$T^2\times S^n$ is parallelizable

Question

This is taken from a UCLA Geometry/Topology qualifying exam.

How would one prove that $T^2\times S^n$ is parallelizable for all $n\geq 1$? Is there a way to find $n+2$ linearly independent vector fields? I am trying to think of the simplest case $n=2$ where $S^2$ is not parallelizable, but $T^2\times S^2$ has to be in some way. I would appreciate a general strategy to treat such problems.

This answer should work almost verbatim, since $T^2 \cong S^1 \times S^1$, and the trick of summing normal and tangent bundles is the same. You've just now got one more trivial line bundle floating around than you need. — , Aug 26 '14 at 04:15
See also http://math.stackexchange.com/questions/665779/what-are-necessary-and-sufficient-conditions-for-the-product-of-spheres-to-be-pa/847263#847263, for products of more spheres. — Jason DeVito - on hiatus, Aug 26 '14 at 17:31

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

There is a similar post a few days ago asking why $$ T(\mathbb{S}^{2}\times \mathbb{S}^{1}) $$ is trivial. The idea is to use the extra dimension coming from $T(\mathbb{S}^{1})$ to fill in the place of the normal bundle. For your question I think it is similar but more subtle, as you have to consider carefully how the pull back bundle splits and how to place the tangent vectors. I am not sure if a long exact sequence+Whitehead's thoerem type of argument would be useful, but you can also give a try.

score 2 · Accepted Answer · answered Aug 26 '14 at 04:28

2

I just want to point out that this is elementary. Consider $S^n \times [1,2] \subseteq \mathbb{R}^{n+1}$ let the vector fields be defined by $\mathbf{e}_i$ the constant fields in the coordinate directions. Now identify the points $(x,1)$ and $(x,2)$ this gives $S^n \times S^1$ the vector fields are equal at the identified points so this gives $n+1$ fields on $S^n \times S^1$

answered Aug 26 '14 at 04:28

Rene Schipperus

39,526

Hopefully dumb question: how do you embed that into $n+1$ dimensions? $S^n$ embeds into nothing smaller, so how can you then tack on an extra dimension (the interval) without embedding into a higher dimension? – zibadawa timmy Aug 26 '14 at 06:46
Yeah so its just like a thick sphere. – Rene Schipperus Aug 26 '14 at 06:56
Would that be called an n+1-dimensional annulus? – zibadawa timmy Aug 26 '14 at 07:00
Yes thats accurate, its all points of radius between $1$ and $2$. – Rene Schipperus Aug 26 '14 at 07:04

score -2 · Answer 3 · answered Aug 26 '14 at 05:22

I may elaborate the reason. Since $\mathbb{S}^1$ has trivial tangent bundle. Denote $\xi$ to be one-dimensional trivial bundle, and thus $$T_{\mathbb{T}^2}=T_{\mathbb{S}^2}=\pi_1^* T_{S^1}\oplus\pi_2^* T_{S^1}=\pi_1^*\xi\oplus\pi_2^*\xi=\xi^2.$$ Also use the same formula, note that the tangent bundle $\tau_n$ os $\mathbb{S}^n$ we have $\tau_n\oplus \xi =\xi^{n+1}$. Thus $$T_{\mathbb{S}^n\times\mathbb{T}^2}=\pi_1^*T_{\mathbb{S}^n}\oplus\pi_2^*\xi\oplus\pi_3^*\xi=\pi_1^*(\tau_n\oplus\xi)\oplus\xi=\pi_1^* \xi^{n+1}\oplus\xi=\xi^{n+2}.$$

$T^2\times S^n$ is parallelizable

3 Answers3