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Problem: Let $M \subset \mathbb{R}^{n+1}$ be a compact submanifold ($\dim M=k$) and $n \geq 2k + 1$.

Show that, for the projection $\pi : \mathbb{R}^{n+1} \longrightarrow H^n$ onto a suitable hyperplane of $\mathbb{R}^{n+1}$, the restriction $\pi|M : M \longrightarrow H$ is a smooth embedding.

I tried to use a corollary from Whitney's Embedding Theorem that says that if $M$ ($\dim M = k$) is a compact manifold with or without boundary and $n \geq 2k + 1$, then every smooth map from $M$ to $\mathbb{R}^n$ can be uniformly approximated by embeddings.

The problem is I don't actually see how this could work here, so any hints and maybe some intuition on what is happening would be really appreciated.

Thanks.

Ken
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1 Answers1

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A series of hints, which should lead you to the solution

1: Try to show that there exists such a hyperplane, by showing that the set of all such hyperplanes has full measure.

2: Note that a hyperplane corresponds to a line and vice versa. We have found a bijection by taking the orthogonal complement!

3: The projection injects iff the corresponding line based at any point hits the embedded manifold at most once.

4: Lines are described by the space $RP^n$. Is this a smooth manifold? If yes which dimension?

5: Try to find all elements of $RP^n$ which correspond to a non-injective map. Try to do so as the image of a smooth map.

6: Try to calculate dimensions to use Sard to show that the image of the above map is a measure zero set.

Finally note that you might want to try to use the hints not in a particular order.

Daniel Valenzuela
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  • I found your hints very useful, thank you. Here's what I've managed to do so far: First, I want to prove that the set of all hyperplanes that satisfy that π|M:M⟶H is a smooth embedding has full measure, i.e, it’s complement has measure zero. For this I used 3 and proved that π|M is invective iff every ortogonal line to Hn intersects M in only one point. –  May 22 '15 at 20:43
  • Then I used 4: I would like to find a smooth function that maps two different points in M into a “direction” of Rn, I can’t seem to find this function; I do understand that by using Sard's theorem, since MxM has dimension 2k and RPn has dimension n (2k<n), then I would get a subset of RPn with measure zero. So it’s complement would have full measure, but it’s not clear how to define such function. –  May 22 '15 at 20:43
  • Also I have a doubt, sorry if it’s obvious. If I finished the proof, wouldn’t I be proving only that the restriction of the projection is invective? I want it to be a smooth immersion and a topological embedding, right? –  May 22 '15 at 20:43
  • Nice! A way to define the map is $M \times M - \Delta(M) \to RP^n$ where you send a pair of points to the line they define. This is smooth as the restriction of $R^{n+1}\times R^{n+1}-\Delta(R^{n+1})$. The domain being compact, we get a proper map, hence an embedding. – Daniel Valenzuela May 23 '15 at 00:44
  • Okay, sorry if I’m repetitive, I just want to know if I understood correctly. This function F maps two different points in M to the class of equivalence of the line that passes through them. F is smooth, and this implies that the image of F has measure zero in RPn. The image of F is exactly the set of of lines that intersect M in more than one point, and each line is in correspondence with a hyperplane, so the set of hyperplanes in which the restriction of the projection is not injective has measure zero. –  May 23 '15 at 19:06
  • We have then proved that the restriction of the projection is injective for almost every hyperplane of Rn+1. I still don’t see how this implies that the differential is also injective for a full measure set of hyperplanes.

    If so, then I could use that since the restriction is injective and it’s differential is injective for al full measure set of hyperplanes, then it’s going to be an injective immersion in the intersection of such hyperplanes, and because M is compact, then the restriction is an embedding.

     –  May 23 '15 at 19:06
  • Everything you said is perfectly right. I have to admit that there remains one step missing, as you said we have to show that we can choose the projection to be an immersion. We can do this similarly as above. Just show that the set of lines where we have an immersion (i.e. lines which are not in any tangent space) has full measure and use that full measure sets are closed under intersections. – Daniel Valenzuela May 23 '15 at 19:27
  • How do you know that the differential will be injective for the set of lines that aren't in any tangent space? –  May 24 '15 at 02:58
  • Another question: intuitively, are we proving that we didn't need n+1 dimensions? I mean, that M with these properties actually "fitted" in Rn? –  May 24 '15 at 03:00