Submanifold of a parallelizable manifold

Question

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.

Answer 1

Here is a summary of the answer written by Jason DeVito in the comments:

A parallelizable manifold is always orientable, therefore $w_1(N)$ vanishes. Since the tangent bundle of $M$ and the normal bundle of the immersion sum to the pullback of $T(N)$ to $M$, we deduce that $w_1(M)+w_1(\mu)=0$, here $\mu$ denotes the normal bundle. Since $M$ is orientable $w_1(\mu)$ must be trivial. Since $\mu$ is a line bundle, this implies it is isomorphic to a trivial bundle.

From this we deduce that $T(M)$ is stably isomorphic to a trivial bundle. By an obstruction theory argument found here, two $2k$-dimensional vector bundles over a $2k$-dimensional manifold that are stably isomorphic are isomorphic, if and only if, their Euler classes agree.

Recalling that the Euler class of the tangent bundle computes the Euler characteristic of the manifold, we deduce that if $M$ is parallelizable, if and only if, its Euler characteristic is 0.

Answer 2

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.