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A cross product can be defined as a bilinear operation on a real vector space with inner product that has the

property of orthogonality: $ \mathbf u \cdot(\mathbf u \times \mathbf v)=\mathbf v \cdot(\mathbf u \times \mathbf v)=0$

and the Pythagorean property $ |\mathbf u \times \mathbf v|^2=u^2v^2-(\mathbf u \cdot \mathbf v)^2$

Starting from such definition we can prove that a non trivial cross product can exists only on $\mathbb R^3$ (where we have two possible cross products with opposite orientation) and on $\mathbb R^7$ (where we can have 480 different cross products. See: https://en.wikipedia.org/wiki/Seven-dimensional_cross_product).

This question is motivated by the fact that I found (see: https://www.jstor.org/stable/2315620) the statement that in seven dimension space the cross product does not satisfy the Jacobi identity.

I know that the Jacobi identity (J).
$$ \mathbf u \times(\mathbf v\times \mathbf w)+ \mathbf w \times(\mathbf u\times \mathbf v)+ \mathbf v \times(\mathbf w\times \mathbf u)=0 $$ can be provided as a consequence of the Lagrange identity (V).

$$\mathbf u \times(\mathbf v\times \mathbf w)=(\mathbf u\cdot\mathbf w)\mathbf v-(\mathbf u\cdot\mathbf v)\mathbf w $$

And the proof of (V)$\Rightarrow$(J) can be done without the use of coordinate, only using te properties that define the cross and dot product (see e.g. Geometric proof for triple vector product Jacobi identity).

I can prove that in the space $\mathbb R^7$, with any one of the 480 possible cross products, the Lagrange identity is not valid. (using an argument that does not use the coordinates, similar to this one: Do the BAC-CAB identity for triple vector product have some intepretation?)

So in $\mathbb R^7$ we cannot use the same argument of $\mathbb R^3$ to proof the Jacobi identity. But, to be sure that the Jacobi identity is not valide in 7-D space we need a proof that, in general, (J)$\Rightarrow$ (V). But I don't see how to find this proof.

Mr.Gandalf Sauron
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Emilio Novati
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    How is the cross product defined in $\Bbb R^7$? – Berci Mar 21 '22 at 21:33
  • As @Berci comments/asks, to what do you refer by "cross product in $7$ dimensions"? In what sense are there $480$ possible ones? Can you clarify? – paul garrett Mar 21 '22 at 21:44
  • You can see https://en.wikipedia.org/wiki/Seven-dimensional_cross_product – Emilio Novati Mar 21 '22 at 22:13
  • Interesting strategy to refute (J), if it can work. Alternatively, use Eq. (15) here. This reduces the issue to octonions not associating. – J.G. Mar 22 '22 at 20:16
  • I am not so surprised that Wiki has "a definition" of such a thing, but it's not really standard. Not that that is a decisive criticism, but... my own diagnostic questions would be about "what it should do?" and "why do we need that functionality?" – paul garrett Mar 22 '22 at 20:25
  • @J.G. Thank you. My idea is to prove the properties of the cross product in 7-D only from the definition, without the use of Octonions. – Emilio Novati Mar 22 '22 at 20:46
  • There are 480 7D cross products that can be obtained from permuting basis vectors in the multiplication table of a given one, but there are infinitely many cross products "between" these ones. The space of all 7D cross products is parametrized by $SO(7)/G_2$. @paulgarrett/Berci The 7D cross product isn't as well-known as the 3D one, but I would say it is standard in the context of literature surrounding / adjacent to real normed division algebras, and at any rate it is a quick google search away. Problem is it's not relevant because it doesn't satisfy the Jacobi identity. – anon Mar 22 '22 at 21:18
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    I recommend renaming (L) the vector triple product identity (V), so we don't confuse it with Lagrange's identity for $|{\bf a}\times{\bf b}|^2$. I will adopt this notation. There is a vacuous sense in which (J) implies (V) in 3D: they are both true, and every true statement "implies" every other true statement, that's how the logical conditional works. Presumably we want algebraic reasoning that leads us from one to the other, though. So we need to figure out how to write out our hypotheses formally so we know what kind of algebraic manipulation is allowed. – anon Mar 22 '22 at 21:36
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    We have an inner product space with a bilinear operation satisfying (J). Should we have it be alternating too, so it's a Lie algebra? We need this operation to play nice with the inner product (otherwise they're just unrelated operations!), so we need to assume (a) the scalar dot product is well-defined, and/or (b) the Binet-Cauchy identity (which implies Lagrange's identity for $|{\bf a}\times{\bf b}|^2$ in 3D but not in higher dimensions). Or if we're talking about a Lie algebra, do we want the inner product to be the Killing form up to a scalar multiple? – anon Mar 22 '22 at 21:41
  • @runway44: Thank you. I changed (L) to (V) as you suggest. Clearly I search a direct proof that (J) (V) so that, proving that (V) is not satisfied we can deduce that also (J) is not defined, starting from the classical axiomatic definition of an inner product vector space. A direct proof that (J) is not true in$ R^7$ is given here :https://www.jstor.org/stable/2315620. But I'm interesting to explore the relation between (J) and (V) in different vector spaces where the cross product can be defined. – Emilio Novati Mar 23 '22 at 09:56
  • You can't get a direct proof that (J) implies (V), or it's contrapositive, just from the definition of the inner product space because the definition doesn't include anything about cross products. And cross products as they're usually defined, by the way, only exist in 3D and 7D so there are no other spaces to explore... – anon Mar 23 '22 at 21:03

1 Answers1

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I found that the proof of (J) $\Rightarrow$ (V) can be done starting from the identity $$ \mathbf u \times (\mathbf v\times \mathbf w)-\mathbf w \times (\mathbf u\times \mathbf v)=2(\mathbf u\cdot \mathbf w)\mathbf v -(\mathbf w\cdot \mathbf v)\mathbf u-(\mathbf u\cdot \mathbf v)\mathbf w $$ ( proved in https://www.jstor.org/stable/2315620 , formula (2))

using this in the first term, the Jacobi identity: $$ \mathbf u \times(\mathbf v\times \mathbf w)+ \mathbf w \times(\mathbf u\times \mathbf v)+ \mathbf v \times(\mathbf w\times \mathbf u)=0 $$ becomes: $$ 2 \mathbf w \times(\mathbf u\times \mathbf v)+2(\mathbf u\cdot \mathbf w)\mathbf v -(\mathbf w\cdot \mathbf v)\mathbf u-(\mathbf u\cdot \mathbf v)\mathbf w+ \mathbf v \times(\mathbf w\times \mathbf u)=0 $$

and using the same identity on the last triple product: $$ 2 \mathbf w \times(\mathbf u\times \mathbf v)+2(\mathbf u\cdot \mathbf w)\mathbf v -(\mathbf w\cdot \mathbf v)\mathbf u-(\mathbf u\cdot \mathbf v)\mathbf w+ \mathbf u \times(\mathbf v\times \mathbf w)+2(\mathbf v\cdot \mathbf u)\mathbf w-(\mathbf u\cdot \mathbf w)\mathbf v-(\mathbf v\cdot \mathbf w)\mathbf u=0 $$ adding the similar terms, $$ 2 \mathbf w \times(\mathbf u\times \mathbf v)+\mathbf u \times(\mathbf v\times \mathbf w)+(\mathbf u\cdot \mathbf w)\mathbf v+(\mathbf v\cdot \mathbf u)\mathbf w-2(\mathbf v\cdot \mathbf w)\mathbf u=0 $$ and using the same identity for the second term we find

$$ 3\mathbf w \times(\mathbf u\times \mathbf v)+3(\mathbf u\cdot \mathbf w)\mathbf v-3(\mathbf v\cdot \mathbf w)\mathbf u=0 $$ that is the lagrange identity (V)

$$ \mathbf w \times(\mathbf u\times \mathbf v)=(\mathbf v\cdot \mathbf w)\mathbf u-(\mathbf u\cdot \mathbf w)\mathbf v $$

And, since it is easy to prove that (V) is true in $\mathbb R^3$ but false in $\mathbb R^7$, this prove also that te Jacobi identity is true in $\mathbb R^3$ and false in $\mathbb R^7$.

Emilio Novati
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  • This can be the algebraic reasoning supposed by @runway44 in a comment. I put this result here as an answer to complete the post. Thanks if anyone wants to test that my calculations are correct. – Emilio Novati Mar 24 '22 at 14:39