You can multiply vectors a and b and get the bivector a∧b, drawn as the pink parallelogram here. Or you can use the vector a×b, pointing at right angles to that parallelogram, with length equal to its area. But that requires a 'right-hand rule' and the concept of angle.

(1/n)

@johncarlosbaez naive question: can the cross product be expressed as a morphism from $$V\wedge V$$ to $$V$$ satisfying some nice property? (some determinant vanishing maybe?)

@thosgood - the short answer is "yes". The longer answer: what you call $$V \wedge V$$, I call $$\Lambda^2 V$$. The wedge product gives a natural map

$\wedge: V \otimes \Lambda^2 V \to \Lambda^3 V$

If we choose a volume form (a nonzero element of $$\Lambda^3 V$$ we get an isomorphism $$\Lambda^3 V \cong \mathbb{R}$$ and thus a map

$V \otimes \Lambda^2 V \to \mathbb{R}$

(1/n)

@thosgood - we can reinterpret the map $$V \otimes \Lambda^2 V \to \mathbb{R}$$ as a map
$\Lambda^2 V \to V^*$

So, a volume form on a 3d vector space $$V$$ lets us identify bivectors (guys in $$\Lambda^2 V$$) with covectors (guys in $$V^\ast$$).

Then, an inner product on $$V$$ gives an isomorphism

$$V \cong V^\ast$$

in the usual way. So a volume form and inner product give an isomorphism

$$\Lambda^2 V \cong V$$

(2/n)

@thosgood - @thosgood - now, the wedge product gives a map

$\wedge : V \otimes V \to \Lambda^2 V$

But when we have a volume form and inner product this gives a map... which is the cross product

$\times : V \otimes V \to V$

Summary: the wedge product of vectors in 3d space becomes the cross product in the presence of an inner product and volume form. But, there is no canonical map $$\times : V \otimes V \to V$$ unless we give $$V$$ extra structure!

(3/n)

@thosgood - All this is stuff every good mathematical physicist should know by heart, since we live in 3d space (more or less) and need to understand how structures on this space interact.

I didn't say it all optimally: for example an inner product and orientation give a volume form, so it would be better to say "inner product and orientation" rather than "inner product and volume form".

(4/n, n = 4)

@johncarlosbaez @thosgood I remember figuring this out for myself after reading your "Rosetta stone" article. The way I like to think of it, knowing functional programming but no Hodge theory, is that "the cross product is the curried form of the determinant" (at least, if you think of the determinant as a function on three column vectors in ℝ³, and identify V → ℝ with V).

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