User:Fly by Night/Sandbox2

Explanation
Let V be an n-dimensional vector space with inner product $$\langle \cdot, \cdot \rangle$$ and basis $$\{ e_1, \ldots, e_n \}$$. For every linear function $$f \in V^*$$, the exists a unique vector $$w \in V$$ such that $$f(v) = \langle v, w \rangle$$ for all $$v \in V$$. (The map $$f \mapsto w$$ is the canonical isomorphism between $$V^*$$ and V.)

Next, for for k with $$0 \le k \le n$$, consider the exterior power spaces $$\bigwedge^k V$$ and $$\bigwedge^{n-k} V$$. Given a k-vector $$\lambda \in \bigwedge^k V$$ and an (n–k)-vector $$\theta \in \bigwedge^{n-k} V$$, the exterior product $$\lambda \wedge \theta$$ will be an n-vector, i.e. $$\lambda \wedge \theta \in \bigwedge^n V$$. However, note that all n-vectors are scalar multiples of $$e_1\wedge \ldots \wedge e_n$$.

Consider a fixed $$\lambda \in \bigwedge^k V$$, for each $$\theta \in \bigwedge^{n-k} V$$, there exists a unique linear function $$f_{\lambda} \in \left(\bigwedge^{n-k} V\right)^*$$ such that $$ \lambda \wedge \theta = f_{\lambda}(\theta)\cdot (e_1\wedge\ldots\wedge e_n)$$. It follows that there exists a unique $$\star \lambda \in \bigwedge^{n-k} V$$ such that $$f_{\lambda}(\theta) = \langle \theta, \star \lambda \rangle$$ for all $$\theta \in \bigwedge^{n-k} V $$. The map $$f_{\lambda} \mapsto \star\lambda$$ is the canonical isomorphism between $$\left(\bigwedge^{n-k} V\right)^*$$ and $$\bigwedge^{n-k} V$$. The (n–k)-vector $$\star\lambda$$ is defined to be the Hodge dual of $$\lambda$$.