I was always bothered by the definition of the cross product given in e.g. a calculus course because it's never made clear how one would go about defining the cross product in a coordinate-free manner. I now know, not one, but two ways of doing this, and I can't quite see how they're related:

  • The cross product is the Lie bracket in the Lie algebra of $\text{SO}(3)$.
  • The cross product is the Hodge star map $\Lambda^2(V) \to V$ where $V$ is an oriented $3$-dimensional real inner product space.

Okay, so there's one obvious relation here: $V$ has automorphism group $\text{SO}(3)$. But for some reason I can't figure out where to go from here. A good starting point would be to exhibit a canonical isomorphism between an oriented $3$-dimensional inner product space $V$ and the Lie algebra of $\text{Aut}(V)$. Maybe this is obvious. In any case, I would appreciate some clarification.

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    $\begingroup$ The Lie algebra $so(V)$ is naturally isomorphic to $\Lambda^2(V).$ $\endgroup$ Jul 30, 2010 at 7:41
  • $\begingroup$ Thanks, Victor. I guess I should've thought about this a little more. $\endgroup$ Jul 30, 2010 at 17:24
  • $\begingroup$ The ways in which I first heard the cross-product defined was coordinate free, albeit deficient in logical rigor: to find $\mathbf a\times\mathbf b,$ let the fingers of your right hand curl in the direction in which you rotate when going from $\mathbf a$ to $\mathbf b$, and then the direction your thumb points tells you which of two directions orthogonal to $\mathbf a$ and $\mathbf b$ is the right one, and then the norm is the area of the rectangle whose sides are $\mathbf a$ and $\mathbf b.$ $\endgroup$ Oct 22, 2017 at 19:46

3 Answers 3


To expand on Victork Protsak's comment, if $V$ is an $n$-dimensional real vector space with inner-product, the inner-product gives an isomorphism $V\to V^*$ and hence $V\otimes V \to \mathrm{End}(V)$. Under this isomorphism, $\Lambda^2(V)$ is identified with skew-adjoint endomorphisms of $V$, which is precisely the Lie algebra $\mathfrak{so}(V)$.

In the case $\dim V =3,$ the Hodge star gives an isomorphism $\Lambda^2(V) \to V$ and so in total we see that $V$ is canonically isomorphic to $\mathfrak{so}(V)$. A more direct way to see this isomorphism is to send the vector $v \in V$ to the generator of the right-handed rotation about the axis in the direction of $v$ with speed $|v|$.

The use of the phrase "right-handed" makes it clear that in order to identify $V$ and $\mathfrak{so}(V)$ we have used an orientation on $V$; indeed, you need that for the Hodge star. What is interesting is that if you reverse the orientation on $V$, the map to $\mathfrak{so}(V)$ changes sign. This means that what ever orientation you chose on $V$, the push-forward to $\mathfrak{so}(V)$ is the same. Conclusion: $\mathfrak{so}(3)$ is naturally oriented. This is analogous to the natural orientation on $\mathbb{C}$. A more prosaic way to describe the orientation is to pick two independent elements $x,y \in \mathfrak{so}(3)$ and then use $[x,y]$ to complete them to an oriented basis. (Of course, you then need to check that this doesn't depend on your choice of $x,y$.)

  • $\begingroup$ No worries. The fact that so(3) has a natural orientation actually came up in something I was working on recently. Some objects I was looking at had a sign associated to them and it turned out that hidden in the background was an isomorphism between two bundles of so(3) lie algebras; the sign was just whether this isomorphism preserved or reversed the orientation. $\endgroup$
    – Joel Fine
    Jul 30, 2010 at 19:04
  • $\begingroup$ Why is this answer accepted? Sure, it does explain how $so(V)$ is canonically isomorphic to $V$ for a 3-dimensional $V$ as vector spaces. But why is the Lie bracket in $so(V)$ the same as the wedge product followed by the Hodge star in $V$? $\endgroup$
    – Kostya_I
    Jul 6, 2018 at 22:31

Let $\varepsilon( )$ be the volume form in $\mathbb R^3$. For given vectors ${\bf p}$ and ${\bf q}$ the function $f:{\bf x}\mapsto\varepsilon({\bf p},{\bf q},{\bf x})$ is a linear functional and so is represented by a vector ${\bf r}\in\mathbb R^3$, i.e., one has $f({\bf x})=\langle{\bf r},{\bf x}\rangle$. This vector ${\bf r}$ depends in a skew bilinear way from ${\bf p}$ and ${\bf q}$ and is called the $vector\ product$ of ${\bf p}$ and ${\bf q}$.

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    $\begingroup$ Yes, I think this answer has the advantage that it uses precisely the assumptions present in a calculus course, and no extra machinery. This means that you can (more or less) explain it to calculus students. I am teaching a multi-variable calc class this summer, and have them derive the formula for $2\times 2$ and $3\times 3$ determinant from signed area and volume. From here, one can give an honest definition of cross product (rather I gave them the computational one earlier, so they could do their physics homework, and now I can revisit that with a proper explanation) $\endgroup$ Jul 30, 2010 at 12:37

Geometric (Clifford) algebra provides yet another way to understand the cross product. If a vector represents a 1D subspace containing the origin, the wedge (outer) product is a bilinear form on 2 vectors ($a \wedge b$) that represents the subspace containing both vectors. The vector has basis $e_1, e_2, ..., e_n$ ($n$ elements), so the wedge product has basis $e_1\wedge e_2, e_1\wedge e_3, \ldots, e_{n-1}\wedge e_n$ ($\binom{n}{2}$ elements). The cross product is a way to represent this subspace (using its dual, the normal vector). This works only for 3D, where the basis sizes match, using $e_1 e_2 \to e_3,\, e_2 e_3 \to e_1,\, e_3 e_1 \to e_2$.


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