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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$?

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  • $\begingroup$ 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 $V^{\ast} \otimes \wedge^p V \to \wedge^{p-1} V$. $\endgroup$ Aug 1 '16 at 4:11
  • $\begingroup$ Well I'm interested in $V^* \times \wedge^p V \to \wedge^{p-1} V$. $\endgroup$
    – WhatIAm
    Aug 1 '16 at 18:32
  • $\begingroup$ In the case of the Hodge dual you have the extra structure of an inner product floating around, so you can identify $V$ and $V^{\ast}$. $\endgroup$ Aug 1 '16 at 18:54
  • $\begingroup$ 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 $\star \left(\mathbf{v}_1 \wedge \mathbf{v}_2\right) = G\Big(\mathbf{v}_1, G\big(\mathbf{v}_2, \mathbf{e}_1\wedge\dots\wedge \mathbf{e}_n\big)\Big)$. $\endgroup$
    – WhatIAm
    Aug 1 '16 at 22:08
  • $\begingroup$ 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? $\endgroup$
    – Muphrid
    Aug 3 '16 at 2:55
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I would not suggest this algorithm in the general case. It is much easier to use an arbitrary orthogonal (or orthonormal) basis and leverage linearity on the arguments to split the expression into terms only involving factorizable terms on the basis elements.

The problem of contractions in general is well understood, however. The operation naturally arises in use of clifford (or geometric) algebras.

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