In the book New foundations for classical mechanics by David Hestenes, the following equation is supposed to be the defining equation multiplicative inverse for a multivector:
$$A^{-1} A = 1$$
Is the $1$ on the RHS a scalar?
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It would no doubt be helpful if you can define what object $A$ is. – Jacob A Sep 03 '21 at 04:48
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Oops I had it initially but I deleted it out when editing @JacobA Added back now – Clemens Bartholdy Sep 03 '21 at 04:52
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Of course it's the identity element in the geometric algebra. So yes, a scalar. – Hans Lundmark Sep 03 '21 at 05:43
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1There is only one identity in geometric algebra, and that's a scalar. – md2perpe Sep 03 '21 at 08:10
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Thanks, that helps. I think I'll accept if that is posted as answer with an example :D – Clemens Bartholdy Sep 03 '21 at 08:11
2 Answers
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Yes, $1$ is meant to be the unit scalar here. For example, assume that you are considering the multivectors generated by a 2D Euclidean vector space with an orthonormal basis $ \left\{ { \mathbf{e}_1, \mathbf{e}_2 } \right\} $: $$ \mathbf{e}_1 \mathbf{e}_1 = 1,$$ $$ \left( { \mathbf{e}_1 \mathbf{e}_2 } \right) \left( { -\mathbf{e}_1 \mathbf{e}_2 } \right) = 1,$$ $$ \left( { \mathbf{e}_1 + \mathbf{e}_2 } \right) \frac{ \mathbf{e}_1 + \mathbf{e}_2 }{2} = 1,$$ $$ \left( { 1 + \mathbf{e}_1 \mathbf{e}_2 } \right) \frac{ 1 - \mathbf{e}_1 \mathbf{e}_2 }{2} = 1.$$ All these inverses follow directly from $ \mathbf{e}_1^2 = \mathbf{e}_2^2 = 1 $, and $ \mathbf{e}_1 \mathbf{e}_2 = -\mathbf{e}_2 \mathbf{e}_1 $.
Peeter Joot
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The $1$ on the right is supposed to be the identity (operator or matrix, depending on what sort of thing $A$ is).
Robert Israel
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