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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. $T(N \times M) \simeq TN \times TM$.2012-09-26
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    @RyanBudney Thanks! I guess that's one way of doing it.2012-09-26

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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"?2012-09-26
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    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 $\mathbb{R}^2$.2012-09-26
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    Thanks, @Neal, for providing so many ways of seeing it2012-09-26
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    You're quite welcome, @henryforever142012-09-27