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

Question

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.

Answer 1

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.

Answer 2

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}

Answer 3

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 $$ a\times(b\times c)=b\langle a,c\rangle-c\langle a,b\rangle. $$

Answer 4

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.