The question is maybe a bit technical, but I find the related construction very beautiful.

In the very famous work - "$C^1$-isometric imbeddings" by J.Nash (1954) the author presented the fundamental theorem (which was especially recognized some years after) about isometric embeddings of Riemannian manifolds.

$\textbf{Theorem of J. Nash:}$ Any Riemannian $n$-manifold has $C^{1}$ an isometric imbedding in $E^{2n+1}$ (Euclidean $2n+1$-dimensional space).

There are also some similair other theorems of J.Nash and N.Kuiper which nowadays are formulated in a bit different form, especially in a view of famous Gromov's H-principle.

I have a question regarding the considerations of J.Nash in the aforementioned paper. I don't detail all the proof of this theorem, but I recall the main ingredient of the construction of such immersion (imbedding) - "short" immersion or imbedding.

$\textbf{Definition}:$ Immersion (imbedding) $z : (M,g) \rightarrow (E^k, h)$ is short if $$ z^*h \leq g \text{ in the sense of quadratic forms}, $$ where $g$ is a metric on $M$, $h$ is euclidean metric on $E_k$.

Having initial short immersion or imbedding J.Nash presents some sequence of immersions (imbeddings) $\{z_n\}_{n=1}^{\infty}$, where all $z_n$ are all short and monotonically increase induced metric $z_n^*h$. This sequence of $z_n$ converges in $C^1$ sense and also gives the desired metric tensor $g$ for induced metric. So we naturally come up to the following question:

$\textbf{Important question:}$ How to construct an initial short immersion (imbedding)?

1) For compact manifolds the answer is simple: use Whitney theorem and multiply the map by a small $\varepsilon > 0$ to decrease the induced metric.

2) For open (non-compact) manifolds it is a bit trickier. The construction is as follows:

Let $\{U_i\}_{i=1}^{\infty}$ locally finite open cover of $M$, $\{\phi_i\}_{i=1}^{\infty}$ is the partition of unity subordinated to this cover. Using Dawker's theorem (or Borsuk theorem) we choose $\{U_i\}$ such that no more than $n$ charts intersect other chart. So we say that the multiplicity of the cover $\{U_i\}$ is $s$.

We divide $\{U_i\}$ into $s$ separate classes: it is easy, we take the charts in order and give them any class that is not yet given to any other neighbor. Next we construct an imbedding in $s(n+2)$ dimensional space by defining the maps: \begin{align} &u_\sigma(x) = \begin{cases} \varepsilon_i \phi_i(x), \text{ if $x$ belongs to vicinity of class } \sigma,\\ 0, \text{ otherwise}, \end{cases}\\ &v_\sigma(x) = \begin{cases} \varepsilon^2_i \phi_i(x), \text{ if $x$ belongs to vicinity of class } \sigma,\\ 0, \text{ otherwise}, \end{cases}, \\ &w_{\sigma j}(x) = \begin{cases} \varepsilon_i \phi_i(x) x_{ij}, \text{ if $x$ belongs to vicinity of class } \sigma,\\ 0, \text{ otherwise}, \end{cases}, \end{align} where $x_{ij}$ stands for the $j$-th coordinate in chart $U_i$. The numbers $\varepsilon_i$ is the sequence of positive constants monotonously decreasing to zero.

One can see that: if two points $x$ and $y$ lie in different charts $U_i, U_j$, then they are separated by maps $u_\sigma, \, v_\sigma$. If they belong to the same chart, then they are separated by $w_{\sigma i}$. All maps are $C^{\infty}$-smooth and, by the latter considerations, are injective. From the form of $w_{\sigma i}$ one can see that the differential is injective, so it is an immersion and as it is injective it is an imbedding. Well, injectivity of differential and global injectivity do not yet imply an embedding, but one can analyze the afforementioned maps and see that it is a homeomorphism onto its image.

Controllong the $\varepsilon_i$ we can make it short. So we've constructed short imbedding in $E^{s(n+2)}$.

Then J.Nash says that since the image of the manifold in $E^{s(n+2)}$ is $n$-dimensional set, "the classical process of generic linear projection can be applied and one can reduce the dimension of surrounding Euclidean space to $2n+1$ without introducing any singularities and self intersections".

$\textbf{Question:}$ What is this "classical process of generic linear projection?"

p.s. I found the latter construction for the case of open manifolds very beautiful. It is sad to miss the last point of such beauty!


1 Answer 1


It seems to be a general position argument.

You need to find a $2{\cdot}n+1$-dimensional subspace of $\mathbb{E}^{s\cdot(n+2)}$ such that (1) it contains no vector orthogonal to your submanifold and (2) no two points of your submanifold are projected to the same point.

All you need is to calculate dimensions correctly and apply Sard's theorem.

  • 1
    $\begingroup$ Yes, you are right. One of my friends told me that it is just how the weak Whitney theorem is proved -- look for "random projections". $\endgroup$ Nov 9, 2017 at 8:56

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