# Proof of bivector multiplication with reciprocal frame vector from Doran, Lasenby?

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.

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.
• 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.) Commented Sep 20, 2023 at 20:03
• The bracketing convention was a bad idea. The close spacing of $e_ie^i$ visually acts like parentheses. Commented Sep 20, 2023 at 20:06