# 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.

• You need to know what $X$ is. en.wikipedia.org/wiki/De_Rham_cohomology Jan 23, 2013 at 5:09
• See section 2 of math.ucsd.edu/~ctiee/math20e-w06/grad_n_curl.pdf Jan 23, 2013 at 5:52
• math.harvard.edu/~knill/teaching/math21a/nash.pdf seems to think the question mark at the end is actually an $8$. Jan 23, 2013 at 5:54
• In the film, when Alicia Larde visits Nash in his office to inform him that she has solved the problem, he rejects her solution, telling her that she could not have solved it because he “did not state whether or not the vector fields were rational.” What do you guys make of his remark? Jan 23, 2013 at 6:07
• @GerryMyerson: The link provided in your first comment doesn't appear to be working; I'm getting a "404" error, and a look through UCSD's math department directory doesn't show a "ctiee" (Chris Tiee?). Jun 4, 2013 at 16:54

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 $$1$$.

• Thanks for your support. I appreciate it a lot. I'm talking about one of your comments... :). Apr 5, 2018 at 19:34
• @OlivierOloa You're welcome! Yours was/is a stunning answer! Apr 6, 2018 at 9:12

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.