TITLE: Submanifold of a parallelizable manifold
QUESTION [6 upvotes]: All manifolds are $ \mathcal{C}^{\infty} $ in the subsequent discussion. A parallelizable manifold of dimension $ k $ is one whose tangent bundle is trivial, or equivalently, there are $ k $ linearly independent smooth global vector fields.
Problem: Let $ N $ be a parallelizable $ (2n+1 )$-dimensional manifold. Suppose $ M $ is a closed orientable $ 2n $-dimensional manifold of Euler characteristic zero which admits a closed immersion into $ N $. Then $ M $ is parallelizable as well.
Of course, the words 'orientable' (every parallelizable manifold is orientable) and 'Euler characteristic zero' (think sphere in $ \mathbb{R}^3 $ for a counterexample) are necessary. Let's try the case where $ N = \mathbb{R}^{2n+1} $. Then $ M $ is the zero locus of a single smooth function $ f : \mathbb{R}^{2n+1} \rightarrow \mathbb{R} $ whose derivative doesn't vanish on $ M $. This shows that the normal bundle of $ M $ in $ N $ is trivial. If $ \theta $ is the trivial line bundle on $ M $, we get an exact sequence $$ 0 \rightarrow TM \rightarrow \theta^{2n+1} = TN \mid_M \rightarrow \theta = N_{M/N} \rightarrow 0 $$
Now I suspect we'll have to use the fact that $ 2n $ is even in some way. The middle cohomology of $ M $ has even dimension by the assumption that the Euler characteristic is $ 0 $. But I'm not really sure where to go from all of this.
Some notes: This problem is from an old Miklos Schweitzer competition. So I expect it to be hard (obviously), but I still feel ashamed being someone who studies algebraic geometry yet cannot make much progress on such a fundamental looking problem. Any hints will be welcome.

REPLY [2 votes]: I don't know whether this is simpler, but here would be my approach.
As in the other answer, the normal bundle of $M$ in $N$ is a trivial line bundle, so we have a commutative diagram
$$\begin{array}{ccc}
& & S^{2n} \\
&  & \downarrow \\
M & \rightarrow & BSO(2n) \\
\downarrow & & \downarrow \\
N & \xrightarrow[*]{} & BSO(2n+1).
\end{array}$$
The horizontal arrows classify the tangent bundles of $M$ and $N$ respectively; in particular, the bottom arrow $N \to BSO(2n+1)$ is nullhomotopic since $N$ is parallelizable.  The right arrow is induced by the standard inclusion $SO(2n) \to SO(2n+1)$, which has homotopy fiber $S^{2n}$.  The commutativity of the square implies that the map $M \to BSO(2n)$ lifts to $M \to S^{2n}$.
Let us focus on the top part of the diagram and extend it to the right:
$$\begin{array}{ccccc}
& & S^{2n} & \xrightarrow{\simeq_{\leq 2n}} & K(\mathbb{Z}, 2n) \\
& \nearrow & \downarrow & & \downarrow \small d \\
M & \rightarrow & BSO(2n) & \xrightarrow{\chi} & K(\mathbb{Z}, 2n)
\end{array}$$
The map $\chi$ is the Euler class in $H^{2n} BSO(2n)$.  The rightmost map is multiplication by some integer $d$ on $K(\mathbb{Z}, 2n)$.  I think we might be able to show that the vertical map $S^{2n} \to BSO(2n)$ classifies the tangent bundle of $S^{2n}$, in which case $d = \chi(S^{2n}) = 1 + (-1)^{2n} = 2$.  In any case, the fact that the tangent bundle of $S^{2n}$ also has the property of being trivialized upon Whitney summing a trivial line bundle implies that $d$ divides $2$.  In particular, $d$ is not zero.
Now consider the outer square.  The bottom composite $M \to K(\mathbb{Z}, 2n)$ represents the Euler class of $M$ in $H^{2n} M$, which is zero by assumption.  This means the upper composite $M \to K(\mathbb{Z}, 2n)$ is $d$-torsion.  Since $H^{2n} M$ is torsion-free, commutativity of the square forces the map $M \to S^{2n}$ is nullhomotopic.  But the map $M \to BSO(2n)$ classifying the tangent bundle factors through this nullhomotopic map, and thus is itself nullhomotopic, which proves that $M$ is parallelizable.