Tangent bundle of manifold with no odd dimensional sub-bundles

Question

First, a preliminary remark:

The Whitney sum of two vector bundles is orientable. I saw this statement somewhere and was wondering if it's true. In particular, it's easy to show that $w_1(\xi\oplus\xi) = 0,$ which implies orientability for reasons outlined here. However, is it true in general? Also, this seems like a somewhat complicated first way of getting orientability.

In Milnor-Stasheff $\S$9, they ask to show that $\gamma^n\oplus\gamma^n$ is orientable, where $\gamma^n$ is the canonical $n$-plane bundle over $G_n(\mathbb{R}^{\infty})$, and the above fact is only touched upon in the obstruction theory chapter much later in the book, so using it is somewhat unsatisfactory. The main question I have is about the following: there is a claim that if $2e(M)\not=0$ for a smooth orientable manifold $M$, then $\tau(M)$ has no odd-dimensional sub-bundles. It's clear that it has no orientable odd-dimensional sub-bundles, but I'm not sure about unorientable ones. Milnor and Stasheff remark that looking at a suitable $2$-fold covering of $M$ works, so maybe taking the standard one where each point is paired with a local orientation and taking the pullback of $\tau(M)$ under the covering map would work?

Answer 1

The Whitney sum of two vector bundles is orientable.

This is certainly not true; for example, one of the vector bundles could be nonorientable and the other could be zero. What's true is that $w_1(V \oplus W) = w_1(V) + w_1(W)$, which shows that the sum of two copies of the same vector bundle is orientable, that the sum of two orientable bundles is orientable, etc.

The main question I have is about the following: there is a claim that if $2e(M) \neq 0$ for a smooth orientable manifold $M$, then $\tau(M)$ has no odd-dimensional sub-bundles.

First use your argument for orientable odd-dimensional subbundles, then take the orientation double cover of a nonorientable odd-dimensional subbundle.