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Doran and Lasenby (Geometric Algebra for Physicists) introduce the reciprocal frame vector and make the below assertion about multiplication with arbitrary bivectors (page 102, eq 4.104):

$e_i e^i \cdot (a \wedge b) = e_i e^i \cdot ab - e_i e^i \cdot ba$

Here $e^i$ is the reciprocal frame vector and $e_i$ is a corresponding basis vector. Where does this expression come from? I thought that the outer product was expressed as $a \wedge b = 0.5(ab-ba)$, so shouldn't there be a factor of 1/2? Any help would be very appreciated as I'm quite new to geometric algebra and might be missing something.

chockeyblocky
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1 Answers1

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You are misunderstanding the bracketing: $$ e_i[e^i\cdot(a\wedge b)] = e_i(e^i\cdot a)b - e_i(e^i\cdot b)a. $$ This follows directly from the fact that the inner product with a vector is an antiderivation over the exterior product, i.e. for any multivectors $A, B$ and vector $v$ $$ v\cdot(A\wedge B) = (v\cdot A)\wedge B + \hat A\wedge(v\cdot B) $$ where $\hat A$ is grade involution.

Nicholas Todoroff
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  • You're missing wedges on the right side. Also, this doesn't work when $A$ or $B$ is a scalar. (The left contraction should be used instead.) – mr_e_man Sep 20 '23 at 20:03
  • The bracketing convention was a bad idea. The close spacing of $e_ie^i$ visually acts like parentheses. – mr_e_man Sep 20 '23 at 20:06
  • @mr_e_man Oops, you're right. And you are correct about needing to use the contractions for the last formula to be true for all grades, but I didn't want to complicate the answer unecessarily; I assumed it would help OP recall whatever version of that formula Doran and Lasenby have. – Nicholas Todoroff Sep 20 '23 at 21:01