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:R3/XR3 so ×F=0}
W={F=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.

Answer

The problem is to find a subset X of R3 such that if V is the vector space of vector fields F on R3\ X with ×F=0 and W is the vector space of vector fields F on R3 \ X satisfying F=g, for some function g on R3 \ X, then V/W has dimension 8.

If F does equal 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

F0=(y(x2+y2), x(x2+y2), 0)

has vanishing curl on R3 \ X but it does not satisfy F0=g, as if we integrate F counterclockwise around the unit circle in the z=0 plane, we will get 2π0.

Now, given any vector field F on R3 \ X with ×F=0, we can make its line integral around the unit circle vanish by subtracting off some multiple of F0.

I claim that this is enough to make F=g. We can try to find a g with F=g by starting at some point x0 not in X and integrating F along a path from x0 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 x0=(1,0,0), and for each x not in X, we pick a reference path Px from x0 to x.

Remaining inside R3 \ X, we can continuously deform any path Px from x0 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 Px 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 Px as along Px. Hence the integral is independent of path and so F=g. This means that for the linear map L from V to the real numbers given by integration around the unit circle, 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 dimV/W=8, we can take X to be any set of 8 non-intersecting lines, for example {x=y=0}{x=0, y=2}{x=0, y=4}{x=0, y=14}

In this case, for i=0,,7 , we can define the vector field Fi in V to be

Fi=((y2i)(x2+(y2i)2), x(x2+(y2i)2), 0)

We can then define a linear map L from V to R8 by setting
L(F)=(I0(F),,I7(F))

where Ij(F) is the line integral of F around the circle {x2+(y2j)2=1, z=0}, taken in the counterclockwise sense.

We have
Ij(Fi)={2πi=j0ij

so L is surjective, and clearly Ker LW.

By an argument similar to the one in the last paragraph, we can prove that Ker LW.

Hence Ker L=W and so L gives an isomorphism from V/W to R8.

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 H1dR(R3X), the dimension 1 de Rham cohomology group of R3X. Its dimension gives the number of dimension 1 holes in R3X.

We can also construct H0dR(R3X).

This is the vector space of functions f on R3X such that f=0, i.e. f is locally constant. Its dimension will equal the number of connected components of R3X.

Finally, we can construct H2dR(R3X).

This will give the number of dimension 0 holes in R3X; for example, if we take X={0}, then H2dR(R3X) will have dimension 1.

Attribution
Source : Link , Question Author : Mlagma , Answer Author : HeatfanJohn

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