# Geometric Algebra: show $A_r \cdot B_s = (-1)^{r(s-1)} B_s \cdot A_r$ and $A_r\wedge B_s = (-1)^{rs} B_s \wedge A_r$ from Hestenes and Sobczyk's book

I'm making a go at self-study from Hestenes and Sobczyk's book Clifford Algebra to Geometric Calculus. I'm stuck on the simple formulas in the first section for reversing the order for the inner and outer product.

The book takes an axiomatic approach to GA and defines the axioms for the geometric algebra and geometric product in section 1.1 p3-4. There's a free pdf of the book available by searching the name if people want to consult it directly.

• It defines the inner for an r-blade and s-blade in terms of the geometric product in eq (1.21) $$A_r \cdot B_s = \langle A_r B_s \rangle_{\lvert r-s\rvert}$$ for $$r, s>0$$ and the inner product is by definition 0, if $$r=0$$ or $$s=0$$ (and where juxtaposition of two elements is the geometric product).
• Similarly, it defines the outer product for blades in (1.22) as $$A_r \wedge B_s = \langle A_r B_s \rangle_{\lvert r+s\rvert}$$.
• It defines the reversion at the bottom of page 5, leading, in particular, to equation (1.19) $$\langle A^{\dagger} \rangle_{r} = (-1)^{\frac{r(r-1)}{2}} \langle A\rangle_{r}$$ and (1.20a) $$\langle A B\rangle_r = (-1)^{\frac{r(r-1)}{2}} \langle B^{\dagger} A^{\dagger} \rangle_{r}$$ No issues so far, and these two equations are clear enough.

My problem is equation (1.23a) $$A_r \cdot B_s = (-1)^{r(s-1)} B_s \cdot A_r$$ for $$r\leq s$$ and (1.23b) $$A_r \wedge B_s = (-1)^{rs} B_s \wedge A_r.$$

The text says these come from using (1.19) and (1.20a), but I'm missing something obvious. Here's what I have for the first. The second is similar. For $$s\ge r$$,

\begin{align} A_r \cdot B_s & = \langle A_r B_s\rangle_{s-r} \newline &= (-1)^{\frac{(s-r)(s-r-1)}{2}} \langle B_s^{\dagger} A_r^{\dagger}\rangle_{s-r} \newline & = (-1)^{\frac{(s-r)(s-r-1)}{2}} (-1)^{\frac{(s)(s-1)}{2}} (-1)^{\frac{(r)(r-1)}{2}} \langle B_s A_r \rangle_{s-r}\newline & = (-1)^{s^2+r^2-sr-s} B_s \cdot A_r \newline & \neq (-1)^{r(s-1)} B_s \cdot A_r \end{align}

My only other thought was that for the inner product not to be zero, $$A_r$$ has to be a subspace of $$B_s$$ (and, similarly, the outer product is non-zero only if the blades are non-overlapping subspaces), but not seeing how that helps much for these formulas.

• Though I’m sure the book is amazing (I haven’t read it), I have found Alan MacDonald’s books to be exceptionally clear. They may be simpler. Commented Oct 15, 2023 at 2:40

$$(-1)^k$$ only depends on the value of $$k$$ modulo 2 (or in other words only on even/oddness). So $$s^2 + r^2 - sr - s \equiv s + r - sr - s \equiv r(1 - s) \equiv r(s - 1) \mod 2.$$