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.
 A: 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 $1$.
A: 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.
