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I read in Wiki (https://en.wikipedia.org/wiki/Bivector) that antisymmetric part of geometric product can be represent as $(a \land b)$ and $(a \cdot b)^2 - (a \land b)^2 = a^2b^2$

I have 2 questions:

  1. How to derive $(a \cdot b)^2 - (a \land b)^2 = a^2b^2$ ?

solved: $(ab)(ba) = (a \cdot b)^2 - (a \land b)^2$

  1. Why $(a \land b)=|a||b||\sin(\phi)|$ if $(a \land b)^2=-|a|^2|b|^2\sin^2(\phi)$?

solved: $B=(a \land b)=|a|b||sin(\phi)|B*$ where $B*$ is unit bivector.

Thanks.

Mike_bb
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    What work have you done in trying to answer these questions by yourself? – Nicholas Todoroff Feb 21 '24 at 18:26
  • @NicholasTodoroff First question is solved. $(ab)(ba) = (a \cdot b)^2 - (a \land b)^2$ – Mike_bb Feb 22 '24 at 05:32
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    There is no way $(a\wedge b)=|a||b|\sin(\phi)$ as the left-hand side is a bivector and the right-hand side is a real number. – Thato Tsabone Feb 22 '24 at 07:22
  • @ThatoTsabone Second question is solved. $B=(a \land b) = |a||b|sin(\phi)B$ where $B$ is unit bivector. – Mike_bb Feb 22 '24 at 09:50
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    @Mike_bb Awesome! Maybe fix your question so that someone with a similar question understands. – Thato Tsabone Feb 22 '24 at 10:07
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    Geometric algebra is unnecessarily esoteric obfuscating how simple their ideas are in the end. When it is combined with Wikipedia it gets even worse. $ab=a\cdot b+(a\wedge b)$ is a new operation that adds together scalars and bivectors similarly to Gauss adding real and imaginary numbers. In the GA case we add something that is symmetric in $a,b$ (the scalar $a\cdot b$) with something that is anti symmetric in $a,b$ (the bivector $a\wedge b,.$) – Kurt G. Feb 22 '24 at 12:09
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    Then $$\require{cancel} (ab)(ba)=(a\cdot b+a\wedge b)(a\cdot b-a\wedge b)=(a\cdot b)^2+\cancel{(a\cdot b)(a\wedge b)}-\cancel{(a\cdot b)(a\wedge b)}-(a\wedge b)^2 $$ – Kurt G. Feb 22 '24 at 12:11

1 Answers1

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This is really two questions, first.

Square.

Given

$\mathbf{a} \cdot \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} } \right),$

$\mathbf{a} \wedge \mathbf{b} = \frac{1}{{2}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right),$

we have

$\begin{aligned}\left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 - \left( { \mathbf{a} \wedge \mathbf{b} } \right)^2 &=\frac{1}{{4}} \left( { \mathbf{a} \mathbf{b} + \mathbf{b} \mathbf{a} } \right)^2 - \frac{1}{{4}} \left( { \mathbf{a} \mathbf{b} - \mathbf{b} \mathbf{a} } \right)^2 \\ &=\frac{1}{{4}} \left( { \left( {\mathbf{a} \mathbf{b}} \right)^2 +\left( {\mathbf{b} \mathbf{a}} \right)^2+ \mathbf{a} \mathbf{b} \mathbf{b} \mathbf{a}+ \mathbf{b} \mathbf{a} \mathbf{a} \mathbf{b} } \right)-\frac{1}{{4}} \left( { \left( {\mathbf{a} \mathbf{b}} \right)^2+\left( {\mathbf{b} \mathbf{a}} \right)^2- \mathbf{a} \mathbf{b} \mathbf{b} \mathbf{a}- \mathbf{b} \mathbf{a} \mathbf{a} \mathbf{b} } \right) \\ &=\frac{1}{4} \left( { 2 \mathbf{a}^2 \mathbf{b}^2} \right)-\frac{1}{4} \left( { - 2 \mathbf{a}^2 \mathbf{b}^2} \right) \\ &= \mathbf{a}^2 \mathbf{b}^2.\end{aligned}$

Assuming a Euclidean vector space, you may introduce a bivector norm defined as

$\left\lVert {\mathbf{a} \wedge \mathbf{b}} \right\rVert = \sqrt{ -\left( { \mathbf{a} \wedge \mathbf{b}} \right)^2 },$

then we may write this identity is a slightly prettier form

$\mathbf{a}^2 \mathbf{b}^2 = \left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 + \left\lVert {\mathbf{a} \wedge \mathbf{b}} \right\rVert{}^2.$

This is a generalization of the 3D identity

$\left\lVert \mathbf{a} \right\rVert{}^2 \left\lVert \mathbf{b} \right\rVert{}^2 = \left( { \mathbf{a} \cdot \mathbf{b} } \right)^2 + \left\lVert {\mathbf{a} \times \mathbf{b}} \right\rVert{}^2.$

Trig relations.

For the second question, let's express one vector in terms of length and direction, say

$\mathbf{a} = \left\lVert {\mathbf{a}} \right\rVert \mathbf{\hat{a}},$

and then express the second vector as a scaled rotation of that unit vector $ \mathbf{\hat{a}} $. That is

$\mathbf{b} = \left\lVert {\mathbf{b}} \right\rVert \mathbf{\hat{a}} e^{i\theta},$

where $ \theta $ is the angle from $ \mathbf{a} $ to $ \mathbf{b} $, and

$i = (\mathbf{a} \wedge \mathbf{b})/\left\lVert {\mathbf{a} \wedge \mathbf{b}} \right\rVert,$

is the unit pseudoscalar for the plane containing $ \mathbf{a}, \mathbf{b} $, and is oriented "from" $ \mathbf{a} $ "to" $ \mathbf{b} $ as sketched below. Also note that we did not need to use half angle sandwiched rotors, since we are rotating in the plane formed by the span of the two vectors.

bivector in plane of two vectors

You can now expand the product $ \mathbf{a} \mathbf{b} $ as

$\begin{aligned}\mathbf{a} \mathbf{b} &= \left\lVert {\mathbf{a}} \right\rVert \mathbf{\hat{a}} \left\lVert {\mathbf{b}} \right\rVert \mathbf{\hat{a}} e^{i\theta} \\ &= \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \mathbf{\hat{a}} \mathbf{\hat{a}} e^{i\theta} \\ &= \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert e^{i\theta} \\ &= \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \left( { \cos\theta + i \sin\theta } \right).\end{aligned}$

Scalar and bivector grade selections show that

$\mathbf{a} \cdot \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \cos\theta,$

and

$\mathbf{a} \wedge \mathbf{b} = \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert i \sin\theta.$

In particular, we have

$\left( { \mathbf{a} \wedge \mathbf{b}} \right)^2 = \left\lVert {\mathbf{a}} \right\rVert^2 \left\lVert { \mathbf{b} } \right\rVert^2 (-1) \sin^2\theta,$

and

$\left\lVert { \mathbf{a} \wedge \mathbf{b}} \right\rVert^2 = \left\lVert {\mathbf{a}} \right\rVert^2 \left\lVert { \mathbf{b} } \right\rVert^2 \sin^2\theta,$

or

$\left\lVert { \mathbf{a} \wedge \mathbf{b}} \right\rVert = \left\lVert {\mathbf{a}} \right\rVert \left\lVert { \mathbf{b} } \right\rVert \left\lvert { \sin\theta } \right\rvert,$

which is what you meant to write.

Peeter Joot
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