The product of sphere and torus is parallelizable. How to prove this? Help solve this question, this is qualifying exams of Hopkins university.
The product of sphere and torus is parallelizable. How to prove this?
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0Namely,S^2×T^2 is parallelizable. – 2011-06-23
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2don't add information important to answering the question in comments: edit the actual question and include it there. – 2011-06-23
2 Answers
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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1That's very nice! – 2011-06-23
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0I 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! – 2011-06-23
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0Now you use natural numbers to denote stuff... Doug's going to have a feast with this! :) – 2011-06-23
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0@Mariano Well, everybody uses natural numbers to denote elements of any (unital) ring, so why a rig $\operatorname{Bun}(X)$ should be any different ;-?.. – 2011-06-24
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0I know, I know... I was joking! – 2011-06-24
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:
As mentioned in the thesis introduction: the parallelizability of $S^2\times T^1$ is a special case of a theorem of M. Kervaire.
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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2In 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. – 2011-06-23
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0Dont' know who M. Paton is, but the general problem of parallelizability of products of spheres was solved completely by Kervaire some 50 years ago. – 2011-06-23
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0@Grigory: I think that the point is that Parton constructs *explicitly* the parallelizations. – 2011-06-23