# Relation between tangent bundle of a manifold and it’s covering manifold

Let $$M$$ be a smooth (or complex) manifold, $$N$$ — topological manifold and $$p: M \to N$$ covering map. Consider the smooth (complex) structure on $$N$$ obtained by the well-known procedure making $$p$$ into smooth (holomorphic) map.

Is that true that from triviality of the (holomorphic) tangent bundle $$TM$$ follows triviality of $$TN$$? Is there any relation between those two whatsoever?

## 2 Answers

Suppose that the dimension of $$N$$ is $$n$$. The tangent space of $$TN$$ is trivial if and only if there exists $$n$$-vector fields $$X_1,...,X_n$$ such that for each $$x\in N, X_1(x),..,X_n(x)$$ are linearly independent. Since $$p:M\rightarrow N$$ is a local diffeomorphism, there exists $$Y_1,...,Y_n$$ vector fields of $$M$$ such that for every $$y\in M$$, $$T_yp(Y_i(y))=X_i(p(y))$$, this implies that $$Y_1,...,Y_n$$ is a trivialization of the tangent space of $$M$$, so the tangent space of $$Y$$ is trivial.

• I think you're answering a different question from the one the OP asked. The OP wanted to know if parallelizability of $M$ implies that of $N$ when $p:M\to N$ is a covering map. – Jack Lee Nov 15 '19 at 23:01
• But rereading the OP, I guess you did answer the second question ("Is there any relation between those two whatsoever?"). – Jack Lee Nov 15 '19 at 23:08

Triviality of $$TM$$ does not imply that of $$TN$$. For example, let $$M=\mathbb R^2$$, and let $$\mathbb Z$$ act on $$M$$ by $$n\cdot (x,y) = (x+n,(-1)^n y).$$ Let $$N$$ be the quotient space $$M/\mathbb Z$$, and let $$p:M\to N$$ be the quotient map. Then $$p$$ is a smooth covering map, and $$N$$ is diffeomorphic to the open Möbius band, which is not even orientable, let alone parallelizable.