Canonical tensors over a nondegenerate quadratic space

Let $(𝑉, ⟨ \cdot, \cdot ⟩)$ be a nondegenerate finite-dimensional quadratic space over $𝕂$ . Consider a basis ${𝑣 𝑖} 𝑛 𝑖 = 1$ , and let ${𝑣' 𝑖} 𝑛 𝑖 = 1$ be the corresponding dual basis. Then the element

𝜔 0 = 𝑛 \sum 𝑖 = 1 𝑣' 𝑖 \otimes 𝑣 𝑖 \in 𝑇 2 𝑉

is independent of the choice of ${𝑣 𝑖} 𝑛 𝑖 = 1$ , #m/thm/geoalg and so is its symmetrization

𝜔 1 = 𝑛 \sum 𝑖 = 1 𝑣' 𝑖 𝑣 𝑖 \in 𝑆 2 𝑉

Alternate representation

Let $𝑖 : 𝑉 * \to 𝑉$ be the linear isomorphism induced by $⟨ \cdot, \cdot ⟩$ and $𝑗 : E n d 𝕂 𝑉 \to 𝑉 * \otimes 𝑉$ be the canonical isomorphism. Then

𝜔 0 = ((𝑖 \otimes 1) \circ 𝑗) (i d 𝑉)

#state/tidy | #lang/en | #SemBr