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All the examples I know of manifolds which are parallelizable are Lie Groups. Can anyone point out an easy example of a parallelizable smooth manifold which is not a Lie Group? Are there conditions on a parallelizable smooth manifold which forces it to be a lie group?

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    $\begingroup$ In the opposite direction, there's a result due to Wolf that $S^7$ is the only exception in the simply-connected case: specifically, for a $X$ compact and simply-connected manifold, $X$ is parallelizable only if it's the product of Lie groups and copies of $S^7$. $\endgroup$
    – anomaly
    Dec 31, 2018 at 16:13
  • $\begingroup$ @anomaly Thanks a lot. I was looking for some result like that. I am not experienced to find this results. Can you please mention where can I find this or may be share a link if possible. $\endgroup$
    – Partha
    Dec 31, 2018 at 16:16
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    $\begingroup$ I don't have a reference to the original paper, but it's mentioned after Proposition 5.15 in Lee's "Introduction to Smooth Manifolds." $\endgroup$
    – anomaly
    Dec 31, 2018 at 17:22
  • $\begingroup$ Noted with thanks. I will definitely look into it. $\endgroup$
    – Partha
    Dec 31, 2018 at 17:23

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The classical example is $S^7$. This comes from the octonions, which are non-associative, so the unit octonions don't form a Lie group.

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  • $\begingroup$ Thanks, I was wondering that $S^7$ still has a group structure which is non-associative, are there examples which can not be given a group structure at all. $\endgroup$
    – Partha
    Dec 19, 2018 at 4:50
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    $\begingroup$ Group operations are by definition associative. The restriction of octonionic multiplication to the unit sphere in the octonions still, however, forms a loop (an operation $S \times S \to S$ with identity and inverses), and the multiplication and inversion maps are smooth with respect to the usual smooth structure. $\endgroup$
    – Travis Willse
    Dec 19, 2018 at 4:55
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    $\begingroup$ Also, this is the only example of parallelizable sphere that admits no Lie group structure. $\endgroup$
    – Travis Willse
    Dec 19, 2018 at 4:56
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Any closed, orientable $3$-manifold $X$ is parallelizable. On the other hand, Lie groups have $\pi_2$ trivial, and there are a lot of such $X$ that don't. (It's not trivial to find them, though. Note that any such $X$ must have $\pi_1 X\not = 0$ by Poincare duality, and aspherical (e.g., hyperbolic) $X$ have $\pi_2 X = 0$ as well. $3$-manifolds are weird.)

(The easiest way to prove the first assertion is to note that the only obstructions to the triviality of $TX$ are the Stiefel-Whitney classes; and $w_1 = 0$ by orientability and $w_2 = 0$ by the Wu formula, since we're conveniently in low dimension. The second is practically a folklore theorem, but it's not that hard to prove directly either via Morse theory or minimal models.)

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    $\begingroup$ You beat me to this answer! Also, we can readily construct examples of oriented, compact $3$-manifolds that do not admit a Lie group structure: Any Lie group $G$ has abelian fundamental group $\pi_1(G)$, but the product of $S^1$ and an orientable, compact surface $\Sigma$ with genus $> 1$ has nonabelian fundamental group $\pi_1(S^1 \times \Sigma) \cong \pi_1(S^1) \times \pi_1(\Sigma) \cong \Bbb Z \times \pi_1(\Sigma)$. (And +1, by the way.) $\endgroup$
    – Travis Willse
    Dec 19, 2018 at 5:09
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    $\begingroup$ Good point! I ignored the hyperbolic case because of $\pi_2$, but $\pi_1$ also presents an obstruction to Lie-groupness. $\endgroup$
    – anomaly
    Dec 19, 2018 at 5:25

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