# How do I see that the tangent bundle of torus is trivial

I've been having a hard time trying to determine if the tangent bundle of a differentiable manifold is trivial. Namely, if there exists a diffeomorphism between the tangent bundle $TM$ of a given manifold $M$ and the product manifold of $M\times \mathbb{R}^n$.

I've managed to build a diffeomorphism from $TS^1$ to $S^1\times \mathbb{R}^1$. But the case with torus $S^1\times S^1$ seems harder, since the dimension is higher.

In general, how do I show that $S^1\times \cdots \times S^1$ has trivial tangent bundle?

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A standard argument is that the tangent bundle of a product is the product of the tangent bundles. i.e. $T(N \times M) \simeq TN \times TM$. –  Ryan Budney Sep 26 '12 at 4:28
@RyanBudney Thanks! I guess that's one way of doing it. –  henryforever14 Sep 26 '12 at 4:47

## 1 Answer

Here are three ways.

1. Take one vector field in the direction of each factor to obtain a trivialization. Work by induction.
2. $\mathbb{S}^1\times\cdots\times\mathbb{S}^1$ is a Lie group. Prove that any Lie group is parallelizable. (Take a left-invariant basis of the tangent space at $(1,1)$ and move it around by the Lie group's self-action.)
3. Prove that the product of two parallelizable manifolds is again parallelizable (see 1). Corollary: Any product of circles is parallelizable.
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what do you mean by taking one vector field in the direction of each "factor"? –  henryforever14 Sep 26 '12 at 4:35
Each $\mathbb{S}^1$ is a factor in the product $(\mathbb{S}^1)^n$. On a product of smooth manifolds, a "vector field in the direction of a factor" is a vector field that is zero when projected to all but one of the factors of the product. For example, $(0,1)$ on $\mathbb{R}^2$. –  Neal Sep 26 '12 at 4:40
Thanks, @Neal, for providing so many ways of seeing it –  henryforever14 Sep 26 '12 at 4:49
You're quite welcome, @henryforever14 –  Neal Sep 27 '12 at 3:36