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Given $k \leq n$, is there a general method that can help us determine whether a given $k-$ dimensional smooth manifold $M$ (let's assume $M$ has empty boundary) can be embedded in $\mathbb{R}^n$?

I know of some ways to know this in a couple of (very particular) cases:

$1)$ If $k = n$ and $M$ is compact, it's never possible (we would have $f(M) = \mathbb{R}^n $, since the embedding must be open in this case).

$2) $If $k = n-1$, then in order for this to be possible $M$ must be orientable (see the answer by Georges Elencwajg to this question).

But of course these are very limited (for example, using them I can't even refute the statement 'every $k-$dimensional manifold embeds into $\mathbb{R}^{k+2}$').

Are there other techniques that can help in more general cases? I'd appreciate any partial answers. Thanks in advance.

EDIT: I also know about Whitney's embedding theorem. The motivation for posting this question came from thinking about why a sharper bound can't be achieved.

Mauro
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    You can use characteristic classes for having some restrictions on the dimension, it is done in the first chapter of the book by Milnor and Stasheff. –  May 23 '17 at 22:21
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    One thing you'd like to look for which is necessary but not sufficient is that the tangent bundle embeds into a trivial rank $n$ bundle. This can be translated to some condition on characteristic classes i believe. – Saal Hardali May 23 '17 at 23:03
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    Regarding your edit/comment on why a sharper bound can't in general be acheived: there are counterexamples of $n$-dimensional manifolds that do not embed into $\mathbb{R}^m$ for any $m < 2n$, i.e. the real projective space of dimension $2^n$... but you might already know that given your interest in this question! – Yousuf Soliman May 23 '17 at 23:38
  • @YousufSoliman Exactly! I wanted to find out what makes the existence of counterexamples (such as the one you mention) possible in general. – Mauro May 23 '17 at 23:55
  • @N.H. Thanks! That sounds like a good example of what I was looking for. I'll check it out. – Mauro May 23 '17 at 23:57
  • Also thanks to @SaalHardali (the site won't let me mention two users in the same comment...). – Mauro May 23 '17 at 23:58
  • The "$k = n-1$" case needs closedness or compactness. (The answer you cite has closedness.) I'm pretty sure I've held a Moebius strip embedded in what amounts to a physical approximation of $\Bbb{R}^3$. – Eric Towers Apr 24 '21 at 22:21
  • @EricTowers indeed. – Mauro Apr 27 '21 at 01:49

2 Answers2

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Here is more details. Let $E \to X$ a vector bundle where $X$ is a manifold. The total Steifel-Whitney class of $E$ is an element $w(E) = 1 + w_1(E) + \dots \in H^*(X, \Bbb Z/2 \Bbb Z)$ where $w_i(E) \in H^i(X)$ such that :

  • If $E$ has rank $r$ then $w_i(E) = 0$ for $i > r$.
  • If $0 \to E \to F \to G \to 0$ is an exact sequence of vector bundles, then $w(F) = w(E)w(G)$ (the multiplication is the cup-product).
  • Other properties such as naturality, functoriality, etc ...

Naturality (i.e $f^*w_i(E) = w_i(f^*(E)$) implies that the trivial bundle $V \times X$ has $w(V \times X) = 1$ as $V \times X = f^*V$ where $f : X \to pt$ is the constant map.

This observation and the fact that for a manifold $M \subset \Bbb R^n$ one has $TM \oplus NM \cong \Bbb R^n$ gives $w(M)w(NM) = 1$ by the second axiom, where $w(M) := w(TM)$ and $NM$ is the normal bundle of $M$. If $r$ is the last index where $w_r(TM) \neq 0$ and $s$ the last index where $w_s(NM) \neq 0$ this implies by the rank axiom that $n \geq r + s$.

As an example, one can use Euler exact sequence for compute $w(\Bbb RP^n) = (1+a)^{n+1}$ where $a$ is the generator of $H^1(\Bbb RP^n, \Bbb Z/2 \Bbb Z)$ (remember that coefficient are taken modulo $2$) for example $w(\Bbb RP^4) = (1+a)^5 =1 + a + a^4 $. Let's assume that $\Bbb RP^4$ is embedded, we have $w(\Bbb RP^4)w(N \Bbb RP^4) = 1 (\star)$.

We can solve $(\star)$ "degree by degree" : the first degree of this equation is $w_1(\Bbb RP^2) + w_1(N\Bbb RP^2) = 0$, i.e $w_1(N\Bbb RP^2) = a$. The second degree gives $w_2(N\Bbb RP^2) = w_1(N\Bbb RP^2)w_1(\Bbb RP^2) + w_2(\Bbb RP^2) = w_1(N\Bbb RP^2)w_1(\Bbb RP^2) = a^2$ and so one. This gives finally that $w(N \Bbb RP^2) = 1 + a + a^2 + a^3$, i.e that $\Bbb RP^4$ can only be embedded (in particular immersed) in $\Bbb R^7$. Indeed, for $k = 2^r$ the same conclusion holds ($\Bbb RP^{2^r}$ can only be immersed in $\Bbb RP^{2^{r+1} - 1}$ and this is the best bound by Whitney's theorem which says that any manifold of dimension $n$ can be immersed into $\Bbb R^{2n-1}$ ($n>1$ here).

Finally let me add you two references recommended by the author on this subject :

  • Smale, S., The classification of immersions of spheres in Euclidean space, Annals of Math. 69 (1959), 327-344.
  • Hirsch, M., Immersions of manifolds, Trans. Amer. Math. Soc. 93 (1959), 242-276.
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    Thanks for the long answer! It's motivated me to look further into this. I will accept it as the answer. Although there is no possible "complete" answer for the question, this is a very useful one. – Mauro Jun 05 '17 at 23:53
  • Thanks, I am glad it helped ! –  Jun 06 '17 at 07:42
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You have Whitney's Embedding Theorem which says that any compact n-dimensional manifold can be embedded into $\mathbb{R}^{2n+1}$.

ADA
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  • Yes, I should have mentioned Whitney's theorem in the original post. Edited it just now. – Mauro May 23 '17 at 23:23
  • A stronger version of Whitney's embedding theorem states that every smooth $n$-dimensional manifold embeds into $\mathbb{R}^{2n}$. Of course this doesn't hold for topological (read not smooth) manifolds -- just think of a nonplanar graph. Topological manifolds of dimension $n$ still embed into $\mathbb{R}^{2n+1}$ – Yousuf Soliman May 23 '17 at 23:29
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    @YousufSoliman: I don't understand your remark about nonplanar graphs. A nonplanar graph is not an example of a topological manifold, and all (compact) topological 1-manifolds embed in $\mathbb{R}^2$. – Lee Mosher May 24 '17 at 18:46