Self-dual vector space

A self-dual vector space is a finite-dimensional $𝕂$ -vector space $𝑉$ equipped with a bilinear form $⟨ ?, ? ⟩ : 𝑉 \times 𝑉 \to 𝕂$ and a distinguished element $𝜔 = \sum 𝑖 𝑣 𝑖 \otimes 𝑣 𝑖 \in 𝑉 \otimes 𝑉$ such that these data are the evaluation and coëvaluation respectively of a duality $𝑉 ⊣ 𝑉$ . #m/def/linalg Explicitly,

\sum 𝑖 𝑣 𝑖 ⟨ 𝑥, 𝑣 𝑖 ⟩ = 𝑥 = \sum 𝑖 ⟨ 𝑣 𝑖, 𝑥 ⟩ 𝑣 𝑖

for any $𝑥 \in 𝑉$ .

Equivalent characterization

Suppose $𝑉$ is a finite-dimensional vector space and $⟨ ?, ? ⟩ : 𝑉 \otimes 𝑉 \to 𝕂$ is a bilinear form. Then there exists an $𝜔$ making $𝑉$ a self-dual vector space iff $⟨ ?, ? ⟩$ is nondegenerate, moreover such a $𝜔$ is unique.

Proof

We adopt the Einstein summation convention.

First suppose there is some $𝜔 = 𝑣 𝑖 \otimes 𝑣 𝑖$ making $𝑉$ a self-dual vector space. Then if $⟨ 𝑥, 𝑉 ⟩ = 0$ then $𝑥 = ⟨ 𝑥, 𝑣 𝑖 ⟩ 𝑣 𝑖 = 0$ , so $⟨ ?, ? ⟩$ is nondegenerate.

Conversely, suppose $⟨ ?, ? ⟩$ is nondegenerate. Since $𝑉$ is finite-dimensional, there exists a basis ${𝑣 𝑖}$ , which has a reciprocal basis ${𝑣 𝑖}$ so that $⟨ 𝑣 𝑖, 𝑣 𝑗 ⟩ = 𝛿 𝑖 𝑗$ . We claim that $𝜔 = 𝑣 𝑖 \otimes 𝑣 𝑖$ makes $𝑉$ self-dual. Indeed, for any $𝑥 = 𝑥 𝑖 𝑣 𝑖 = 𝑥 𝑖 𝑣 𝑖 \in 𝑉$ we have

⟨ 𝑥, 𝑣 𝑖 ⟩ 𝑣 𝑖 = 𝑥 𝑗 ⟨ 𝑣 𝑗, 𝑣 𝑖 ⟩ 𝑣 𝑖 = 𝑥 𝑖 𝑣 𝑖 = 𝑥 = 𝑥 𝑖 𝑣 𝑖 = 𝑥 𝑗 ⟨ 𝑣 𝑖, 𝑣 𝑗 ⟩ 𝑣 𝑖 = ⟨ 𝑣 𝑖, 𝑥 ⟩ 𝑣 𝑖

as required. Uniqueness follows from uniqueness of duals.

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