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I recently learned that the only parallelizable spheres are $\mathbb{S}^1$, $\mathbb{S}^3$, and $\mathbb{S}^7$. This led me to wonder:

What is $T\mathbb{S}^2$? Is it diffeomorphic to a more familiar space? What about $T\mathbb{S}^n$ for $n \neq 1, 3, 7$?

EDIT (for precision): Is $T\mathbb{S}^2$ diffeomorphic to some finite product, connected sum, and/or quotient of spheres, projective spaces, euclidean spaces, and linear groups?

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3 Answers 3

up vote 7 down vote accepted

Here's how I think about it. (Ryan Budney posted his answer while I was typing this. One can think of this as a fleshing out of his answer).

First, we need to understand the unit tangent bundle $T^1 S^2$. Once we know this, we product this with $[0,\infty)$ and then quotient all points of the form $(u, 0)\in T^1S^2 \times\mathbb{R}$ somehow to get the 0 section $S^2$. (This is precisely the mapping cylinder construction Ryan mentions).

Before we can talk about the "unit tangent bundle", we must have a notion of length of vectors. So in the background, pretend like I picked a Riemannian metric so lengths make sense.

I claim $T^1 S^2$ is diffeomorphic to $SO(3)$ (the collection of 3 x 3 orthogonal matrices of determinant 1) which is diffeomorphic to $\mathbb{R}P^3$.

The map from $T^1 S^2$ to $SO(3)$ sends $(u,v)$ to the matrix with columns $u, v, u\times v$. Here, I'm thinking of a unit tangent vector $v\in T_u S^2$ as a vector in $\mathbb{R}^3$ orthogonal to the vector $u$.

The easiest way to see $SO(3)$ and $\mathbb{R}P^3$ are diffeomorphic is to note they are both quotients of $S^3 = SU(2)$ by the same quotienting map.

So, we understand $T^1 S^2$, the unit length vectors in $TS^2$.

To allow for length, we product with $[0,\infty)$. Now, the only problem is the 0 section should be an $S^2$, and it's currently an $\mathbb{R}P^3$, so some quotienting must happen.

What quotienting must happen? Well, all the unit vectors at a given point must collapse to the point. Well, there is the action of a circle on $T^1S^2$ given by rotation vectors clockwise (say) as seen from the normal vector to the sphere. This action is clearly free. Now, it's a fact that if you translate this circle action into the $SO(3)$ picture, the circle action is the Hopf action. This implies that we identify $\mathbb{R}P^3$ with $S^2$ by quotienting by the Hopf action: Two points in $\mathbb{R}P^3$ iff they are in the same Hopf orbit.

Incidentally, I just learned a few days ago that $TS^2$ is not homeomorphic to $S^2\times \mathbb{R}^2$, though I'm still not sure how to prove it ;-). (Of course, it's clear that they are not bundle isomorphic, but they could still be abstractly homeomorphic). I don't know about the other nonparallizable tangent bundles, though.

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"Of course, it's clear that they are not bundle isomorphic, but they could still be abstractly homeomorphic)" -- If they were abstractly homeomorphic, wouldn't it follow that they were isomorphic as bundles? I.e. trivializations of each bundle give local homeomorphisms, which we can combine with the abstract homeomorphisms to get a bundle isomorphism? –  Gunnar Magnusson Feb 21 '11 at 6:22
    
@Gunnar: A bundle map must send fibers to fibers and an abstract homeomorphism doesn't need to do this. To contrast this, the manifold $S^3\times S^2$ is a bundle over $S^2\times S^2$ in infinitely many different ways (i.e., by inspecting the bundles $S^1 \rightarrow S^3\times S^2\times S^2\times S^2$, one can find infinitely many Pontrjagin classes). –  Jason DeVito Feb 21 '11 at 15:42
2  
The total spaces aren't homeo/diffeomorphic since $TS^2$ has a sphere whose normal Euler class is $2$. $S^2 \times \mathbb R^2$ does not -- the idea is that $S^2 \times \mathbb R^2$ embeds in $\mathbb R^4$ -- but $2$-knots in $\mathbb R^4$ have Seifert surfaces and so trivial normal bundles. The mechanics of this argument are Serre/Thom era -- check out Ch III of Kervaire and Weber's paper on knots in Springer LNM 685 if you'd like details. –  Ryan Budney Feb 21 '11 at 18:27
    
@Ryan: Thanks for that. I'm not at all familiar with 2-knots (and only vaguely aware of regular knots). The argument given a couple of days ago involved the phrase "no proper homotopy equivalences" due to "simple homotopy type". Of course, a homeomorphism is a proper homotopy equivalence, but I'm not sure how to compute the simple homotopy type of either of these spaces (nor am I sure what the simple homotopy type is, except that it's homotopy type where you restrict the homotopies to be "nice" somehow). –  Jason DeVito Feb 22 '11 at 3:53

Your question is perhaps too vague. The tangent bundle has a definition and that's what it is -- presumably you're asking for more than this but you don't specify with any precision. It would be good to edit your question to make it more precise.

One description of $TS^2$ is you take the mapping cylinder $SO_3 \to S^2$ where this map is the orbit of a single point in $SO_3$'s natural action on $S^2$ by isometries. Technically, $TS^2$ is the above mapping cylindre after you erase the $SO_3$ boundary (to make it non-compact).

Another description of $TS^2$ would be the configuration space of two points in $S^2$. Precisely,

$C_2 S^2 = \{ (x,y) \in S^2 \times S^2 : x \neq y \}$

You can identify the two by a stereographic projection map construction. There's many more such constructions. But you really ought to say what you're looking for because the list is endless.

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OK, I've edited accordingly. –  Jesse Madnick Feb 21 '11 at 2:28

A slightly random answer, but if we concretely identify $TS^{n-1}$ as the embedded real manifold $\{ (x, v) \in \mathbb{R}^n \times \mathbb{R}^n : \| x \| = 1, \langle x, v \rangle = 0 \}$, then it is diffeomorphic to the complex affine quadric $\{ (z_1, \ldots, z_n) \in \mathbb{C}^n : z_1^2 + \cdots + z_n^2 = 1 \}$. (This was an amusing homework exercise I did yesterday.)

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