# Is the tangent bundle of a $S^2$ flat?

A vector bundle is said to be flat if it is endowed with an linear connection with vanishing curvature, ie. a flat connection.

Is the tangent bundle $TS^2$ of a $S^2$ flat? My question is also about how do we know if our bundle is flat or not? What are the obstacles to this?

Let we have a flat connection $\nabla$. This is equivalent to the fact that we have a parallel transport $P_{\gamma (t)}$. Fundamental group $\pi_1 (S^2) = 0$. Then parallel transport is path-independent (because connection is flat). We can take an arbitrary vector $X \in T_pS^2$ and spread it around the manifold by the parallel transport. Thus, we obtain a non-trivial nondegenerate global vector field $X(q)$, where $q \in M$. But this contradicts the Hairy ball theorem.

Is my reasoning correct?

• Do you mean the tangent bundle of the tangent bundle of a $2$-sphere? – Arnaud Mortier Jun 3 '18 at 23:46
• @Arnaud Mortier, Sorry, no, just the tangent bundle to the $S^2$ – Ann Jun 3 '18 at 23:47

Your reasoning indeed shows that No, there is no flat connection on $$TS^2$$.
The distinction is not just nit-pickiness, it does matter. For example, the tangent bundle to the torus admits a flat connection, but when most of us think of the torus as the surface of a donut sitting in $$\mathbb R^3$$, we are actually thinking of it as being equipped with a non-flat connection.
It turns out that there are some vector bundles where you simply cannot find any connection which is flat everywhere, and I think this is what your original question was probably about. Chern-Weil theory is the precise version of that statement. It says roughly that some vector bundles (like the tangent bundle to $$S^2$$) have certain topological quantities (like the Euler characteristic of $$S^2$$) that can be computed from the curvature of a (metric) connection on that bundle. When these topological quantities are nonzero (like $$\chi(S^2)=2$$) then flat connections are forbidden by the topology of the bundle. I mentioned Chern-Weil theory, but really the Chern-Gauss-Bonnet theorem is sufficient to argue that $$TS^2$$ does not admit a flat connection.