In class, i learn Adam's Theorem.
Which states the tangent bundle on $S^n$ is trivial only $n=1,3,7$.
I arise some question about $n$. Why $n=1,3,7$ are so special?
The professor give simple explanation about $n=1$.
$i.e$ for $n=1$, we can introduce complex coordinate such that $S^1$ contain $\mathbb{R}^2=\mathbb{C}$.
This is a very difficult problem that thwarted algebraic topologists a long time. The reference is:
That you can find on JSTOR. Adams provides this helpful diagram:
In particular it has also to do with the existence of a divison algebra structure on $\mathbb{R}^n$, on the existence of Hopf invariant one maps...
The proof in that paper goes through secondary cohomology operations. The idea itself is very simple (once you know it), it's the details that are insanely difficult. For example it's easy to see with ordinary Steenrod squares that $n$ has to be a power of $2$ for $S^{n-1}$ to have a trivial tangent bundle, thanks to the fact that the four conclusions in this theorem are equivalent for a given $n$:
It's because $\operatorname{Sq}^n$ is decomposable in terms of $\operatorname{Sq}^i \operatorname{Sq}^j$ with $i,j < n$ when $n$ is not a power of $2$. To prove that $n$ must be one of $1,2,4,8$, you go through secondary operations and apply the same idea (try to decompose $\operatorname{Sq}^n$ with lower terms).
Note that the converse is much simpler: if $n=1,2,4,8$ it's very easy to show that $S^{n-1}$ has trivial tangent bundle. The answer comes from the real line $\mathbb{R}$, the complex plane $\mathbb{C}$, the quaternions $\mathbb{H}$ and the octonions $\mathbb{O}$; they're all division algebras, and $S^{n-1}$ is the unit sphere of these division algebras. It's then easy to show that this imply what we want.