I would like to prove the formula $ a\cdot A_r = \frac{1}{2}(aA_r - (-1)^r A_r a)$ for a grade-1 vector $a$ and a simple r-blade $A_r$. How?

In Geometric Algebra for Physicists (Doran & Lasenby), they don't prove the formula. They treat it as a definition, and then find various other results as theorems.

In Clifford Algebra to Geometric Calculus (Hestenes & Sobczyk), they purport to prove the formula but my impression is that they use circular reasoning (they state the desired result (1.27a), then state (1.27a) implies (1.30), and (1.30) implies (1.34). But the only thing we know about (1.34) is that it has the correct $(r-1)$ grade).

(1.27a): $$a\cdot A_r = \langle aA_r\rangle_{r-1} = \frac{1}{2}(aA_r - (-1)^rA_ra)$$

(1.29): Decompose A to even and odd grades: $$A=A_+ + A_-$$

(1.30a, b, c, d): $$ a\cdot A_+ = \textstyle{\frac{1}{2}}(aA_+ - A_+a) \\ a\wedge A_+ = \textstyle{\frac{1}{2}}(aA_+ + A_+a) \\ a\cdot A_- = \textstyle{\frac{1}{2}}(aA_- + A_-a) \\ a\wedge A_- = \textstyle{\frac{1}{2}}(aA_- - A_-a) $$

(1.34): $$ a\cdot(a_1a_2\dots a_r) = \textstyle{\frac{1}{2}}(aa_1\dots a_r - (-1)^ra_1\dots a_r a) $$

They show that

$$ a(a_1\dots a_r) - (-1)^r(a_1\dots a_r) a = 2\sum_{k=1}^r (-1)^{k+1} a\cdot a_k a_1\dots \check{a_k}\dots a_r $$

but they don't show that $$ a\cdot (a_1\dots a_r) = \frac{1}{2}\left(a(a_1\dots a_r) - (-1)^r(a_1\dots a_r) a\right) $$ They only proved that $\frac{1}{2}\left(a(a_1\dots a_r) - (-1)^r(a_1\dots a_r) a\right) $ has the correct grade of (r-1).

They need to prove the "=" in $a\cdot A_r \equiv \langle aA_r \rangle_{r-1} = \sum_{k=1}^r (-1)^{k+1} a\cdot a_k a_1\dots \check{a_k}\dots a_r$ but this is missing.


1 Answer 1


The reasoning is not presented clearly in the book at all, but here's how you can prove it.

They've derived the equation $$ aa_1\dots a_r = 2 \sum_{k=1}^r (-1)^{k+1}a\cdot a_k a_1\dots \check{a}_k \dots a_r + (-1)^r a_1 \dots a_r a \quad (*)$$ for arbitrary vectors $a,a_1,\dots, a_r$.

Consider an $r$-blade (or ``simple $r$-vector") $E_r = e_1\dots e_r$. By definition, $a\cdot E_r = \langle aE_r\rangle_{r-1}$, so applying $(*)$ to $a_i = e_i$ and taking the $(r-1)$-grade component of both sides gives $$ \langle a E_r \rangle_{r-1} = 2 \sum_{k=1}^r (-1)^{k+1} a\cdot e_k \langle e_1\dots \check{e}_k \dots e_r\rangle_{r-1} + (-1)^r \langle E_r a\rangle_{r-1}$$ using distributivity of the grade operator. Now we use the fact that $e_1\dots \check{e}_k \dots e_r$ has grade $r-1$, which follows from the assumption that $e_1 \dots e_r$ is a blade. We also use (1.20d) from the book to get $\langle E_r a\rangle_{r-1} = (-1)^{(r-1)^2}\langle a E_r\rangle_{r-1}$. Thus, $$ \langle a E_r\rangle_{r-1} = 2 \sum_{k=1}^r (-1)^{k+1} a\cdot e_k e_1\dots \check{e}_k \dots e_r + (-1)^r (-1)^{(r-1)^2}\langle a E_r\rangle_{r-1}.$$ Noting that $(-1)^{r}(-1)^{(r-1)^2} = (-1)^{r(r-1) + 1} = -1$, we rearrange this and apply $(*)$ again to obtain $$2\langle a E_r\rangle_{r-1} = a e_1 \dots e_r - (-1)^r e_1\dots e_r a,$$ i.e., $a\cdot E_r = \frac{1}{2}(aE_r - (-1)^r E_ra)$, which is (1.27a) for $r$-blades. This can be extended to arbitrary $r$-vectors using the fact that any $r$-vector is a sum of $r$-blades, in conjunction with the distributivity of the dot product (1.24a).


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .