The formula is $(A_r\land b)\cdot C_s=A_r\cdot (b\cdot C_s)$, where $0<r<s$. Hestenes suggests to expand $(A_rb)C_s=A_r(bC_s)$ and extract the $s-r-1$-vector part. But this method requires one to know the formula for the Clifford product of two arbitrary blades, and as far as I can see, the book only gives formulas for the Clifford product of a vector and a blade. One can prove this formula using formulas for the exterior and interior product in Grassmann algebra. But this is not the method Hestenes suggests. If somebody knows the proof that Hestenes means, please let me know too.

  • $\begingroup$ @mr_e_man What is the symbol that looks like an upside down r? $\endgroup$ – alexanderyaacov Dec 11 '18 at 10:08
  • $\begingroup$ That $\lrcorner$ "\lrcorner" is the contraction product, an asymmetric dot product. en.wikipedia.org/wiki/… , av8n.com/physics/clifford-intro.htm#sec-contractions $\endgroup$ – mr_e_man Dec 15 '18 at 0:19
  • $\begingroup$ I remember this identity by comparing it with the analogous one in logic: $(A\land B)\to C\equiv A\to(B\to C)$, where $A,B,C$ are statements which are either true or false. In words, "$A$ and $B$ together imply $C$" is equivalent to "If $A$ is true, then $B$ implies $C$". $\endgroup$ – mr_e_man Dec 15 '18 at 0:25
  • $\begingroup$ @mr_e_man why is it then a special case? It seems to me to be exactly the same. $\endgroup$ – alexanderyaacov Dec 16 '18 at 7:24
  • $\begingroup$ It's a special case because the grades are restricted. The identity I gave is always true; $B$ doesn't need to be a vector, nor $A$ lower-grade than $C$. $\endgroup$ – mr_e_man Dec 16 '18 at 23:43

Here's a proof from the appendix of Geometric Algebra for Electrical Engineers, with the variable names changed to match your question. The order of the variables is reverse of your question, but you can apply the reverse operator to each side and make a change of variables to prove the formula as stated in NFCM.

Theorem: Distribution of inner products Given two blades $C_s, A_r$ with grades subject to $s > r > 0$, and a vector $\mathbf{b}$, the inner product distributes according to $$C_s \cdot \left( { \mathbf{b} \wedge A_r } \right) = \left( { C_s \cdot \mathbf{b} } \right) \cdot A_r.$$


The proof is straightforward, relying primarily on grade selection, but also mechanical. Start by expanding the wedge and dot products within a grade selection operator $$C_s \cdot \left( { \mathbf{b} \wedge A_r } \right)={\left\langle{{C_s (\mathbf{b} \wedge A_r)}}\right\rangle}_{{s - (r + 1)}}=\frac{1}{{2}} {\left\langle{{C_s \left( {\mathbf{b} A_r + (-1)^{r} A_r \mathbf{b}} \right) }}\right\rangle}_{{s - (r + 1)}}.$$

Solving for $C_r \mathbf{b}$ in $$2 \mathbf{b} \cdot A_r = \mathbf{b} A_r - (-1)^{r} A_r \mathbf{b},$$ we have $$\begin{aligned}C_s \cdot \left( \mathbf{b} \wedge A_r \right) &= \frac{1}{2} {\left\langle C_s \mathbf{b} A_r + C_s \left( \mathbf{b} A_r - 2 \mathbf{b} \cdot A_r \right) \right\rangle}_{s - (r + 1)} \\ &= {\left\langle C_s \mathbf{b} A_r \right\rangle}_{s - (r + 1)} - {\left\langle C_s \left( \mathbf{b} \cdot A_r \right) \right\rangle}_{s - (r + 1)} \\ &= {\left\langle C_s \mathbf{b} A_r \right\rangle}_{s - (r + 1)}.\end{aligned}$$

The last term in the second step is zero since we are selecting the $s - r - 1$ grade element of a multivector with grades $s - r + 1$ and $s + r - 1$, which has no terms for $r > 0$. Now we can expand the $C_s \mathbf{b}$ multivector product, for $$C_s \cdot \left( { \mathbf{b} \wedge A_r } \right)={\left\langle{{ \left( { C_s \cdot \mathbf{b} + C_s \wedge \mathbf{b}} \right) A_r }}\right\rangle}_{{s - (r + 1)}}.$$

The latter multivector (with the wedge product factor) above has grades $s + 1 - r$ and $s + 1 + r$, so this selection operator finds nothing. This leaves $$C_s \cdot \left( { \mathbf{b} \wedge A_r } \right)={\left\langle{{\left( { C_s \cdot \mathbf{b} } \right) \cdot A_r+ \left( { C_s \cdot \mathbf{b} } \right) \wedge A_r}}\right\rangle}_{{s - (r + 1)}}.$$

The first dot products term has grade $s - 1 - r$ and is selected, whereas the wedge term has grade $s - 1 + r \ne s - r - 1$ (for $r > 0$), which completes the proof.


This is a special case of the identity $(A\wedge B)\,\lrcorner\,C=A\,\lrcorner\,(B\,\lrcorner\,C)$. It can be proven first for blades, then extended to all multivectors by linearity.

Suppose $A$ has grade $r$, $B$ has grade $k$ (in your case $k=1$), and $C$ has grade $s$. Then

$$AB=\langle AB\rangle_{r+k}$$

$$+\langle AB\rangle_{r+k-2}$$

$$+\langle AB\rangle_{r+k-4}$$


$$+\langle AB\rangle_{-r+k+2}$$

$$+\langle AB\rangle_{-r+k}$$

The first term is $A\wedge B$, and the last term is $A\,\lrcorner\,B$. We could also write this as $$\langle AB\rangle_{r+k}+\cdots+\langle AB\rangle_{|r-k|}=A\wedge B+\cdots+A\cdot B$$ because the last term is the same if $r\leq k$, and otherwise we're only adding $0$'s after $A\cdot B$. (For example, with $r=3$ and $k=1$, the expression is $\langle Ab\rangle_4+\langle Ab\rangle_2+\langle Ab\rangle_0+\langle Ab\rangle_{-2}=\langle Ab\rangle_4+\langle Ab\rangle_2$.)

Multiply by $C$, expanding each term in the same way, to get

$$ABC=\langle\langle AB\rangle_{r+k}C\rangle_{r+k+s}+\langle\langle AB\rangle_{r+k}C\rangle_{r+k+s-2}+\cdots+\langle\langle AB\rangle_{r+k}C\rangle_{-(r+k)+s}$$

$$+\langle\langle AB\rangle_{r+k-2}C\rangle_{r+k-2+s}+\langle\langle AB\rangle_{r+k-2}C\rangle_{r+k-2+s-2}+\cdots+\langle\langle AB\rangle_{r+k-2}C\rangle_{-(r+k-2)+s}$$


$$+\langle\langle AB\rangle_{-r+k}C\rangle_{-r+k+s}+\langle\langle AB\rangle_{-r+k}C\rangle_{-r+k+s-2}+\cdots+\langle\langle AB\rangle_{-r+k}C\rangle_{-(-r+k)+s}$$

Each row has successively fewer terms ($2$ less each time), so the rightmost term should be indented, and the expression above would look something like this trapezoid:

$$\begin{matrix} x&x&x&x&x&x&x&x&x \\ x&x&x&x&x&x&x \\ x&x&x&x&x \\ x&x&x \end{matrix}$$

The grade decreases going right or going down, so all the terms with a given grade are on a diagonal line. The leftmost diagonal (the highest grade) has only a single term, $\langle ABC\rangle_{r+k+s}=\langle\langle AB\rangle_{r+k}C\rangle_{r+k+s}=(A\wedge B)\wedge C$. The rightmost diagonal (the lowest grade) also has only a single term, $\langle ABC\rangle_{-(r+k)+s}=\langle\langle AB\rangle_{r+k}C\rangle_{-(r+k)+s}=(A\wedge B)\,\lrcorner\,C$.

Now write it the other way, expanding $BC$ first:

$$BC=\langle BC\rangle_{k+s}+\langle BC\rangle_{k+s-2}+\cdots+\langle BC\rangle_{-k+s}$$

and multiplying by $A$:

$$ABC=\langle A\langle BC\rangle_{k+s}\rangle_{r+k+s}+\langle A\langle BC\rangle_{k+s-2}\rangle_{r+k+s-2}+\cdots+\langle A\langle BC\rangle_{-k+s}\rangle_{r-k+s}$$

$$+\langle A\langle BC\rangle_{k+s}\rangle_{r-2+k+s}+\langle A\langle BC\rangle_{k+s-2}\rangle_{r-2+k+s-2}+\cdots+\langle A\langle BC\rangle_{-k+s}\rangle_{r-2-k+s}$$


$$+\langle A\langle BC\rangle_{k+s}\rangle_{-r+k+s}+\langle A\langle BC\rangle_{k+s-2}\rangle_{-r+k+s-2}+\cdots+\langle A\langle BC\rangle_{-k+s}\rangle_{-r-k+s}$$

This time, each row has the same number of terms, so it's just a rectangle instead of a trapezoid. Again, all the terms with a given grade are on a diagonal, and the leftmost diagonal has only $\langle ABC\rangle_{r+k+s}=\langle A\langle BC\rangle_{k+s}\rangle_{r+k+s}=A\wedge(B\wedge C)$. The rightmost diagonal also has only one term, $\langle ABC\rangle_{-r-k+s}=\langle A\langle BC\rangle_{-k+s}\rangle_{-r-k+s}=A\,\lrcorner\,(B\,\lrcorner\,C)$.

Comparing the two expansions, we can see both identities at once:

$$(A\wedge B)\wedge C=A\wedge(B\wedge C),\qquad(A\wedge B)\,\lrcorner\,C=A\,\lrcorner\,(B\,\lrcorner\, C).$$


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