# Contraction of multivectors

I am trying to understand a way to contract 1-vectors with p-vectors. So let this be a function $G$ $$G:V\times \wedge^p V\to \wedge^{p-1} V$$ We'd desire the following properties

$$G(\hat{\mathbf{x}}, \hat{\mathbf{x}}\wedge\hat{\mathbf{y}}) = -G(\hat{\mathbf{x}}, \hat{\mathbf{y}}\wedge\hat{\mathbf{x}}) =\pm\hat{\mathbf{y}}$$ $$G(\hat{\mathbf{x}}, \hat{\mathbf{y}}\wedge \hat{\mathbf{z}}) = 0$$

I am wondering how we might elegantly generalize and compute this function. I suppose we could algorithmically calculate $G(\mathbf{u}, \mathbf{v}_1\wedge\dots\wedge\mathbf{v}_p)$ in some manner like this

1. Express $\mathbf{v}_1\dots\mathbf{v}_n$ in a basis including $\mathbf{u}$.
2. Collect terms.
3. Send all terms without a $\mathbf{u}$ to $\mathbf{0}$.
4. Permute and sum to produce a single multivector term with $\mathbf{u}$ first.
5. Remove the leading $\mathbf{u}$ to produce a vector in $\wedge^{p-1} V$

I am not sure if this is a good or right way to think about it. Any explanation or reference suggestion would be appriciated. Are there elegant ways to express the result in terms of $\mathbf{u}$ and $\mathbf{v}_1\dots\mathbf{v}_p$?

• No function satisfying this property can be bilinear, since the LHS of your first condition would be quadratic in $x$ but $x$ doesn't even appear on the RHS. I am reasonably sure no nonzero bilinear function of the form you describe exists at all, with no other constraints, which is invariant under change of coordinates. What does exist are functions $V \otimes \wedge^p V \to \wedge^{p+1} V$ and $V^{\ast} \otimes \wedge^p V \to \wedge^{p-1} V$. Aug 1 '16 at 4:11
• Well I'm interested in $V^* \times \wedge^p V \to \wedge^{p-1} V$. Aug 1 '16 at 18:32
• In the case of the Hodge dual you have the extra structure of an inner product floating around, so you can identify $V$ and $V^{\ast}$. Aug 1 '16 at 18:54
• Yes, I'm sorry if I should have specified that $V$ was an inner product space. So my question is [how] can we express the Hodge dual as a series of steps. So we would like $\star \left(\mathbf{v}_1 \wedge \mathbf{v}_2\right) = G\Big(\mathbf{v}_1, G\big(\mathbf{v}_2, \mathbf{e}_1\wedge\dots\wedge \mathbf{e}_n\big)\Big)$. Aug 1 '16 at 22:08
• I've given an answer, but it sounds like you have a bigger question in mind? Is that comment what you really want to solve? Aug 3 '16 at 2:55