I'm reading through Tu's Introduction to manifolds and today I learned about tangent bundles and vector bundles. I was surprised to learn that $TS^2$, tangent bundle of the 2-sphere, isn't trivial (i.e. $TS^2 \not \simeq S^2 \times \mathbb R^2$. I learned that it could be seen as a corollary to the Hairy ball theorem.

I know the question is extremely vague, but how does $TS^2$ look like then? What do we know about it's topological/differential properties? Is there some way we can visualize it?

What are the other ways to see the non-triviality of $TS^2$?

I know I'm little out of my depth here and that I probably won't understand all the answers, but I hope they could motivate me to learn more of the differential geometry. Also, it's always good to get a little taste of what's ahead before you see your first definition.

  • $\begingroup$ It's twisted. Why is the Möbius strip not $S^1\times [-1,1]$? $\endgroup$ Aug 24, 2014 at 22:26
  • $\begingroup$ One way to visualize this is by glueing two copies of $\mathbb{C}$ together to get $S^2$, which gives charts $\cong\mathbb{C}^2$ on the the tangent space, with transition map $(z,w)\mapsto(1/z,-w/z^2)$. $\endgroup$
    – yoyo
    Aug 25, 2014 at 0:00
  • $\begingroup$ see also people.ucsc.edu/~lewis/Math208/hairyball.pdf for an analytic proof of the hairy ball theorem $\endgroup$
    – yoyo
    Aug 25, 2014 at 1:08

2 Answers 2


As a corollary of the hairy ball theorem, $S^2$ is not a parallelizable manifold, that is, it does not have a set of $2$ globally non-vanishing vector fields that span its tangent space at every point. The parallelizable vector fields can be used to introduce a trivialization.

  • $\begingroup$ Smooth* vector fields is missing I believe. In fact it cannot even have a continuous nonvanishing field assigning one tangent vector to each point on the 2-sphere $\endgroup$ Apr 1 at 0:07

Maybe a nice excersise to help visualizing the tangent spaces of the spheres is the following:

$TS^n=S^n\times S^n - Δ$

where $Δ$ is the diagonal $Δ=\{(x,x)|(x,x) \in S^n\times S^n \}$.

To see that, imagine what would hapen if you tried to close up each tangent space to a sphere minus a point, using the stereographic projection.

(For a more explicit explanation as to what $TS^2$ is and what is not, see the introduction of Moore's lecture notes on Seiberg-Witten theory. Beware though that he uses quite a lot of things,like the $H^*(S^2)$, to characterise the tangent bundle)

  • $\begingroup$ To be clear, in this case $\Delta$ is the set of all "points at infinity" for each tangent space right? So the map $S^n \times S^n \to S^n \times S^n - \Delta$ is performing a stereographic projection on each tangent space. It's quite an elegant picture $\endgroup$
    – Kai
    Dec 22, 2020 at 5:09
  • $\begingroup$ I suppose this goes for $n \geq 2$, right? Since for $n = 1$ the tangent bundle is trivial, but the product minus the diagonal is homeomorphic to the Möbius strip $\endgroup$ May 11 at 9:54

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