Manuscript XVI-O1-ST@5 · XVI-O1-ST-v5.tex · section · dd45e0d1

Descent without non-feedback

Retained exact synthetic examples › Descent without non-feedback · lines 1859–1884

Take

T = 101001102130, L_{T} = (- 1 0 - 1 0 1001), R = (2, - 1)^{T} .

The upper-left minor is one and $L_{T}$ is complete. For

F = 10102130, F_{d} = 10112132,

L_{T} F = 0, F R = (0, - 1, - 1, 0)^{T} \neq = 0,

L_{T} F_{d} = (0102) \neq = 0, L_{T} F_{d} R = 0.

Thus $F_{d}$ 's obstruction descends but does not vanish. For $F_{b} = (10110110)$ , one has $L_{T} F_{b} R = (0, 2)^{T} \neq = 0$ , so descent fails.

Displayed formulas in this passage

unlabeled · line 1862
$T = 101001102130, L_{T} = (- 1 0 - 1 0 1001), R = (2, - 1)^{T} .$
unlabeled · line 1869
$F = 10102130, F_{d} = 10112132,$
unlabeled · line 1873
$L_{T} F = 0, F R = (0, - 1, - 1, 0)^{T} \neq = 0,$
unlabeled · line 1876
$L_{T} F_{d} = (0102) \neq = 0, L_{T} F_{d} R = 0.$