In the interview of John Nash taken by Christian Skau and Martin Gaussen, in EMS Newsletter, September, 2015 when asked 



> 
> Is it true, as rumours have it, that
>  you started to work on the embedding problem as a result
>  of a bet?
> 
> 
> 


Nash answered 



> 
> I began to work on it. Then I got shifted onto the
>  $^1$ case. It turned out that one could do it in this case
>  with very few excess dimensions of the embedding space
>  compared with the manifold. I did it with two but then
>  Kuiper did it with only one. But he did not do it smoothly,
>  which seemed to be the right thing—since you are given
>  something smooth, it should have a smooth answer.
> 
> 
> 


and 



> 
> But a few years later, I made the generalisation to
>  smooth. I published it in a paper with four parts. There
>  is an error, I can confess now. Some forty years after the
>  paper was published, the logician Robert M. Solovay from
>  the University of California sent me a communication
>  pointing out the error. I thought: “How could it be?” I
>  started to look at it and finally I realized the error in
>  that if you want to do a smooth embedding and you have
>  an infinite manifold, you divide it up into portions and
>  you have embeddings for a certain amount of metric on
>  each portion. So you are dividing it up into a number of
>  things: smaller, finite manifolds. But what I had done was
>  a failure in logic. I had proved that—how can I express
>  it?—that points local enough to any point where it was
>  spread out and differentiated perfectly if you take points
>  close enough to one point; but for two different points
>  it could happen that they were mapped onto the same
>  point.
> 
> 
> 


My question is: 


1. What did Nash mean by very few excess dimensions of the embedding space compared with the manifold? It means that the they can do it in the case the dimension of the embedding space is a little bit greater than the dimension of the manifold, doesn't it?
2. What did Nash mean by his generalization to smooth?
3. What did Nash mean by a certain amount of metric on each portion? Does this mean that each portion has some different metrics?
4. What did Nash mean by "I had proved that that points local enough to any point where it was spread out and differentiated perfectly if you take points close enough to one point"? How can a point be spread out and differentiated?


Well, I don't want to make a cross post but this question were posted two days on [MSE](https://math.stackexchange.com/questions/1822918/questions-on-j-f-nashs-answer-about-his-errors-in-the-proof-of-embedding-theo) but there is no answer so I decide to post it here.


Please explain for me. 


Thanks. 



Questions on J. F. Nash's answer about his errors in the proof of embedding theorem
Igor already answered questions 1 and 2. 


3. What Nash wrote is an attempt to describe to a non-expert audience the solution scheme he had for *non*-compact manifolds. In the noncompact case he proceeds by a reduction process to the compact case. The process involves decomposing the manifold into smaller neighborhoods each diffeomorphic to disks/balls. 


Very roughly speaking, Nash then took a partition of unity adapted to this system of neighborhoods, and used that to cut-off the Riemannian metric on the original manifold to these disks. Now, the truncated objects are no longer Riemannian metrics (since they are not everywhere positive definite); but since the problem is not really that of geometry but that of PDEs in this setting, this is not an obstruction. 


When two neighborhoods overlap, the cut-off function is less than 1 for both of the pieces. So each of the neighborhood inherits "a portion of the metric". 


In other words, the phrase that you are asking about is just Nash trying to describe the idea of a "partition of unity" without using exactly those words.
4. This is in regards to the second portion of the scheme. After dividing into disks and keeping track of their overlaps (using what Nash called "classes"; that's also the sense of the word in Solovay's message), it is possible to (very roughly speaking) replace the disks by spheres. You do this using that the disks can be topologically identified with the sphere with a closed disk removed, and that the "portion of the metric" on the disc is cut-off so that it approaches 0 smoothly on the boundary. This is just largely playing with cut-off functions. 


Now, each sphere has an isometric embedding into some Euclidean space by the theorem for compact manifolds (proven earlier in the paper). So all remains is to somehow reassemble all these spheres into one larger Euclidean space while guaranteeing that there are no self-intersections. 


(Remark: the method of assembly works. Full stop. So if you are happy with an isometric immersion then you are done. The problem that Solovay pointed out has to do with the embedding (non-self-intersecting) part.)


The method of assembly goes something like this. 


a. The choice of the disks earlier means that the disks can be divided into some finitely many classes, such that two disks of the same class cannot intersect. 


b. For each class, construct a smooth map from $M$ such that points within the discs of that class are sent to the image of the embedding of the corresponding sphere. But points outside the disks of that class are sent to the origin. 


c. Take the cartesian product over the classes. This guarantees an immersion. 


To get non-self-intersection Nash tried to exploit the fact that his isometric embedding theorem in the compact case allows one to squeeze the manifold into arbitrarily small neighborhoods. So within each class he can arrange for the different disks to be almost disjointly embedded. He claims that this is enough. Solovay showed that there is a hole in the argument. See below the cut for more info.




---


Incidentally, Nash's paper is available [here](http://www.jstor.org/stable/1969989?origin=crossref&seq=40); the portion that concerns questions 3 and 4 are all in part D, which is decidedly in the "easy" part of the paper. (The hard analytic stuff all happened in part B; here it is essentially combinatorics.)




---


To reveal the "logic problem" with the embedding proof, we remove all unnecessary portions of the proof and focus on the argument that "ensures" non-self-intersection. This operates entirely on the level of sets and does not require any geometry or analysis. 


What Nash did boils down to:


Given a set $M$, he showed that we can decompose $M$ as the union of a bunch of sets $U^{(i)}\_j$. The index $i$ runs from $1, \ldots, n+1$ (a finite number). The index $j$ can be infinite. Each $U^{(i)}\_j$ has a subset within called $V^{(i)}\_j$, and the union of all these subsets $V^{(i)}\_j$ is also assumed to cover $M$. 


For each $i$ Nash finds a space $X^{(i)}$, and a point $x^{(i)}\in X^{(i)}$, and for each $j$ there is an injective mapping $\psi^{(i)}\_j: U^{(i)}\_j \to X^{(i)}\setminus \{x^{(i)}\}$. Since $U^{(i)}\_j$ and $U^{(i)}\_k$ do not intersect unless $j = k$, for each $i$ we can extend this to a map


$$ \psi^{(i)}: M \to X^{(i)}$$


by requiring


$$ \psi^{(i)}(p) = \begin{cases} \psi^{(i)}\_j(p), &p\in U^{(i)}\_j \\
x^{(i)}, &\text{otherwise}\end{cases} $$


Let $X = X^{(1)} \times X^{(2)} \cdots \times X^{(n+1)}$. We are interested in the map
$$ \psi: M \to X = \psi^{(1)}\times \psi^{(2)}\times \cdots \times \psi^{(n+1)}.$$


We want to show that $\psi$ is injective. 


Nash's idea:



> 
> We can assume that $\psi^{(i)}(V^{(i)}\_j) \cap \psi^{(i)}(U^{(i)}\_k) = \emptyset$ if $k > j$. (This he achieves by the fact that isometric embedding can be "made small".) 
> 
> 
> 


He claims this is enough to show injectivity of $\psi$, because:


1. If $p,q\in U^{(i)}\_j$ for the same $i,j$, then we are done because $\psi^{(i)}\_j$ is by construction injective.
2. If $p \in U^{(i)}\_j$ and $q \not\in U^{(i)}\_k$ for for any $k$, then we know $\psi^{(i)}(p) \neq x^{(i)} = \psi^{(i)}(q)$ by construction, and so we are done.
3. So the main worry is that $p\in U^{(i)}\_j$ and $q \in U^{(i)}\_k$ for some different $j,k$. To deal with this, Nash argued thus (I paraphrase)



> 
> Since the $V$'s cover $M$, $p$ is in some $V^{(i)}\_j$ and similarly $q$. So either $q$ does not belong to any $U^{(i)}\_\*$ in which case we are done by point 2, or $q$ belongs to some $U^{(i)}\_k$. If $k > j$ by Nash's idea their images are disjoint, so we get injectivity. If $k = j$ we are done by point 1. If $k < j$ we **swap the roles of $p$ and $q$**. 
> 
> 
> 


The bold phrase was not stated as such in Nash's original paper, but it was his intent. Stated in this form, however, it becomes clear what the problem is: in the argument $p$ and $q$ is not symmetric! One cannot simply swap the roles of $p$ and $q$. It could easily be the case that $q \in U^{(i)}\_k \setminus V^{(i)}\_k$ and then the idea of making the images of $U^{(i)}\_\*$ "almost disjoint" fails to yield anything useful. 


Solovay's message instantiates this observation by setting up a situation where


$$ p \in (U^{(1)}\_1 \setminus V^{(1)}\_1) \cap V^{(2)}\_2 $$


and


$$ q \in (U^{(2)}\_1 \setminus V^{(2)}\_1) \cap V^{(1)}\_2 $$


and both not in any other $U^{(i)}\_j$. 




---


So in terms of the turn of phrase Nash used:


"points local enough to any point" == we take $p \in U^{(i)}\_j$ and $q \in U^{(k)}\_l$ (they are in neighborhoods of some point)


"it was spread out and differentiated perfectly" == the map $\psi(p) \neq \psi(q)$ (is injective; "differentiated" in the sense of "to tell apart")


"if you take points close enough to one point" == if $i = k$ and $j = l$ (in the same neighborhood)


"but for two different points it could happen that they were mapped onto the same point." == injectivity may fail otherwise. 


