Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
Namely,S^2×T^2 is parallelizable. – henry Jun 23 '11 at 9:29
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

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).

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.