Wedge product = set intersection?

Question

In a research article [1] I found the following formulation:

The wedge product may be considered as set intersection. For example, surfaces of constant $f(x,y,z)$ and surface of constant $g(x,y,z)$ intersects along the lines given by $df \wedge dg$. The notion of interpreting the wedge product as set intersection is appealing from a topological standpoint.

The article almost does not make use of exterior differential systems or other sophisticated math. It also gives no precise reference for the above statement.

Questions:

1. Does this viewpoint (wedge product = set intersection) make sense at all?
2. What is meant with "the lines given by $df \wedge dg$"? Edit: E.g. When I set $f(x,y,z) = \frac{1}{2}(x^2 + y^2 + z^2) $ and $g(x,y,z) = ax + by + cz$ with real constants $a,b,c$. Then we have $df\wedge dg = (x dx + y dy + z dz)\wedge(a dx + b dy + c dz)$. But what line does correspond to this 2-form. (I would suspect that we obtain circles.)

[1] http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=990890

Answer 1

The statement in the article has to be taken with a huge grain of salt. Even for $1$-forms, there's no consistent way to interpret all wedge products of $1$-forms as intersections.

If you restrict attention to nonvanishing decomposable forms (ones that can be written locally as wedge products of $1$-forms), and add a few extra restrictions, then there is something that can be said.

If $\omega$ is any differential form on an $n$-manifold $M$, let us define the kernel (sometimes called the radical) of $\omega$ at a point $p\in M$ to be the set of vectors $v\in T_pM$ such that $\omega_p(v,\cdot,\dots,\cdot) = 0$. If $\omega$ is a nonvanishing decomposable $k$-form, its kernel at $p$ is an $(n-k)$-dimensional subspace of $T_pM$ (the common kernel of $\eta_1,\dots,\eta_k$ when we write $\omega_p = \eta_1\wedge\dots\wedge\eta_k$), and these subspaces fit together to form a smooth $(n-k)$-dimensional distribution on $M$.

Now if both $\omega$ and $\theta$ are nonvanishing decomposable forms, and their distributions are transverse to each other (i.e., the two distributions span the tangent space at each point), then $\omega\wedge\theta$ is a nonvanishing decomposable form whose distribution is the intersection of the distributions determined by $\omega$ and $\theta$.

If the distributions in question happen to be integrable, then this can be rephrased in terms of the leaves of the corresponding foliations: Each leaf determined by $\omega\wedge\theta$ is a (connected component of) an intersection of a leaf determined by $\omega$ and one determined by $\theta$.

If the forms in question are allowed to vanish, or the distributions happen not to be transverse somewhere, then all of this falls apart because $\omega\wedge\theta$ will be zero at all such points.

Answer 2

At a non-critical point $x$ of $f$, the kernel of the $1$-form $\text{d}f$ is the tangent space to the surface $X := f^{-1}(f(x))$ at $x$; similarly, for a non-critical point $x$ of $g$, $\text{d}g_x$ has the tangent space of $Y := g^{-1}(g(x))$ at $x$ as its kernel. Putting both together, at non-critical points $x$ for both $f$ and $g$ the radical of the skew-symmetric $2$-form $\text{d}f\wedge\text{d}g$ contains the intersection $\text{T}_x X\cap\text{T}_x Y$, which at a point of transversal intersection of $X$ and $Y$ is just $\text{T}_x (X\cap Y)$. In this sense, the $2$-form $\text{d}f\wedge\text{d}g$ can be viewed as 'cutting out' the intersection of level sets of $f$ and $g$.

Also, you might want to look at Poincaré Duality between the $\cup$/$\wedge$-product in cohomology and the intersection product in homology.

Edit concerning the added example

Wlog we take $g(x,y,z) := z$, so that $\text{d}g = \text{d}z$, pick any ${\mathbf x}_0=(x_0,y_0,z_0)\in {\mathbb R}^3$ and put $r_0 := f({\mathbf x}_0)=\frac{1}{2}(x_0^2+y_0^2+z_0^2)$. Building on the intuition above, we expect $(\text{d}f\wedge\text{d}g)_{{\mathbf x}_0}$ to contain the tangent space of the intersection $\{z = z_0\}\cap\{\frac{1}{2}(x^2+y^2+z^2)=r_0\}$ at ${\mathbf x}_0$ in its radical, with equality at points ${\mathbf x}_0$ of transversal intersection, which are precisely those ${\mathbf x}_0$ where $(x_0,y_0)\neq (0,0)$.

Now, $(\text{d}f\wedge \text{d} g)_{{\mathbf x}_0} = (x_0 \text{d}x + y_0\text{d}y + z_0\text{d}z)\wedge \text{d}z = x_0\text{d}x\wedge\text{d}z + y_0\text{d}y\wedge\text{d}z$. Let's call this form $\omega$; on a tangent vector ${\mathbf v} := v_x\frac{\partial}{\partial x} + v_y\frac{\partial}{\partial y} + v_z\frac{\partial}{\partial z}$ its restriction is given by $\omega({\mathbf v},-) = (v_x x_0 + v_y y_0)\text{d} z - v_z x_0 \text{d} x - v_z y_0\text{d}y$. For $(x_0,y_0)=(0,0)$, this vanishes completely, while for $(x_0,y_0)\neq (0,0)$ the radical is given by those $\mathbf v$ where $v_z=0$ and $(v_x,v_y)\perp (x_0,y_0)$. This is precisely the expected tangent line.

Answer 3

Define the characteristic vector (fields) of a form as $\iota_v\omega=0$. By the properties of the interior product if both $v_1$ and $v_2$ are c.v., then so are their linear combinations, $\text{span}(v_1,v_2)$. Furthermore, if the rank of the form $r$, the number of linearly independent characteristic vectors is $n-r$.

One can describe a tangent plane to $f=\text{const}$ at a regular point $p$ as a linear space $\{v\}$ such that $\iota_vdf=df(v)=f_xv^1+f_yv^2+f_zv^3=0$.

Next, using $\iota_v(\beta\wedge\gamma) = (\iota_v\beta)\wedge\gamma+(-1)^p\beta\wedge(\iota_v\gamma)$ we look for $\{v\}:\iota_v(df\wedge dg)=(\iota_vdf)\wedge dg-(\iota_vdg)\wedge df=0$. This is generally satisfied only if the vector is in the tangent spaces to both level set surfaces at points of intersection, giving us the tangent lines of the intersection. For your example, you'd have to solve simultaneously $xv^1+yv^2+zv^3=0$ and $av^1+bv^2+cv^3=0$ to get the tangent line. The intersection points, at least initial, are still to be found from $f=g=\text{const}.$