Manuscript IX-C@1 · IX-C.tex · theorem · 03ce8e6b

Theorem: Exact scalar factorization tests

Actual-functional descent of the second stage › The calibrated scalar relation tests · lines 878–909 · proof included or linked

IX-C@1#thm:scalarfactor, IX-C@1#eq:leftscalarcondition, IX-C@1#eq:rightscalarcondition

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A calibrated bilinear map

ℓ_{2, c} : L_{2} \times B \to C 1, ℓ_{2, c} (Q_{f, g}, B_{f^{'}}) = N_{c} (f, g, f^{'}) 1,

exists algebraically if and only if

n_{c}^{L} ((ker q_{L}) \otimes T) = 0.

The analogous right map

r_{2, c} : B \times R_{2} \to C 1

exists if and only if

n_{c}^{R} (T \otimes (ker q_{R})) = 0.

When they exist, both maps are unique on these domains.

Proof. Apply Theorem thm:descent to $q_{L}$ and $q_{R}$ , using injectivity of the sharp test map on the other argument. Constants belong to $P_{Σ}$ , so the scalar target has no additional admission issue.

Displayed formulas in this passage

IX-C@1#eq:leftscalarcondition · line 888
$n_{c}^{L} ((ker q_{L}) \otimes T) = 0.$
IX-C@1#eq:rightscalarcondition · line 897
$n_{c}^{R} (T \otimes (ker q_{R})) = 0.$
unlabeled · line 881
$ℓ_{2, c} : L_{2} \times B \to C 1, ℓ_{2, c} (Q_{f, g}, B_{f^{'}}) = N_{c} (f, g, f^{'}) 1,$
unlabeled · line 893
$r_{2, c} : B \times R_{2} \to C 1$

Theorem: Exact scalar factorization tests

Displayed formulas in this passage

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