# Why is it clear that $A\times(B\times C) = (B\wedge C)(L(A),\cdot)?$

As motivation, this exercise claims to be a way to gain intuition for the $$BAC-CAB$$ identity, but I'm gaining a sum total of zero intuition by doing it.

First and foremost, I have to lay out notation. In the title of this post, $$A,B,C\in\mathbb{R}^3$$ and $$L:\mathbb{R}^3\rightarrow\mathbb{R}^{3*}$$. Furthermore, I'm using the dual metric for all of this: $$L(A)(\cdot) = (A|\cdot)$$. The empty slot in $$(L(A),\cdot)$$ is to emphasize that this is partial application of the $$(0,2)$$ tensor $$B\wedge C$$, defined by $$(B\otimes C)(L(A),\cdot) = B(C(L(A))) = B(L(A)(C)) = B(A|C)$$.

Now, having laid all of this out, and using the Euclidean metric $$(A|B) = A\cdot B$$, how am I supposed to see why $$A\times(B\times C) = (B\wedge C)(L(A),\cdot)$$? Carrying out the RHS, it's direct to show that

$$(B\wedge C)(L(A),\cdot) = (B\otimes C - C\otimes B)(L(A),\cdot) = B(A\cdot C) - C(A\cdot B)$$

thus yielding the familiar $$BAC-CAB$$ identity, but I fail to see why I would be able to make this association with that triple-cross in the first place. The route that would be clear to me would be the following. By considering an $$\alpha\in\Lambda^2\mathbb{R}^3$$ and defining $$J:\Lambda^2\mathbb{R}^3\rightarrow\mathbb{R}^3$$ to have it's i'th component as $$(J(\alpha))^i = \frac{1}{2}\epsilon^i_{jk}\alpha^{jk}$$, it follows that $$A\times B = J(A\wedge B)$$, so for this problem I would be more inclined to do

$$A\times(B\times C) = J(A\wedge J(B\wedge C))$$

and if, from there, I could somehow see that this is equivalent to $$(B\wedge C)(L(A),\cdot)$$, then I would gain the "intuition" that this exercise claims to be the motivation. Just showing both sides to be $$BAC-CAB$$ leaves me empty.

• There are different conventions on the meaning of $B\wedge C$: either it's $(B\otimes C-C\otimes B)$; or it's $(B\otimes C-C\otimes B)/2$; or it's not a tensor at all, but a different type of object. Jan 1, 2020 at 21:54

Write $$B\times C$$ as $$D$$, then $$D$$ is a blade. Let $$I$$ be the unit pseudoscalar for right-hand rule. Then $$J(B\times C)=-ID$$. The equation is now written as $$A\times(-ID)=D(L(A),\cdot)$$.

I think the interpretation here is that both side compute the same quantity. This quantity is a form of dot product between vector $$A$$ and blade $$D$$, which is the result of canceling out the component of $$D$$ the direction along $$A$$ leaving only the scalar, while retaining the direction orthogonal to $$A$$. But they do it in different manner: the left hand side do it by representing $$D$$ with a vector using the right hand rule representation, the right hand side do it by projecting $$A$$ down onto $$D$$ and canceling.

So now to prove it. The claim is clearly true if $$D=0$$ so assume not, so $$D$$ represent an unique plane. By linearity, it's sufficient to check the claim for: $$A$$ orthogonal to $$D$$ plane, and $$A$$ parallel to $$D$$ plane. If $$A$$ is orthogonal to $$D$$ then both side are trivially $$0$$. So we assume $$A$$ is parallel to $$D$$ and not orthogonal, or in other word $$A$$ lie on $$D$$ plane and is nonzero.

Let $$u=\frac{A}{|A|}$$ then we can write $$D=uv=u\wedge v$$ for some $$v$$ orthogonal to $$u$$ and is on the plane represented by $$D$$; specifically $$v=uD$$.

Then $$-ID=u\times v$$. So $$A\times(-ID)=A\times(u\times v)=|A|u\times(u\times v)=-|A|v$$ because left multiplication by unit vector $$u$$ induce a right angle rotation on the plane orthogonal to it according to right hand rule. And $$(u\wedge v)(L(A),\cdot)=|A|(u\otimes v-v\otimes u)(u,\cdot)=|A|u(u|v)-|A|v(u|u)=-|A|v$$

One key property we are using here is how flexible it is to represent a blade using vectors on it, which is why since the beginning we are treating this as an equation between a vector $$A$$ and a blade $$D$$, ignoring $$B$$ and $$C$$ completely. Once you take wedge product, the only thing that matter is what blade does $$B$$ and $$C$$ form and not they are in particular (and same for cross product which is the result of wedge product times unit pseudoscalar).

I don't know exactly what you're looking for, but here are my thoughts anyway. I'll use multiplication (as defined here) instead of function notation which I find confusing and unnecessary.

With the unit trivector $$I=e_1e_2e_3$$, the cross product can be defined as

$$A\times B=-(A\wedge B)I=-(A\wedge B)\,\lrcorner\,I=-A\,\lrcorner\,(B\,\lrcorner\,I)=-A\,\lrcorner\,(BI).$$

(I have here a proof of the middle equation.) Thus the triple product is

$$A\times(B\times C)=-A\,\lrcorner\,\Big(\big(-(B\wedge C)I\big)I\Big)=A\,\lrcorner\,\big((B\wedge C)I^2\big)=-A\,\lrcorner\,(B\wedge C).$$

Contraction of a vector with a bivector is anticommutative, so this is $$(B\wedge C)\,\llcorner\, A$$.

To see that this is $$(A\cdot C)B-(A\cdot B)C$$, I would just verify that it's true for the basis vectors $$e_i$$ (which is easier because they're orthogonal, and the products reduce to the geometric product or $$0$$), and then it's true for all vectors because the expressions are multilinear. Indeed, I used this method for most identities when I was first learning geometric algebra. But it may be unsatisfying.

Alternatively, we can break $$A$$ into parts parallel and perpendicular to the $$B\wedge C$$ plane, as in @calcstudent's answer; or write the products as symmetric and antisymmetric parts of the geometric product:

$$B\wedge C=\frac12(BC-CB)$$

$$(B\wedge C)\,\llcorner\,A=\frac12\big((B\wedge C)A-A(B\wedge C)\big)$$

$$=\frac14(BCA-CBA-ABC+ACB)$$

$$=\frac14\big(B(AC+CA)+(AC+CA)B-C(AB+BA)-(AB+BA)C\big)$$

$$=\frac12\big(B(A\cdot C)+(A\cdot C)B-C(A\cdot B)-(A\cdot B)C\big)$$

$$=(A\cdot C)B-(A\cdot B)C.$$