Manuscript IX-Free@1 · IX-Free.tex · proposition · 33b20e96

Proposition: Exact mixed relation equations

F1: concrete presentation-relation kernels › Mixed degree-two relations · lines 557–585 · proof included or linked

IX-Free@1#prop:relations, IX-Free@1#eq:leftrelation, IX-Free@1#eq:rightrelation

For $v \in E_{L}$ ,

q_{L} (v) = 0 ⟺ \partial_{t} (v K) = 0 in D^{'} (M_{+}^{2}) .

For $v_{R} \in E_{R}$ , the right equation is

q_{R} (v_{R}) = 0 ⟺ \partial_{t^{'}} (v_{R} (y, x^{'}) K (x^{'}, y)) = 0.

Proof. The scalar slot and the $00$ -metric slot are independent. Differentiating the mixed polynomial once in each field direction recovers its coefficient distribution on product tests, with the normalization $2$ in (eq:qkernel). Finite sums of product tests are dense in the joint test space, so these pairings determine the distribution. The converse is immediate. The right argument is identical.

Displayed formulas in this passage

IX-Free@1#eq:leftrelation · line 560
$q_{L} (v) = 0 ⟺ \partial_{t} (v K) = 0 in D^{'} (M_{+}^{2}) .$
IX-Free@1#eq:rightrelation · line 568
$q_{R} (v_{R}) = 0 ⟺ \partial_{t^{'}} (v_{R} (y, x^{'}) K (x^{'}, y)) = 0.$

Proposition: Exact mixed relation equations

Displayed formulas in this passage

Nearby in this section