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The product of sphere and torus is parallelizable. How to prove this? Help solve this question, this is qualifying exams of Hopkins university.

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Namely,S^2×T^2 is parallelizable. –  henry Jun 23 '11 at 9:29
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don't add information important to answering the question in comments: edit the actual question and include it there. –  Mariano Suárez-Alvarez Jun 23 '11 at 14:47

2 Answers 2

Let $\tau$ be the tangent bundle of $S^2$. Observe that $\tau\oplus 1\cong 3$ (indeed, if we add to $\tau$ the normal bundle (for the standard embedding in $\mathbb R^3$) -- which is trivial -- we get a trivial vector bundle; $n$ denotes trivial $n$-dimensional bundle) and tangent bundle of $\mathbb T^2$ is trivial. Now let $\pi_1\colon S^2\times\mathbb T^2\to S^2$ and $\pi_2\colon S^2\times\mathbb T^2\to\mathbb T^2$ be natural projections; then $$T_{S^2\times \mathbb T^2}=\pi_1^*T_{S^2}\oplus\pi_2^*T_{\mathbb T^2}=\pi_1^*\tau\oplus\pi_2^*2=\pi_1^*\tau\oplus 2=\pi_1^*(\tau\oplus1\oplus1)=\pi_1^*4=4.$$


Can't resist adding a bonus: a short and elementary proof that a product of spheres is parallelizable if one of them is odd (E.B.Staples, 1966).

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That's very nice! –  t.b. Jun 23 '11 at 18:29
    
I had typed up a similar answer but couldn't figure out the (now obvious) $\pi_1^*(\tau)\oplus 2 = \pi_1^*(\tau\oplus 1\oplus 1)$ step. Thanks for writing this up! –  Jason DeVito Jun 23 '11 at 22:03
    
Now you use natural numbers to denote stuff... Doug's going to have a feast with this! :) –  Mariano Suárez-Alvarez Jun 23 '11 at 22:39
    
@Mariano Well, everybody uses natural numbers to denote elements of any (unital) ring, so why a rig $\operatorname{Bun}(X)$ should be any different ;-?.. –  Grigory M Jun 24 '11 at 8:40
    
I know, I know... I was joking! –  Mariano Suárez-Alvarez Jun 24 '11 at 12:26

It is sufficient to prove that $S^2\times T^1$ is parallelizable. This will imply the parallelizability of $S^2\times T^2$ being the product of two parallelizable manifolds.

The general problem of parallelizability of products of spheres was considered in Maurizio Parton's thesis:

  1. As mentioned in the thesis introduction: the parallelizability of $S^2\times T^1$ is a special case of a theorem of M. Kervaire.

  2. In the thesis, a new parallelizability proof was given for the more general case of $S^n\times T^1$ by explicit construction (Proposition 2.1.2).

The main idea is as follows:

Let $x = (x_i)$ be the Eucledian coordinates of $R^{n+1}$ and the sphere $S^n$ be given by:

$ |x|^2 = \sum_i x_i^2 = 1$

$S^n\times T^1$ is diffeomorphic to the quotient manifold $(R^{n=1}-0)/\Gamma$, where the group $\Gamma$ is generated by the map $x \mapsto e^{2\pi} x$. Then the projection:

$ R^{n=1}-0 \rightarrow S^n\times T^1$

$x \mapsto (x/|x|, \log|x| \mod 2\pi)$

is $\Gamma$ equivariant, thus defines a parallelization.

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In other words, the universal covering space of $S^n\times S^1$ is $S^n\times\mathbb R$, and the group of covering transformations, which is $\mathbb Z$, acts on it by translations in the second factor. In particular, it preserves the vector field of unit vectors tangent to the $\mathbb R$ direction, which therefore passes down to the quotient. –  Mariano Suárez-Alvarez Jun 23 '11 at 14:30
    
Dont' know who M. Paton is, but the general problem of parallelizability of products of spheres was solved completely by Kervaire some 50 years ago. –  Grigory M Jun 23 '11 at 18:39
    
@Grigory: I think that the point is that Parton constructs explicitly the parallelizations. –  Mariano Suárez-Alvarez Jun 23 '11 at 19:37

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