Finite coverings and Tangent bundles

Question

Motivation: Apparently the tangent bundle of real projective space is trivial if and only if the tangent bundle of it's universal cover (the sphere) is trivial. That is, $ n=1,3,7 $. Does that follow from a general fact trivial tangent bundle of universal cover implies triviality if the covering is finite?

Let $ M $ be a manifold and $ M' $ it's universal cover. Then what is the relationship between the tangent bundle $ T(M) $ and the tangent bundle $ T(M') $? First it seems that $ T(M') $ should be the universal cover of $ T(M) $. Indeed the tangent bundle is homotopy equivalent to the base space so $ T(M') $ must be simply connected.

What else can we say? Is there any relationship between the characteristic classes of $ T(M) $ and $ T(M') $? And how about triviality? Every flat or hyperbolic manifold has a contractible universal cover and thus $ T(M')$ is the trivial bundle. Yet many flat and hyperbolic manifolds are not even orientable let alone parallellizable.

Answer 1

Let $\pi : N \to M$ be a smooth covering map. Then $\pi$ is a local diffeomorphism so it follows that $\pi^*TM \cong TN$. The characteristic classes of $N$ are therefore the pullbacks of the corresponding classes of $M$ by $\pi$. Another corollary is that if $TM$ is trivial, then so is $TN$, but the converse is not true as you've observed. Even if $N = M'$ and the covering is finite, $TM$ may not be trivial, e.g. $S^2\times S^5$ has trivial tangent bundle, but $S^2\times\mathbb{RP}^5$ does not since $w_2(S^2\times\mathbb{RP}^5) \neq 0$.

Regarding triviality of $T\mathbb{RP}^n$, from the above we know that if $T\mathbb{RP}^n$ is trivial, then so is $TS^n$ and hence $n \in \{1, 3, 7\}$. However, it is not clear if all three values of $n$ yield a trivial bundle. This turns out to be true. One way to see this is to give a basis of vector fields $\{V_i\}_{i=1}^n$ on $S^n$ for which $V_i(-x) = -V_i(x)$, so they descend to a basis of vector fields on $\mathbb{RP}^n$; see this answer for details.