Lemma: Gram matrix and inverse pairing

The Gram matrix is

Let $B^i=(G^{-1}{)}^{ij}{B}_j$. Then

as a contravariant tensor on the fiber.

Proof. A diagonal basis matrix has pairing with itself equal to one. An off-diagonal $B_{(ab)}$ contributes two entries, giving $2{\eta }^{aa}{\eta }^{bb}$. Distinct basis matrices have disjoint component support. This gives (eq:Gram). The remaining statements are the defining inverse-matrix identities.