# Coordinate free proof for $a\times (b\times c) = b(a\cdot c) - c(a\cdot b)$

The vector triple product is defined as $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$. This is often re-written in the following way: \begin{align*}\mathbf{a}\times (\mathbf{b}\times \mathbf{c}) = \mathbf{b}(\mathbf{a}\cdot\mathbf{c}) - \mathbf{c}(\mathbf{a}\cdot\mathbf{b})\end{align*} This is a very useful identity for integrating over vector fields and so on (usually in physics).

Every proof I have encountered splits the vectors into components first. This is understandable, because the cross product is purely a three dimensional construct. However, I'm curious as to whether or not there is a coordinate free proof of this identity. Although I don't know much differential geometry, I feel that tensors and so on may form a suitable framework for a coordinate free proof.

• By ”components“ do you mean coordinates relative to some basis or a decomposition of $a$ into components parallel to and perpendicular to the plane of $b$ and $c$? The latter can be without reference to coordinates by using orthogonal projection/rejection. The identity can be derived via these components of $a$.
– amd
Apr 11 '17 at 5:38
• I was referring to the former. The second approach sounds interesting. I feel the overall nature is similar as they both involve splitting things into components but I'll give that a go when I have a bit more time on my hands! Apr 11 '17 at 6:33
• Coordinate-free proofs like this can "easily" be computed using differential forms and particularly the Hodge map. Of course, by "easily", I mean you need a knowledge and understanding of differential forms in the first place. Apr 11 '17 at 7:05
• Could this be what you're after ?math.stackexchange.com/questions/305285/… Apr 11 '17 at 7:43

Since $b\times c$ is normal to the plane $b,\,c$ span, $a\times (b\times c)$, which is orthogonal to this vector, is in said plane. The coefficients $B,\,C$ for which the result is $Bb+Cc$ are invariant under rotations, and clearly $B$ must be linear in $a,\,c$ while $C$ is linear in $a,\,b$, so constants $B',\,C'$ exist with $a\times (b\times c) =B' (a\cdot c) b + C' (a\cdot b) c$. Since the left-hand side is antisymmetric, $C'=-B'$. Since $B'$ must be a constant (since both sides are linear in each vector), we can use any vectors we like for which both sides are non-zero to compute $B'$. Example: $a=b=i,\,c=j$ so $a\times (b\times c) = i\times k = -j$ and $(a\cdot c) b - (a\cdot b) c = -j$ as required.

• two questions: 1: why are B,C invariant under rotation? Jun 11 at 10:49
• sorry, i wanted to edit this. i start with a more findamental question: i have questions about the argument about the coefficients B and C. a priori they are scalar functions A(a,b,c) and B(a,b,c). why are the individually linear in two variables and independent of the third? a priori only the entire expression needs to be linear, not the individual summands. and even if we get that the summand are individually linear, why the independence of the third variable? at first glance i see linear only dictate homogeneity of degree 0. Jun 11 at 10:56
• @peter The reason $a\times(b\times c)=Bb+Cc$ requires $B,\,C$ to be invariant scalars is because any equation of the form $v=Bb+Cc$ with $v$ a vector requires coefficients independent of your coordinate system viz.$$\left(\begin{array}{c} v\cdot b\\ v\cdot c \end{array}\right)=\left(\begin{array}{cc} b^{2} & b\cdot c\\ b\cdot c & c^{2} \end{array}\right)\left(\begin{array}{c} B\\ C \end{array}\right).$$By Cauchy-Schwarz, the matrix is invertible if $b,\,c$ aren't parallel or antiparallel.
– J.G.
Jun 11 at 12:08
• you seem to be saying that B,C are determined by fixed v. that is certainly so. but now we have v(a,b,c)=B(a,b,c)b+C(a,b,c)c. why do the dependencies simplify/disappear? Jun 11 at 12:17
• @peter I think you're conflating "independent of the vectors themselves" (which isn't how the coefficients I derived work) with "independent of the coordinate system used" (which is true for the aforementioned reason).
– J.G.
Jun 11 at 12:59

Okay, I really hate the sign issues with the Hodge star, so I am gonna assume that $\star\star=1$, the end result will be good.

The fundamental relationship is that for $k$-vectors/forms, we have $\langle\omega,\eta\rangle\mu=\omega\wedge\star\eta$, where the angle brackets are the inner product on the exterior algebra and $\mu$ is the volume form/multivector.

Let's start with $x$ being an arbitrary vector and taking a look at $$\langle x,a\times(b\times c)\rangle\mu=\langle x,\star(a\wedge\star(b\wedge c))\rangle\mu= \\=x\wedge(a\wedge\star(b\wedge c))=(x\wedge a)\wedge\star(b\wedge c)= \\=\langle x\wedge a,b\wedge c\rangle\mu=\det\left(\begin{matrix}\langle x,b\rangle & \langle x,c\rangle \\ \langle a,b\rangle & \langle a,c\rangle\end{matrix}\right)\mu= \\=(\langle x,b\rangle \langle a,c\rangle-\langle x,c\rangle \langle a,b\rangle)\mu,$$ comparing the LHS with the RHS we can "divide" (ofc not really divide) by $\mu$ since the coefficients need to agree, and because $x$ was arbitrary, and the inner product is nondegenerate, we can "" divide "" by $x$ and we have $$a\times(b\times c)=b\langle a,c\rangle-c\langle a,b\rangle.$$

• I wonder why the inner product of wedge products is a determinant? Apr 11 '17 at 7:32
• Also, may cancel be a more appropriate substitute for "divide"? Apr 11 '17 at 7:33
• @FrenzyLi Maybe "cancel" would be ok. In the first case, it was because 3-forms in a 3-dim space form a one dimensional space, so equality of 3-forms $\Longleftrightarrow$ equality of the single coefficient, in the second case, I have essentially viewed the inner product as the action of a 1-form (due to Riesz isomorphism and all), and 1-forms agree if their actions on an arbitrary vector agree. As for the determinant, the inner product on the exterior algebra is defined as, if $\alpha=\alpha_1\wedge...\wedge\alpha_k$ and $\beta=\beta_1\wedge... \wedge\beta_k$ then... Apr 11 '17 at 7:38
• @FrenzyLi $\langle \alpha,\beta\rangle=\det(\langle \alpha_i,\beta_j\rangle)$. See for example en.wikipedia.org/wiki/…. Apr 11 '17 at 7:39
• @FrenzyLi Generally $\star\star=\pm 1$. It depends on the dimension of the space, the signature of the metric/inner product and also on the degree of the form/multivector you apply it on. A hassle to deal with it, fortunately here, even if I did make a sign error, the sign errors cumulatively cancelled out. Apr 11 '17 at 7:42

Let $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ be vector fields on $\mathbb{R}^{3}$ (we could extend to $\mathbb{R}^{n}$ if we wish!), considered as a Riemannian manifold equipped with metric $g$ and induced Hodge map $\star$. Let $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ have corresponding vector field representations $U,V,W$ on $\mathbb{R}^{n}$ respectively. Then

\begin{align} \mathbf{a}\times(\mathbf{b}\times \mathbf{c}) \,\equiv\, \star(\widetilde{U} \wedge \star(\widetilde{V} \wedge \widetilde{W})) \end{align}

where $\widetilde{X}$ denotes the metric dual of $X$ (i.e. $\widetilde{X}=g(X,-)$) and the equivalence is up to metric dual. Then

\begin{align} \star(\widetilde{U} \wedge \star(\widetilde{V} \wedge \widetilde{W})) &\,=\, \star(\widetilde{U} \wedge i_{W}\star \widetilde{V}) \,=\, \star ( i_{W}\widetilde{U} \wedge \star \widetilde{V} - i_{W} (\widetilde{U}\wedge \star \widetilde{V})) \\ &\,=\, (i_{W}\widetilde{U})\star\star \widetilde{V} - \star i_{W}(\widetilde{U}\wedge\star \widetilde{V}) \\ &\,=\, g(U,W)\widetilde{V} - \star(\widetilde{U}\wedge\star \widetilde{V}) \wedge \widetilde{W} \\ &\,=\, g(U,W)\widetilde{V} - g(U,V)\widetilde{W} \end{align}

where $i_{X}$ denotes the interior derivative with respect to $X$ and we have used the identities:

\begin{align} \star\star \alpha &\,=\, \alpha \\[0.2cm] \star(\widetilde{X}\wedge\star\widetilde{Y}) &\,=\, g(X,Y) \\[0.2cm] \star(\alpha \wedge \widetilde{X}) &\,=\, i_{X}\star \alpha \\[0.2cm] \widetilde{X} \wedge \star\alpha &\,=\, (-1)^{p+1}\star i_{W}\alpha \end{align}

for any $p$-form $\alpha$ and vector fields $X,Y$ (note that the first and second identities are specific to $\mathbb{R}^{3}$). Then note that

$$g(U,W)\equiv \mathbf{a}\cdot\mathbf{c}\quad\text{and}\quad g(U,V)\equiv\mathbf{a}\cdot\mathbf{b}$$

and so then the result follows after taking the metric dual of our expression.

• For now I have no idea what this means, but it's absolutely beautiful. Apr 11 '17 at 22:20
• If you are interested, I found the following application-oriented introduction to differential forms extremely useful and it's where the notation can be found: A primer on exterior differential calculus by David A. Burton Apr 12 '17 at 12:28
• it's been over a year, but i've spent the past half year primarily learning differential geometry, and I just came back to this and realised I understood it all :) Jun 19 '18 at 11:26

I just want to add a proof using my favourite method for this kind of thing: Penrose graphical notation. I think it is as coordinate free as you can go.

Every piece of the diagram has a meaning. Tensors are shapes with a lines going upwards or downwards, depending on the type of tensor. For example, vectors have 1 line going upwards, and covectors one line going downwards. Contracion is represented joining the lines.

For example: the first line on the right side, over the word "proof", is the statement $(b\times c)^{a} = b^{b}c^{c}g_{bd}g_{ce}\epsilon^{dea} = b^{b}c^{c}\epsilon_{bcd}g^{da}$ where the indices are proper abstract indices: they do not represent components, but the slots of the tensors.

At the bottom I repeat the properties I use. 1. Antisymmetry of $\epsilon^{abc}$
2. $\epsilon^{abc}\epsilon_{def} = \delta^{abc}_{def}$
3. $\delta^{abc}_{dec} = \delta^{ab}_{de}$
4. $\delta^{ab}_{cd}A_{ab} = 2A_{[cd]} = A_{cd} - A_{dc}$

Adapted from my previous proof of $\nabla \times (\vec{A} \times \vec{B})$: \begin{align} \vec a \times (\vec b \times \vec c) & = a_l \hat{e}_l \times (b_i c_j \hat{e}_k \epsilon_{ijk}) \\ & = a_l b_i c_j \epsilon_{ijk} \underbrace{ (\hat{e}_l \times \hat{e}_k)}_{(\hat{e}_l \times \hat{e}_k) = \hat{e}_m \epsilon_{lkm} } \\ & = a_l b_i c_j \hat{e}_m \underbrace{\epsilon_{ijk} \epsilon_{mlk}}_{\text{contracted epsilon identity}} \\ & = a_l b_i c_j \hat{e}_m \underbrace{(\delta_{im} \delta_{jl} - \delta_{il} \delta_{jm})}_{\text{They sift other subscripts}} \\ & = a_j (b_i c_j \hat{e}_i) - a_i (b_i c_j \hat{e}_j) \\ & = (b_i \hat{e}_i) (a_j c_j) - (c_j \hat{e}_j) (a_i b_i) \\ & = \vec b (\vec a\cdot\vec c) - \vec c(\vec a\cdot \vec b) \end{align}

• It uses tensors but in the index notation. Not sure if this is what the OP wants. Apr 11 '17 at 4:49
• Yes I've seen a very similar proof before and it's not exactly what I wanted. However, due to what the cross product is, I wouldn't be surprised if all proofs involved index notation or explicit components. (Unless there are "suitable" generalisations!) Apr 11 '17 at 5:11
• @Shanye2020 I would actually like to see this proof using more compact notation. I do hope someone could enlighten us. Apr 11 '17 at 5:13
• Hmm... An almost identical proof is also seen here. Apr 11 '17 at 5:28
• @FrenzyLi Ok, I'm done, posted an answer. Apr 11 '17 at 7:28

We can do this by using the definition of the cross product as an bilinear map $$V \times V \to V$$. We also take as a basic property that it is orthogonal to each of its arguments, $$a \cdot (a \times b) = b \cdot (a \times b) = 0 \tag{Orth}$$ We need another property to fix the scaling; the easiest one is that the magnitude of $$a \times b$$ is the area of the parallelogram spanned by $$a$$ and $$b$$. This effectively means that $$(a \times b) \cdot (a \times b) = (a \cdot a)(b \cdot b) - (a \cdot b)^2 \tag{Area}$$ (If this looks opaque, you might prefer to start with perpendicular $$a$$ and $$b$$, then extending with bilinearity.)

(A) and $$v \cdot v = 0 \implies v = 0$$ force that $$a \times a = 0, \tag{Alt}$$ (alterating) and expanding $$(a+b) \times (a+b)$$ and using this implies that $$a \times b = - b \times a : \tag{AntiSym}$$ the cross product must be antisymmetric.

The scalar triple product is $$[a,b,c] = a \cdot (b \times c)$$. It is trilinear since both products are bilinear, and it is also zero when two arguments are equal, by (Orth) and (Alt): $$[a,a,b] = [a,b,a] = [b,a,a] = 0$$ Using linearity on $$[a+b,a+b,c]$$ implies that further $$[a,b,c] = -[b,a,c]$$, and combining this with the inherited antisymmetry in the last two arguments from $$\times$$, we find that $$[a,b,c] = [b,c,a] = [c,a,b] = -[b,a,c] = -[a,c,b] = -[c,b,a] ,$$ and in particular, $$a \cdot (b \times c) = (a \times b) \cdot c :$$ we can swap the position of the dot and the cross.

We can now derive a restricted form of the triple product identity: using linearity on (Area), we have $$0 = ((a + b) \times c) \cdot ((a+b) \times c) + ((a+b) \cdot c)^2 - (a+b) \cdot (a+b))(c \cdot c) \\ = \dotsb = 0 + 0 + (a \times c) \cdot (b \times c) + (a \cdot c)(b \cdot c) - (a \cdot b)(c \cdot c) .$$ Switching the dot and the cross in the first term, we find that $$0 = a \cdot ( c \times (b \times c) + (b \cdot c)c - (c \cdot c)b )$$ But $$a$$ is arbitrary here, so $$c \times (c \times b) = (b \cdot c)c - (c \cdot c)b . \tag{R}$$

We can derive a couple more identities that are true in general, but now we can cheat and use that we are in three dimensions, so we can expand $$a = \lambda b + \mu c + \nu (b \times c)$$. Then using linearity, (R) and linearity again, $$a \times (b \times c) = \lambda (b \times (b \times c)) + \mu ( c \times b \times c ) \\ = \lambda( (c \cdot b)b - (b \cdot b)c ) + \mu ( (c \cdot c)b - (b \cdot c)c ) \\ = ((\lambda b + \mu c) \cdot c)b - ((\lambda b + \mu c) \cdot c)b$$ Finally, we can add the $$\nu (b \times c)$$ part back into both dot products, since it is orthogonal to $$b$$ and $$c$$, which gives $$a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$$ as expected.