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Intuitively, it seems the only space covered by $S^n$ that admits immersion into $\mathbb{R}^{n+1}$ is itself. A necessary condition is that the product of the tangent bundle with $\mathbb{R}$ be parallelizable. Is there a rigorous proof of the general case? (or a counter-example) .

Moishe Kohan
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user3257842
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    I mean, there's a quotient map from $S^n$ to the product $[0,1]^{n-k}$ for any non-negative $k$, so do you have more restrictions other than just 'quotient map'? – Dan Rust Aug 28 '17 at 14:26
  • I meant spaces that have $S^n$ as universal covering space. (eg. real projective space) . Maybe the question was not properly formulated. – user3257842 Aug 28 '17 at 14:34
  • Yeah, that's a pretty different question. – Randall Aug 28 '17 at 14:37
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    Are there that many spaces up to homeomorphism whose cover is the $n$-sphere? Presumably one can't have a space with an infinite fundamental group covered by $S^n$ because it's compact, so you're left with classifying finite free and properly discontinuous group actions on spheres, for which I doubt there are many classes, and which have probably been well studied. – Dan Rust Aug 28 '17 at 14:44
  • In fact, for $n$ even there is precisely one non-trivial such action on the sphere up to homotopy (homotopic to the antipodal map) whose quotient is real projective space. For odd $n$ it might be more interesting. – Dan Rust Aug 28 '17 at 14:57
  • The lens spaces have been studied in "Einlagerung von Mannigfaltigkeiten in euklidische Raume" where it's been proven that no $3$-dimensional lens space embeds into $\mathbb{R}^{4}$ . Or so I've heard. It's hard to follow since my German is lacking. It's still not the general case. Its existence would correspond to an immersion of the sphere 'folded-in' upon itself, as each point on the immersed quotient space corresponds to several on the sphere. Which is why I have such a hard time believing it possible at all. – user3257842 Aug 28 '17 at 15:59
  • @DanRust Look up the classification of "spherical space forms" to find all the examples where the covering is a local isometry. There are a few of these. – PVAL-inactive Aug 30 '17 at 19:19

1 Answers1

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Every oriented 3-dimensional manifold $M^3$ admits a (smooth) immersion in $E^4$. This is a corollary of two things:

  1. $TM^3$ is trivial, see e.g. here. Hence, $T(M\times (0,1))$ is also trivial.

  2. Hirsch-Smale theory which shows that a smooth parallelizable $n$-dimensional manifold admits an immersion in $E^n$. Apply this to $M^3\times (0,1)$.

Now, take any manifold (necessarily orientable!) which is covered by the 3-sphere, say a lens space or the Poincare homology sphere.

Moishe Kohan
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  • That doesn't seem to be correct. Specifically, I believe there are more conditions required for the Hirsch-Smale theorem . If it were true, you could apply to $S^3$ as a smooth, parallelize, $3$-dimensional manifold to say it admits immersion in $E^3$ . This is obviously false. – user3257842 Aug 31 '17 at 07:49
  • @user3257842: No, it is correct as written. This particular application is about immersions of open manifolds. A good reference (which I find more readable than the original proof) is "Submersions of open manifolds" by Anthony Phillips, http://www.sciencedirect.com/science/article/pii/0040938367900341 – Moishe Kohan Sep 01 '17 at 05:00
  • I'm dumb. I've looked up the definition of an immersion to find that it allows for self-intersections. I've proved that the real projective space $P^3$ (which can be obtained by identifying the antipodal points of the $3$-sphere) admits no embedding in $R^4$, however, that's not the same as immersion. The word I was looking for was embedding. Thanks for answering the question anyway. – user3257842 Sep 05 '17 at 15:33