notation for vector product in the space

The notation for vector (a.k.a. cross) product in $\mathbb{R}^3$ I usually see is $\times$.

However, some places use $\wedge$ instead, which IMHO creates a lot of confusion, as $\wedge$ usually is used for multiplication in the exterior algebra.

E.g. I saw the following: let's prove the equivalence of linear dependence of three vectors $u_1,u_2,u_3$ in the space and linear dependence of their pairwise products $u_i\wedge u_j$:

take

$$u_k\wedge \sum_{i<j} a_{ij} u_i\wedge u_j = 0$$

and derive from this that $a_{ij}=0$, as $u_1\wedge u_2\wedge u_3=0$ means linear dependence of the $u_i$'s. (which would be fine if we talked about exterior product, but totally wrong for vector product, as it isn't even associative to begin with...)

Are there any arguments for using $\wedge$ for cross product?

• See here en.wikipedia.org/wiki/Exterior_algebra for the wedge product. In three dimensions, it is (identified with) the cross product. In other dimensions, not. – Gerald Edgar Oct 13 '15 at 18:13
• Exterior product is associative, and the vector product is not. Thus I don't get how they can be identified... – Dima Pasechnik Oct 13 '15 at 18:57
• of course in the exterior algebra of $\mathbb{R}^3$ one has that $a\wedge b\wedge c$ is a scalar, so this identification cannot go too far. So you tell undergrads that they shouldn't think of $a\wedge b\wedge c$, and then in multivariate calculus in $\mathbb{R}^n$ you all of a sudden start talking about $a\wedge b\wedge c$? – Dima Pasechnik Oct 13 '15 at 19:03
• Using the Euclidean metric to identify contra- and co-variant vectors, and to construct the hodge dual, you have $v\times w = *(v\wedge w)$. The insertion of the extra $*$ makes it no longer associative. // Another argument for using $\wedge$ is that sometimes we do like to have $\times$ retain either its original meaning (multiplication of scalars) or its categorical meaning (cartesian product). In terms of collision of symbols, you are damned either way. – Willie Wong Oct 13 '15 at 20:23
• In terms of the argument you quoted: since $*$ is a vector space isomorphism linear dependence of some number of objects of the form $v_i\wedge w_i$ is equivalent to the linear dependence of some number of objects of the form $v_i \times w_i$. // Of course the onus falls on the writer of the passage you quoted to explain what he means by $\wedge$. – Willie Wong Oct 13 '15 at 20:28