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.

Thanks in advance!

  • 1
    $\begingroup$ 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. $\endgroup$
    – NicNic8
    Commented Oct 15, 2023 at 2:40

1 Answer 1


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

On an unrelated note I would not recommend this book as learning material; not only have I found various typos that are not at all obvious, it is very terse in most dervations often to the point of being inscrutable. I've only really found it useful as a reference.

For learning GA I would instead recommend MacDonald's Linear and Geometric Algebra (as NicNic8 mentioned in the comments), Doran and Lasenby's Geometric Algebra for Physicists, or Lounesto's Clifford Algebra's and Spinors.

  • $\begingroup$ You wouldn't happen to have a list of these typos in GCGA, would you ? I remember that Lounesto had found some, but don't know the list. $\endgroup$
    – lmsteffan
    Commented Oct 15, 2023 at 15:40
  • $\begingroup$ Also, I would add Geometric Algebra for Computer Science to the list of books to be recommended for their clarity (geometricalgebra.org) $\endgroup$
    – lmsteffan
    Commented Oct 15, 2023 at 15:43
  • $\begingroup$ @lmsteffan Saddly no, it doesn't seem I've written any down anywhere. $\endgroup$ Commented Oct 15, 2023 at 18:48
  • $\begingroup$ @NicholasTodoroff Thank you. I knew it was something simple. Should have been obvious (it was obvious, I suppose, just not to me). $\endgroup$ Commented Oct 17, 2023 at 0:54
  • $\begingroup$ Also, thanks for the book recommendations. @NicholasTodoroff I've glanced at Geometric Algebra for Physicists and I like the look of it. I may veer off the Hestenes book if I'm willing to pay for it. $\endgroup$ Commented Oct 17, 2023 at 1:18

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