 · proof included or linked

Em01-C@3#cor:rank, Em01-C@3#eq:rank-relation, Em01-C@3#eq:quadratic-relation

Put $h = β - α > 0$ . Along $N$ ,

d q_{R} = 4 β d u, d q_{1} = 4 α d v, d q_{2} = 4 h (d u - d v) .

The three differentials have rank two there, and

d q_{2} = \frac{h}{β} d q_{R} - \frac{h}{α} d q_{1} .

Throughout the coordinate chart,

\frac{q _{1}}{α} + \frac{q _{2}}{h} = \frac{q _{R}}{β} - \frac{β}{α h} ∣ ξ ∣^{2} .

On $S_{resp}$ , simultaneous vanishing of $q_{1}, q_{2}$ is equivalent to $v = ξ = 0$ . Their restricted differentials have rank one along $N$ , with the remaining constraint supplied by the definite quadratic form in (eq:quadratic-relation).

Proof. Differentiate the exact equations at $u = v = ξ = 0$ . Adding their divided forms gives the nonlinear relation. On $S_{resp}$ , $q_{R} = 0$ ; positivity of $β / (α h)$ forces $ξ = 0$ , and then $v = 0$ .

Displayed formulas in this passage

Em01-C@3#eq:rank-relation · line 1020
$d q_{2} = \frac{h}{β} d q_{R} - \frac{h}{α} d q_{1} .$
Em01-C@3#eq:quadratic-relation · line 1025
$\frac{q _{1}}{α} + \frac{q _{2}}{h} = \frac{q _{R}}{β} - \frac{β}{α h} ∣ ξ ∣^{2} .$
unlabeled · line 1014
$d q_{R} = 4 β d u, d q_{1} = 4 α d v, d q_{2} = 4 h (d u - d v) .$

Corollary: Rank and definite quadratic defect

Displayed formulas in this passage

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