Manuscript XV-B@2 · XV-B.tex · proposition · 4d3f12e8

Proposition: Separate value extensions need not form a linear map

What is still required for the full map › Relations among different reconstructed values · lines 2073–2101 · proof included or linked

XV-B@2#prop:relations

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If reconstructed kernels satisfy

j = 1 \sum^{n} λ_{j} k_{j} = 0,

a linear joint map must satisfy

j = 1 \sum^{n} λ_{j} M (k_{j}) = 0.

Separate extensions of the individual canonical values do not ensure this identity.

Proof. The first statement is linearity. For separately chosen extensions $t_{j}$ , canonical agreement only forces

r (j \sum λ_{j} t_{j}) = 0;

the sum may be nonzero in $Y$ .

For example, let $k_{2} = - k_{1}$ and let both canonical values be zero. If a point $p$ lies outside $O$ , the scalar extensions $t_{1} = δ_{p}$ , $t_{2} = 0$ have the required restrictions but violate $t_{1} + t_{2} = 0$ . This is an abstract interface example, not a permitted graph prescription.

Displayed formulas in this passage

unlabeled · line 2076
$j = 1 \sum^{n} λ_{j} k_{j} = 0,$
unlabeled · line 2080
$j = 1 \sum^{n} λ_{j} M (k_{j}) = 0.$
unlabeled · line 2091
$r (j \sum λ_{j} t_{j}) = 0;$

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section: Relations among different reconstructed values