# 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

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 $$XX$$ of $$R3\mathbb{R}^3$$ such that if $$VV$$ is the vector space of vector fields $$FF$$ on $$R3\mathbb{R}^3$$\ $$XX$$ with $$∇×F=0\nabla\times F = 0$$ and $$WW$$ is the vector space of vector fields $$FF$$ on $$R3\mathbb{R}^3$$ \ $$XX$$ satisfying $$F=∇gF = \nabla g$$, for some function $$gg$$ on $$R3\mathbb{R}^3$$ \ $$XX$$, then $$V/WV / W$$ has dimension $$88$$.

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

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

In this case the vector field

$$F0=(−y(x2+y2), x(x2+y2), 0)F_0 = \left(\frac{−y}{(x^2 + y^2)},\ \frac{x}{(x^2 + y^2)},\ 0\right)$$

has vanishing curl on $$R3\mathbb{R}^3$$ \ $$XX$$ but it does not satisfy $$F0=∇gF_0 = \nabla g$$, as if we integrate $$FF$$ counterclockwise around the unit circle in the $$z=0z=0$$ plane, we will get $$2π≠02\pi \neq 0$$.

Now, given any vector field $$FF$$ on $$R3\mathbb{R}^3$$ \ $$XX$$ with $$∇×F=0\nabla\times F = 0$$, we can make its line integral around the unit circle vanish by subtracting off some multiple of $$F0F_0$$.

I claim that this is enough to make $$F=∇gF = \nabla g$$. We can try to find a $$gg$$ with $$F=∇gF = \nabla g$$ by starting at some point $$x0x_0$$ not in $$XX$$ and integrating $$FF$$ along a path from $$x0x_0$$ to some other point not in $$XX$$, $$xx$$, say.

$$g(x)g(x)$$ 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 $$PP$$ as along a path $$QQ$$ as long as the gap between $$PP$$ and $$QQ$$ can be filled in by a surface which avoids $$XX$$.

Suppose we take $$x0=(1,0,0)x_0 = (1,0,0)$$, and for each $$xx$$ not in $$XX$$, we pick a reference path $$PxP_x$$ from $$x0x_0$$ to $$xx$$.

Remaining inside $$R3\mathbb{R}^3$$ \ $$XX$$, we can continuously deform any path $$P′xP'_x$$ from $$x0x_0$$ to $$xx$$ into a path that:

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

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

Since the line integral of $$FF$$ around the unit circle vanishes, this means that $$FF$$ has the same integral along $$PxP_x$$ as along $$P′xP'_x$$. Hence the integral is independent of path and so $$F=∇gF = \nabla g$$. This means that for the linear map $$LL$$ from $$VV$$ to the real numbers given by integration around the unit circle, $$Ker L=W\text{Ker}\ L = W$$.
Therefore $$V/WV/W$$ is isomorphic to the real numbers and so is $$11$$-dimensional.

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

To get an example with $$dimV/W=8\text{dim} V/W = 8$$, we can take $$XX$$ to be any set of $$88$$ non-intersecting lines, for example $${x=y=0}∪{x=0, y=2}∪{x=0, y=4}∪…∪{x=0, y=14}\{x=y=0\} \cup \{x=0,\ y=2\} \cup \{x=0,\ y=4\} \cup \ldots \cup \{x=0,\ y=14\}$$

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

$$Fi=(−(y−2i)(x2+(y−2i)2), x(x2+(y−2i)2), 0)F_i = \left(\frac{−(y−2i)}{(x^2 + (y−2i)^2)},\ \frac{x}{(x^2 + (y−2i)^2)},\ 0\right)$$

We can then define a linear map $$LL$$ from $$VV$$ to $$R8\mathbb{R}^8$$ by setting
$$L(F)=(I0(F),…,I7(F))L(F) = (I_0(F), \ldots, I_7(F))$$

where $$Ij(F)I_j(F)$$ is the line integral of $$FF$$ around the circle $${x2+(y−2j)2=1, z=0}\{x^2 + (y−2j)^2 = 1,\ z = 0\}$$, taken in the counterclockwise sense.

We have
$$Ij(Fi)={2πi=j0i≠jI_j(F_i) = \begin{cases} 2\pi & i = j \\ 0 & i ≠ j \end{cases}$$

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

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

Hence $$Ker L=W\text{Ker}\ L = W$$ and so $$LL$$ gives an isomorphism from $$V/WV/W$$ to $$R8\mathbb{R}^8$$.

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.

$$V/WV/W$$ has the name $$H1dR(R3∖X)H^1_{dR}(\mathbb R^3\setminus X)$$, the dimension $$11$$ de Rham cohomology group of $$R3∖X\mathbb R^3\setminus X$$. Its dimension gives the number of dimension $$11$$ holes in $$R3∖X\mathbb R^3\setminus X$$.

We can also construct $$H0dR(R3∖X)H^0_{dR}(\mathbb R^3\setminus X)$$.

This is the vector space of functions $$ff$$ on $$R3∖X\mathbb R^3\setminus X$$ such that $$∇f=0\nabla f = 0$$, i.e. $$ff$$ is locally constant. Its dimension will equal the number of connected components of $$R3∖X\mathbb R^3\setminus X$$.

Finally, we can construct $$H2dR(R3∖X)H^2_{dR}(\mathbb R^3\setminus X)$$.

This will give the number of dimension $$00$$ holes in $$R3∖X\mathbb R^3\setminus X$$; for example, if we take $$X={0}X = \{0\}$$, then $$H2dR(R3∖X)H^2_{dR}(\mathbb R^3\setminus X)$$ will have dimension $$11$$.