Intuitively when to use the wedge product?

Question

When I first learned the dot product and the cross product in $\mathbb{R}^3$ I spent some time understanding when I would like to use them. After some time I understood that the dot product usefulness was its ability to detect orthogonality and its relationship with projections. In the same manner that the cross product gives means to deal with areas.

Now in multilinear algebra I've learned about the wedge product of vectors defined in the tensor algebra. The definition I have is that if $T(V)$ is the tensor algebra of the space $V$, we let $J$ be the ideal generated by all elements of the form $T-\operatorname{sgn}(\sigma)f_{\sigma}(T)$ for each $T\in T^k_0(V)$, $\sigma\in S_k$ and $k\in \mathbb{N}$ and we set $\Lambda(V) = T(V)/J$. The wedge product is then the product of this algebra, given by means of $(T+J)\wedge(S+J)=T\otimes S + J$ from where it follows bilinearity, skew-commutativity and so on.

This is hardly used on differential geometry and Physics, so that there should be some intuition on when to use the wedge product. I've also been able to see that if we are in $\mathbb{R}^2$, then $v\wedge w = \det(A)e_1\wedge e_2$ with $A$ being the matrix whose columns are $v$ and $w$. In that case, $v\wedge w$ is the area spanned by $v$ and $w$ times this $e_1\wedge e_2$ so we could think of it as a piece of plane.

However, in other cases this doesn't work apparently. So, when we use the wedge product? What it is supposed to represent?

Answer 1

A simple point of view on the matter is that wedge products of vectors represent subspaces. Each subspace has a weight and an orientation. Think of vectors for a moment: a vector can be taken to represent a 1d subspace. Such a 1d subspace has two unit vectors--some unit vector $\hat v$ and its additive inverse $-\hat v$. A vector representing this subspace is then $\pm \alpha v$, for some positive scalar $\alpha$, which we can call a weight or magnitude.

Wedge products of vectors in general can be thought of in the same way: a wedge product of two vectors $u \wedge v$ represents some 2d subspace. Just as you can add two vectors to get another vector (and thus, another subspace), you can add two bivectors to get another bivector. It's not always true that this yields another subspace--in 3d and below it always does; in 4d, you could have two planes that are mutually orthogonal in all respects and thus can't be added to form a subspace.

I submit to you that adding or subtracting bivectors is no less intuitive than adding or subtracting vectors; you merely need to develop your intuition. Think, for example, about the rectangular faces of a triangular prism. Each face could be seen as a bivector, and you can add two of the faces to get the third face. This is no different than adding two vectors in a triangle to get the third.

You can extend this on and on, for trivectors and so on, to develop a useful algebra of general multivectors. The discipline of geometric (or clifford) algebra is ideally suited for thinking about such operations in a transparent geometric manner. It has a well-developed literature for incorporating wedge products and other useful tools into your everyday life as a physicist or differential geometer.

Answer 2

If you take $k$ vectors $v_1,\ldots, v_k$ in $\mathbb R^n$ then their wedge product vanishes precisely when they are linearly dependent. This generalizes what you know about the vector product in $\mathbb R^3$ and is just as useful when making calculations in geometry and physics. You can also give a geometric meaning to $v_1\wedge\ldots\wedge v_k$ when it doesnt vanish. For $k=n$, $v_1\wedge\ldots\wedge v_n$ gives the volume of the parallelepiped whose edges are the $v_i$ (the determinant of the matrix whose columns are the $v_i$'s). For general $k\leq n$, the ${n \choose k}$ coordinates of $v_1\wedge\ldots\wedge v_n$ measure the (signed) volume of the projections of the $k$-parallelepiped generated by the $v_i$'s onto the $k$-dim coordinate planes (each one given by choosing $k$ of the $n$ coordinates and equating the rest to $0$).