In the context of Geometric Algebra, in $ \mathbb{R}^3 $: Let $A$ and $B$ be bivectors (representing planes). Show that $ (\langle AB \rangle_2)^* $ is a vector in the line of intersection of the two planes, where the $ ^* $ represents the dual.

My approach: $ A^*, B^* $ are vectors orthogonal to $A$, $B$; $ (A^* \wedge B^*) $ represents the plane orthogonal to both $A$ and $B$, and $ (A^* \wedge B^*)^* $ is a vector orthogonal to this plane, thus lying in the line of intersection of $A$ and $B$. Using the duality of the inner and outer products: $$ (A^* \wedge B^*)^* = A^*\cdot B^{**} = A^*.(-B) = - A^*\cdot B$$

But not quite sure how to proceed from here. Is there a general commutativity property of the dot product of a vector and a bivector? Also, what is the geometric interpretation of this result?

  • $\begingroup$ What do you mean by "the plane orthogonal to both $A$ and $B$"? I presume you mean the plane orthogonal to their intersection? What is the definition of $\langle AB\rangle_2$? $\endgroup$ – Ted Shifrin Jun 30 '13 at 19:17
  • $\begingroup$ I suspect 997712 means "the plane that intersects $A$ and $B$ both at right angles". $\langle AB \rangle_2$ is a standard notation for "the bivector (that is, $2$-vector) part of the geometric product $AB$". $\endgroup$ – Muphrid Jun 30 '13 at 20:02
  • $\begingroup$ Yes, that's what i meant. Is my language incorrect? $\endgroup$ – user997712 Jul 1 '13 at 20:24

In general, vector-bivector contractions anti-commute. You can use this to break the contraction down into geometric products. You have, within a minus sign,

$$(IA) \cdot B = \frac{1}{2} (IAB - BIA)$$

where $I$ is the 3d pseudoscalar. You should use at this point that $I$ commutes with all blades in 3d, which allows you to factor it out, and you should then be able to recognize the rest of the expression is in fact $\langle AB \rangle_2 = A \times B$, the commutator product (MacDonald has the commutator product defined, I hope, right?).

Edit: remember, the dot product of a vector $c$ and a bivector $D$ gives the vector in $D$ that is orthogonal to $c$.

Edit edit: as always, if you don't know whether a given product commutes, you can use grade projection and associativity instead.

$$\langle IAB \rangle_1 = \langle (IA) B \rangle_1 = (IA) \cdot B$$

but using different associative grouping,

$$\langle IAB \rangle_1 = \langle I(AB) \rangle_1 = I \langle AB \rangle_2$$

$AB$ has only grade-0 and grade-2 terms, so this is exhaustive.

  • $\begingroup$ Thank you Muphrid, helpful as always. <br/> Macdonald defines the cross product: $ u\times v = (u\wedge v)^* $ (which then would be the dual of the commutator product?). <br/> I do have a few follow up questions: <br/> $\endgroup$ – user997712 Jul 1 '13 at 19:25
  • $\begingroup$ 1) What is the general rule regarding commutativity/anti-commutativity of dot products? Is there one? 2) How can i prove that the pseudoscalar commutes with every vector in 3d? I asked a generalized form of this question here: math.stackexchange.com/questions/433920/… $\endgroup$ – user997712 Jul 1 '13 at 19:57
  • $\begingroup$ 3) Geometrically, with vector $ c $ and bivector $ D $, $ c \cdot D = (c_{||} + c_{\bot})\cdot D = c_{||}\cdot D + 0 = c_{||}D $. Now let $ c_{||} = e_1, D = e_1e_2 $. Then $ C_{||}D = e_1e_1e_2 = e_2 $. If $ e_1 e_2 $ is an orthogonal basis, we get a vector which is perpendicular to $ c $. But what if $ e_1, e_2 $ are not perpendicular, i.e not a standard basis? I should get the same answer regardless of the basis, so what did i go wrong here? $\endgroup$ – user997712 Jul 1 '13 at 20:17
  • $\begingroup$ 1) In general, the dot product of a vector and a $k$-vector alternates in commutativity. Vector dot vector is symmetric, vector dot bivector is antisymmetric, vector dot trivector is symmetric, and so on. 3) $D$ must be composed of some number of orthogonal vectors in a geometric product, so your question is moot. $\endgroup$ – Muphrid Jul 1 '13 at 20:27

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