21

I am trying to show that the tangent bundle of $S^2$ not diffeomorphic to $S^2\times \mathbb{R}^2$. This is from an exam, where there is a hint stating that this is more than showing that $TS^2$ is non-trivial.

I know how to show the hairy ball theorem, according to which $TS^n$ is non-trivial iff n is even.

I also know that a vector bundle $\pi:E\rightarrow M$ of rank $m$ on a smooth manifold $M$ is trivial (by definition) iff there exists a diffeomorphism $f:E\rightarrow M\times \mathbb{R}^m$ such that for every $p\in M$, $f$ induces a vector space isomorphism $f:\pi^{-1}(p)\rightarrow \{p\}\times \mathbb{R}^m$.

So I see that showing that $TS^2$ is non-trivial only guarantees that $TS^2$ is not diffeomorphic to $S^2\times \mathbb{R}^2$ via a diffeomorphism satisfying the property above, but it is not enough to conclude that there isn't any diffeomorphism.

How can I show this then?

user54631
  • 737
  • 4
  • 16
  • 5
    What you state as the hairy ball theorem is not true! In fact, $TS^n$ is trivial iff $n=0, 1, 3$, or $7$. What you mean to say is that $S^n$ has a nonvanishing vector field iff $n$ is odd. –  Aug 11 '14 at 00:12
  • 2
    Relevant link: http://mathoverflow.net/questions/92337/total-spaces-of-ts2-and-s2-times-r2-not-homeomorphic – PVAL-inactive Aug 11 '14 at 00:40
  • 3
    Euler characteristic computes the self-intersection number of the image of the zero section which is a generator of 2nd homology group. The self-intersection number is a topological invariant. – Moishe Kohan Aug 11 '14 at 02:44

3 Answers3

2

Here's a proof that relies on a theorem of Kervaire with a somewhat-involved proof (I think the particular case of $n=k=2$ is probably much easier, but I couldn't find such a proof to confirm my suspicions.)

Theorem: The normal bundle to an $n$-sphere embedded inside $\Bbb R^{n+k}$ is trivial if $k> \frac{n+1}2$

This implies any 2-sphere embedded in $S^2 \times \Bbb R^2$ has a tubular neighborhood isomorphic (as fiber bundles) to $ S^2 \times \Bbb R^2$, since $S^2 \times \Bbb R^2$ embeds as an open manifold of $\Bbb R^4$. If $f:TS^2\rightarrow S^2 \times \Bbb R^2$ is a diffeomorphism, then the image of the zero section has a trivial tubular neighborhood $N$ by the above disscussion, but then $f^{-1}(N)$ is a trivial tubular neighborhood of the zero section in $TS^2$.

References:

  • Massey's On the Normal Bundle of a Sphere Imbedded in Euclidean Space
  • Kervaire's An Interpretation of G. Whitehead's Generalization of H. Hopf's Invariant
PVAL-inactive
  • 8,799
-2

I think you can try this.

$S^2\times\mathbb{R}^2$ is trivial, while $TS^2$ is not trivial because only the spheres of odd dimension have non-trivial tangent bundle (Hopf's theorem). So they can't be diffeomorphic.

Andrea
  • 323
  • This is exactly what mike said in the comment. – Anubhav Mukherjee Nov 30 '17 at 19:14
  • 1
    This doesn't necessarily follow. Two different bundle structures can have diffeomorphic total spaces. – Jason DeVito - on hiatus Mar 02 '21 at 16:43
  • @JasonDeVito Could you give an example of such a situation? – John Mars Mar 10 '23 at 13:13
  • @JohnMars: Sure. Consider $S^1$ bundles over $S^2\times S^2$. They are classified by their Euler class, an element $e\in H^2(S^2\times S^2)\cong \mathbb{Z}\oplus \mathbb{Z}$. If we select $(a,b)\in \mathbb{Z}^2$ with $\gcd(a,b) = 1$, then the total space $E_{a,b}$ is a simply connected $5$-manifold with $H^2(E)\cong \mathbb{Z}$. From Barden-Smale, there are only two diffeomorphism types of such $5$-manifolds (determined by their second Stiefel-Whitney class). So, countably many bundles fall into at most $2$ total space types. – Jason DeVito - on hiatus Mar 10 '23 at 13:59
  • 1
    @JohnMars: The is also a classification of $S^3$-bundles over $S^4$ by Crowley and Escher https://arxiv.org/abs/math/0004147. It turns out that $S^7$ appears as the total space for multiple distinct bundles. I didn't go through their congruences in detail, but I think it actually appears infinitely many times. – Jason DeVito - on hiatus Mar 10 '23 at 14:14
-4

I think that if suppose that exist a diffeomorphism $F$ then there is an correspondence between sections in $S^{2}\times\mathbb{R}^{2}$ and vector fields on $TS^{2}$. Take the constant section $s(x)=(x,(1,0))$ and by diffeomorphism there is $X\epsilon\mathfrak{X}(S^{2})$ such that $dF(s)=X$, but this section vanishing in south pole under stereographic proyection, that means $dF$ isnt isomorphism

Donyarley
  • 65