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?
Example of a parallelizable smooth manifold which is not a Lie Group
differential-geometrylie-groupssmooth-manifoldstangent-bundle
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(Disclaimer. The following describes some cohomological conditions that formally answer the question. But they are 1) usually too hard to compute; 2) don't really explain what is the class of parallelizable manifolds geometrically.)
The problem of existence of a section is answered (well, in a sense) by obstruction theory. Namely, there is the first obstruction $o_1\in H^1(X,\pi_0(F))$ and there is a section on $sk_1(X)$ iff $o_1=0$; if $o_1=0$ each section on $sk_1(X)$ defines an obstruction $o_2\in H^2(X,\pi_1(F))$ and so on (and if all obstructions are trivial, the bundle has a section).
(Well, actually one should be careful with $H^1(X;\pi_0(F))$: in general, $\pi_0(F)$ is not a group, so this $H^1$ just doesn't make sense, and the story starts a step (or two, if $\pi_1(F)$ is not abelian) later. But in the cases we're interested in, $o_1$ is well-defined.)
In the case of frame bundle of $n$-dimensional vector bundle, the fiber is $O(n)$, so obstructions lie in groups $H^i(M,\pi_{i-1}O(n))$. In the stable range homotopy groups of orthogonal groups are given by Bott periodicity. If we're talking about tangent bundle, we care only about $\pi_{i-1}(O(n))$ for $i\leq n$ and for $i<n$ these groups lie in the stable range.
A (toy) example: for $S^3$ only nontrivial (reduced) cohomology group is $H^3(S^3)$; but $\pi_2 O(n)=0$ — so any vector bundle on $S^3$ is trivial (well, not the simplest proof of the fact, but still).
In case of vector bundles these obstructions can be also described more geometrically (in the spirit of characteristic classes theory).
- First obstruction $o_1\in H^1(M;\pi_0 O(n))=H^1(M;\mathbb Z/2\mathbb Z)$ is nothing else but $w_1$, the first Stiefel-Whitney class. It gives the obstruction to orientability — i.e. to reducing the structure group of the bundle from $O(n)$ to $SO(n)$.
- If the bundle is oriented, second obstruction $o_2\in H^2(M;\pi_1 O(n))=H^1(M;\mathbb Z/2\mathbb Z)$ is defined. It coincides with $w_2$ and gives the obstruction to the existence of a spin struction — i.e. to lifting structure group of the bundle from $SO(n)$ to its universal cover, $Spin(n)$.
- Next obstruction is defined for a spin bundle; $\pi_2O(n)=0$, so first non-trivial obstruction here is $o_4\in H^4(M;\pi_3 O(n))=H^4(M;\mathbb Z)$. In fact, it coincides with $\frac12p_1$ (where $p_1$ is the first Pontryagin class of oriented bundle). And it is the obstruction to lifting the structure group from $Spin(n)$ to (infinite-dimensional) topological group $String(n)$.
...And so on: the sequence of obstructions corresponds to the Postnikov tower $$ O(n)\gets SO(n)\gets Spin(n)\gets String(n)\gets FiveBrane(n)\gets... $$ (this is a kind of duality: one can think either about sequence of extensions of the section through the filtration of $M$ by skeleta, or about sequence of lifts through the Postnikov tower of $O(n)$).
Some references. Obstruction theory in general is discussed in the section 4.3 of Hatcher — but Hatcher uses the Postnikov-towers-approach (like in the second part of the answer), AFAIR. And more classical approach + obstruction-theoretic POV on characteristic classes is explained e.g. in section 12 of Milnor-Stasheff, I believe.
One more remark. As it is explained in the other answer, ordinary ("primary") characteristic classes can't answer the question, since they coincide for stably equivalent vector bundles. What obstruction theory gives is, in a sense, a theory of higher characteristic classes: secondary class (a priori) defined only if primary one is zero and so on.
$S^2$ is not parallelizable. This fact is famously known as the Hairy ball theorem.
On any coordinate patch $U$ with coordinates $(x^1, x^2, \dots, x^n)$, the vector fields $V_i = \frac{\partial}{\partial x^i}$ are linearly independent. But when you cover a manifold with multiple coordinate patches, there may not be a way to extend those vector fields to the entire manifold and keep them linearly independent (or nonzero).
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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.