$TS^2$ is not trivial , seeing explicitely what might fail when we try and construct a trivialization

Question

I just did an exercise, that was around the fact that if we have an orientable $1$-dimensional vector bundle over a paracompact space then it's trivial.

It's a well know fact that $TS^2$ is not trivial do to the hairy ball theorem, and this is a $2$-dimensional vector bundle so the latter result does not apply but I was wondering what failed if I tried to construct a trivialization for this, that is , we have local trivializations of $TS^2$ and then we can glue them together using partitions of unity . What would fail for the fact that we do not have that this gives an isomorphism of vector bundles to $S^2\times \mathbb{R}^2$. I don't know if anyone has tried to do this or knows a reference where I could look something like this up but any discussion regarding what may fail at the fiber levels, since we are using partitions of unity and I guess we will need to use formulas for the transition functions trough the tangent vector represented in different bases. Any comment is appreciated. Thanks in advance.

Answer 1

The issue is that a convex combination of invertible linear maps may not be invertible. Consider what your partition of unity construction would do on a point $p$ that is in two different open sets $U$ and $V$ on which you have a trivialization. As a part of your trivialization over the open set $U$, you have chosen an isomorphism $f_U:T_pS^2\to\mathbb{R}^2$, and similarly the trivialization over $V$ gives an isomorphism $f_V:T_pS^2\to \mathbb{R}^2$. Your partition of unity would then choose nonnegative functions $\varphi_U$ and $\varphi_V$ supported on $U$ and $V$ such that $\varphi_U(p)+\varphi_V(p)=1$, and would combine the two trivializations by taking $\varphi_U(p)f_U+\varphi_V(p)f_V:T_pS^2\to\mathbb{R}^2$. However, $\varphi_U(p)f_U+\varphi_V(p)f_V$ might not be invertible! For instance, it might be the case that $\varphi_U(p)=\varphi_V(p)=1/2$ and $f_V=-f_U$, and then $\varphi_U(p)f_U+\varphi_V(p)f_V$ would be $0$.

It is worthwhile to contrast what's going on here with, for instance, the construction of a Riemannian metric on any manifold using a partition of unity. The difference there is that a Riemannian metric is positive definite, and any convex combination of positive definite bilinear forms is still positive definite. That's how you can be sure that when you combine local Riemannian metrics using a partition of unity, the result is still positive definite (and in particular, nondegenerate). Similarly, in the case of an orientable rank $1$ vector bundle, the orientation gives a notion of "positivity" such that a convex combination of two "positive" isomorphisms to $\mathbb{R}$ is still positive (and thus in particular, it is still an isomorphism).

Answer 2

The standard way of gluing up the Riemann sphere is using gluing to copies of $\mathbb{C}$ using $z^{-1}$. Hence the way of gluing up the tangent bundle is by its derivative $\frac{-1}{z^2}$. The asymptotics of these gluing functions give a reasonably intuitive reason why the vector bundle is not trivial (even why the first chern class is $2$).

Another good exercise is to pick a Riemannian metric on $S^2$ compute $\pi_{1}$ of the $3$-manifold $$U = \{(p,v) \in TS^2 : |v| = 1 \} .$$

If the bundle where trivial then one may see that we should have $U = S^2 \times S^1$, but actually computing then fundamental group of $U$ shows this cannot be the case.