Manuscript XIV-E@1 · XIV-E.tex · proposition · e89fd2a5

Proposition: Exact transverse quadratic defect

The elementary product phase is not a common symbol calculus › An exact quadratic identity for the invisible directions · lines 1047–1084 · proof included or linked

XIV-E@1#prop:quadratic, XIV-E@1#eq:quadratic

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On this affine transverse slice,

- \frac{q _{R}}{β} + \frac{q _{1}}{α} + \frac{q _{2}}{β - α} = - \frac{∣ α v - β w ∣ ^{2}}{α β ( β - α )} .

There are two independent relative transverse components in $α v - β w$ . They are not detected by the first differentials of the three cone equations at the base point.

Proof. Orthogonality to $n$ and nullity of $ℓ_{n}$ give

q_{R} = - ∣ v ∣^{2}, q_{1} = - ∣ w ∣^{2}, q_{2} = - ∣ v - w ∣^{2} .

Therefore the left side of (eq:quadratic) equals

\frac{∣ v ∣ ^{2}}{β} - \frac{∣ w ∣ ^{2}}{α} - \frac{∣ v - w ∣ ^{2}}{β - α} .

Expanding the last square gives exactly the right side.

All three expressions have zero first derivative at $v = w = 0$ . The linear map $(v, w) \mapsto α v - β w$ has rank two, which identifies the two relative components invisible to that linearization.

Displayed formulas in this passage

XIV-E@1#eq:quadratic · line 1050
$- \frac{q _{R}}{β} + \frac{q _{1}}{α} + \frac{q _{2}}{β - α} = - \frac{∣ α v - β w ∣ ^{2}}{α β ( β - α )} .$
unlabeled · line 1067
$q_{R} = - ∣ v ∣^{2}, q_{1} = - ∣ w ∣^{2}, q_{2} = - ∣ v - w ∣^{2} .$
unlabeled · line 1073
$\frac{∣ v ∣ ^{2}}{β} - \frac{∣ w ∣ ^{2}}{α} - \frac{∣ v - w ∣ ^{2}}{β - α} .$

Proposition: Exact transverse quadratic defect

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