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I was reading these notes and in the fifth page it is said that:

Given a (smooth) manifold of dimension $n$, $M$, $M$ is parallelizable iff $M$ is the product of a Lie group and some number of copies of $\mathbb{S}^7$.

If $M$ is the product of a Lie group and some number of copies of $\mathbb{S}^7$, then $M$ is parallelizable since it is the product of parallelizable manifolds. However, I don't see the "only if" part of the statement. So any help would be appreciated, both a reference or an explicit argument or idea on how to prove that.

Just for the sake of completeness, I say that the manifold $M$ is parallelizale if $TM \cong M \times R^n$, or, equivalently, if it admits $n$ linearly independent vector fields.

Just for future reference, the statement is false. See, for example this question.

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    $\begingroup$ That seems quite false to me. Any 3-manifold is parallelizable. $S^n \times \Bbb R$ is parallelizable. etc... $\endgroup$ Apr 26, 2017 at 19:46
  • $\begingroup$ Then we agree because I was looking for some comment in that lines. $\endgroup$
    – D1811994
    Apr 26, 2017 at 19:55
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    $\begingroup$ I've added a notice on my website about the error: math.uchicago.edu/~chonoles/expository-notes/… $\endgroup$ Apr 30, 2017 at 0:41
  • $\begingroup$ Great! By the way, congratulations @ZevChonoles for the amazing material on your website! $\endgroup$
    – D1811994
    Apr 30, 2017 at 13:06

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I would like to give a simple counterexample. Notice that if $M$ is the product of a Lie group and some number of copies of $\mathbb{S}^7$ then $\pi^1(M)$ is abelian, because $\pi^1$ of Lie groups are always abelian.

Now take $M = \mathbb R^2 \setminus \{(0,0),(0,1)\}$, a plane with two holes. It is clear that $M$ is parallelizable and $\pi^1(M)=Z \ast Z$, as consequence of van Kampen theorem, which is not abelian. So, the result must be false.

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