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I am looking for a proof that $$X\times (Y\times Z)+Y\times (Z\times X)+Z\times (X\times Y)=0 \qquad \textrm{for all } X, Y, Z\in \mathbb{R^3} .$$

I know that as the left-hand side is a summation of all the even permutations, so it should be zero. However, I am looking for some clearer and understandable proof. Any comments?

Travis Willse
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Majid
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    What does $X\times Y$ mean for $X$, $Y\in\Bbb R^n$? – Angina Seng Aug 09 '19 at 02:33
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    Binary cross products only exist on $\Bbb R^n$ for $n=0,1,3,7$. The equation is true for $n=0,1$ (vacuously) and $n=3$ (Jacobi identity) but not $n=7$ (since octonions are nonassociative). – anon Aug 09 '19 at 02:37
  • In $\mathbb{R}^n$, you may write $(X\times Y)i=\epsilon{ijk}X_jY_k$. $\epsilon_{ijk}$ is antisymmetric with respect to swap of any two indices. – mike Aug 09 '19 at 02:38
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    @mike The Levi-Civita symbol $\epsilon_{ijk}$ is defined for $i,j,k\in{1,2,3}$, no? – anon Aug 09 '19 at 02:39
  • Thanks for the comments, I just edited the question. – Majid Aug 09 '19 at 03:12
  • This identity is sometimes called the Jacobi identity, owing to the fact that we can identify $(\Bbb R^3, \times)$ with the Lie algebra $\mathfrak{so}(3, \Bbb R)$ with its usual Lie bracket (or more concretely, the Lie algebra of antisymmetric $3 \times 3$ matrices endowed with the matrix commutator). – Travis Willse Aug 09 '19 at 03:39
  • https://math.stackexchange.com/questions/924121/geometric-proof-for-triple-vector-product-jacobi-identity – cqfd Aug 09 '19 at 04:11

2 Answers2

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I suspect this question is a duplicate, but I cannot find another instance of it. In any case:

Hint Rewrite the left-hand side of the identity using the vector triple product identity that expresses the iterated cross product in terms of the dot product: $${\bf x} \times ({\bf y} \times {\bf z}) = ({\bf x} \cdot {\bf z}) {\bf y} - ({\bf x} \cdot {\bf y}) {\bf z} .$$

Travis Willse
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From the fact $X\times Y$ is alternating (i.e. $Y\times X=-X\times Y$), deduce

$$ (X,Y,Z):=X\times(Y\times Z)+Y\times(Z\times X)+Z\times(X\times Y) $$

is alternating. Moreover, $\times$ is rotationally invariant, i.e. $(RX)\times(RY)=R(X\times Y)$ for all rotation matrices $R$ (this is because $X\times Y$ is uniquely determined by $X$ and $Y$ and the geometry of space), so the multilinear form $(X,Y,Z)$ must be as well. Thus, WLOG we can consider $X=e_1$, then from the fact it's alternating and linear WLOG $Y=e_2$ and $Z=e_3$ and then it's a single calculation.

anon
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