Manuscript XVI-K@1 · XVI-K.tex · lemma · aa7553bb

Lemma: Inverse fiber pairing

The selected source and differentiated graph kernels › Why the edge uses the dual basis · lines 742–777 · proof included or linked

XVI-K@1#lem:dual, XVI-K@1#eq:edgefiber

Back to manuscript

The Gram matrix and its inverse are

G = diag (1, - 2, - 2, - 2, 1, 2, 2, 1, 2, 1),

G^{- 1} = diag (1, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, 1, \frac{1}{2}, \frac{1}{2}, 1, \frac{1}{2}, 1) .

For $B^{i} = (G^{- 1})^{ij} B_{j}$ , one has

b (B^{i}, B_{j}) = δ^{i}_{j} .

After removal of its scalar factor $S$ , the edge fiber tensor is

E = (- P \otimes id) b^{- 1} = i \sum Z_{i} \otimes B_{i}, Z_{i} = - P B^{i} .

Proof. A diagonal basis element has squared pairing one. An off-diagonal $B_{(ab)}$ contributes two entries, giving $2 η^{aa} η^{bb}$ . Distinct basis matrices have disjoint component support. This proves the Gram matrix. The inverse-pairing identity $b^{- 1} = \sum_{i} B^{i} \otimes B_{i}$ then gives (eq:edgefiber).

Displayed formulas in this passage

XVI-K@1#eq:edgefiber · line 760
$E = (- P \otimes id) b^{- 1} = i \sum Z_{i} \otimes B_{i}, Z_{i} = - P B^{i} .$
unlabeled · line 745
$G = diag (1, - 2, - 2, - 2, 1, 2, 2, 1, 2, 1),$
unlabeled · line 748
$G^{- 1} = diag (1, - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, 1, \frac{1}{2}, \frac{1}{2}, 1, \frac{1}{2}, 1) .$
unlabeled · line 755
$b (B^{i}, B_{j}) = δ^{i}_{j} .$

Nearby in this section

section: Why the edge uses the dual basis