What is the solution to Nash's problem presented in "A Beautiful Mind"?

I was watching the said movie the other night, and I started thinking about the equation posed by Nash in the movie. More specifically, the one he said would take some students a lifetime to solve (obviously, an exaggeration). Nonetheless, one can't say it's a simple problem.

Anyway, here it is

$$V = \{F:\mathbb{R^3}/X\rightarrow \mathbb{R^3} \text{ so } \hspace{1mm}\nabla \times F=0\}$$ $$W = \{F = \nabla g\}$$ $$\dim(V/W) = \; 8$$

I haven't actually attempted a solution myself to be honest, but I thought it would be an interesting question to pose. I have done a quick search on this site and Google, but there were surprisingly few results.

In any case, I was curious if anyone knew the answer aside from the trivial.

The problem is to find a subset $X$ of $\mathbb{R}^3$ such that if $V$ is the vector space of vector fields $F$ on $\mathbb{R}^3$\ $X$ with $\nabla\times F = 0$ and $W$ is the vector space of vector fields $F$ on $\mathbb{R}^3$ \ $X$ satisfying $F = \nabla g$, for some function $g$ on $\mathbb{R}^3$ \ $X$, then $V / W$ has dimension $8$.

If $F$ does equal $\nabla g$, then the line integral of $F$ along a path will be independent of the path, so the line integral of $F$ around any closed curve must vanish.

Suppose we take $X$ to be the (infinite) line $x=y=0$.

In this case the vector field

$$F_0 = \left(\frac{−y}{(x^2 + y^2)},\ \frac{x}{(x^2 + y^2)},\ 0\right) $$

has vanishing curl on $\mathbb{R}^3$ \ $X$ but it does not satisfy $F_0 = \nabla g$, as if we integrate $F$ counterclockwise around the unit circle in the $z=0$ plane, we will get $2\pi \neq 0$.

Now, given any vector field $F$ on $\mathbb{R}^3$ \ $X$ with $\nabla\times F = 0$, we can make its line integral around the unit circle vanish by subtracting off some multiple of $F_0$.

I claim that this is enough to make $F = \nabla g$. We can try to find a $g$ with $F = \nabla g$ by starting at some point $x_0$ not in $X$ and integrating $F$ along a path from $x_0$ to some other point not in $X$, $x$, say.

$g(x)$ can then be set equal to the value of this integral.

By Stokes's Theorem, we will get the same result integrating along a path $P$ as along a path $Q$ as long as the gap between $P$ and $Q$ can be filled in by a surface which avoids $X$.

Suppose we take $x_0 = (1,0,0)$, and for each $x$ not in $X$, we pick a reference path $P_x$ from $x_0$ to $x$.

Remaining inside $\mathbb{R}^3$ \ $X$, we can continuously deform any path $P'_x$ from $x_0$ to $x$ into a path that:

$(1)$ moves around the unit circle in either a counterclockwise or a clockwise direction a number of times, possibly zero and then

$(2)$ goes along $P_x$ to $x$. (This is because we are free to move the path around as we choose as long as we don't intersect the line $x=y=0$.)

Since the line integral of $F$ around the unit circle vanishes, this means that $F$ has the same integral along $P_x$ as along $P'_x$. Hence the integral is independent of path and so $F = \nabla g$. This means that for the linear map $L$ from $V$ to the real numbers given by integration around the unit circle, $\text{Ker}\ L = W$. Therefore $V/W$ is isomorphic to the real numbers and so is $1$-dimensional.

In general, we will get one extra dimension of $V/W$ for each independent element of $X$ which stops us from continuously deforming paths into other paths. Placing a point or ball into $X$ will not increase the dimension of $V/W$ as we can simply move the paths around it.

To get an example with $\text{dim} V/W = 8$, we can take $X$ to be any set of $8$ non-intersecting lines, for example $$\{x=y=0\} \cup \{x=0,\ y=2\} \cup \{x=0,\ y=4\} \cup \ldots \cup \{x=0,\ y=14\}$$

In this case, for $i=0, \ldots, 7$ , we can define the vector field $F_i$ in $V$ to be

$$F_i = \left(\frac{−(y−2i)}{(x^2 + (y−2i)^2)},\ \frac{x}{(x^2 + (y−2i)^2)},\ 0\right)$$

We can then define a linear map $L$ from $V$ to $\mathbb{R}^8$ by setting $$L(F) = (I_0(F), \ldots, I_7(F))$$

where $I_j(F)$ is the line integral of $F$ around the circle $\{x^2 + (y−2j)^2 = 1,\ z = 0\}$, taken in the counterclockwise sense.

We have $$I_j(F_i) = \begin{cases} 2\pi & i = j \\ 0 & i ≠ j \end{cases}$$

so $L$ is surjective, and clearly $\text{Ker}\ L \supseteq W$.

By an argument similar to the one in the last paragraph, we can prove that $\text{Ker}\ L \subseteq W$.

Hence $\text{Ker}\ L = W$ and so $L$ gives an isomorphism from $V/W$ to $\mathbb{R}^8$.

This problem is a special case of what is called de Rham cohomology, where people construct vector spaces of differential forms on a space in such a way that their dimension yields topological information about the space.

$V/W$ has the name $H^1_{dR}(\mathbb R^3\setminus X)$, the dimension $1$ de Rham cohomology group of $\mathbb R^3\setminus X$. Its dimension gives the number of dimension $1$ holes in $\mathbb R^3\setminus X$.

We can also construct $H^0_{dR}(\mathbb R^3\setminus X)$.

This is the vector space of functions $f$ on $\mathbb R^3\setminus X$ such that $\nabla f = 0$, i.e. $f$ is locally constant. Its dimension will equal the number of connected components of $\mathbb R^3\setminus X$.

Finally, we can construct $H^2_{dR}(\mathbb R^3\setminus X)$.

This will give the number of dimension $0$ holes in $\mathbb R^3\setminus X$; for example, if we take $X = \{0\}$, then $H^2_{dR}(\mathbb R^3\setminus X)$ will have dimension $1$.

As far as remind this is identical with the calculation of a de Rham cohomology of $\Bbb R^3\setminus X$. I guess it makes only sense if $\Bbb R^3\setminus X$ can be somehow captured as manifold, that is not possible for arbitrary $X$. For closed $X$ the set is a manifold and following the de Rham Theorem the problem turns to calculate the singularity cohomology of $\Bbb R^3\setminus X$ - i.o.w. number of holes in $\Bbb R^3\setminus X$. As far as I remember the problem can only be solved if an $X$ is given or certain property conditioning $\Bbb R^3\setminus X$.

I hope this is good answer in brief.

PS: The 2D version for $\Bbb R^2\setminus X$ is there as well.