- Fact: the only parallelizable spheres are 
$$
S^0, S^1, S^3, S^7
$$ 
corresponding to $\RR, \CC, \mathbb{H}, \mathbb{O}$.

:::{.proposition title="?"}
$S^1$ is parallelizable.
:::

:::{.proof title="?"}
Let $M = S^1$, then the tangent bundle yields a rank 1 vector bundle
$$
\RR \to TM \to M
$$
We can find a nonzero global section: the vector field 
$$
s(x, y) := -y\dx + x\dy \quad\in \Gamma( T\RR^2 )
$$ 
restricted to $S^1$:

![](attachments/Pasted%20image%2020210510004301.png)

:::

:::{.proposition title="?"}
$S^2$ is not parallelizable.
:::

:::{.proof title="?"}
Let $M = S^2$, which is associated to the rank 2 vector bundle
$$\RR^2 \to TM \to M$$

Then $TM$ is trivial iff there are 2 independent global sections. Since there is a zero section, a second independent section must be everywhere-nonzero - however, this would be a nowhere vanishing vector field on $S^2$, which by the Hairy Ball theorem does not exist.
:::

:::{.proof title="Alternative"}
Any such a vector field would allow a homotopy between the identity and the antipodal map on $S^2$. 
But this is a contradiction: the identity map $\id_{S^n}$ has degree 1 but the degree of the antipodal map $S^n\to S^n$ has degree $(-1)^{n+1}$, since it is the composition of $n+1$ reflections, all of which have degree $(-1)$, and the degree is a group morphism $([S^n, S^n], \circ) \to (\ZZ, \cdot)$.
:::