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.

  • 2
    $\begingroup$ Unrelatedly, this is the first I've heard of the Miklos Schweitzer competition, and I think it's cool that the there's a competition at the undergrad level that's more representative of the types of math that mathematicians actually work on, rather than contest-math or recreational-math problems. Thanks for the information. $\endgroup$
    – anomaly
    Commented Mar 31, 2022 at 16:20
  • $\begingroup$ Does the fact that $M$ admits a nowhere-vanishing vector field help? $\endgroup$
    – anomaly
    Commented Mar 31, 2022 at 16:23
  • $\begingroup$ @anomaly I forgot to write that down. That follows from the description of the Euler class of the tangent bundle, right? But I'm not sure how to use that either. $\endgroup$ Commented Mar 31, 2022 at 17:02
  • $\begingroup$ Right, though the more interesting direction involves Poincare-Hopf or looking at something like the Thom isomorphism, rather than just using the relation $e(X) = \chi(X) [X]$ for reasonable $X$. It's not clear how to proceed from there, though. The Euler class is sensitive enough to distinguish between triviality and stable triviality, at least, which is useful if you're sitting inside a parallelizable manifold; but that's a long way from a proof. Maybe there's something to be said about the normal bundle of $M$. $\endgroup$
    – anomaly
    Commented Mar 31, 2022 at 17:24
  • $\begingroup$ I have two very naive questions (although a positive answer to both implies the solution, I can’t help but be suspicious of such a path). I think obstruction theory can be relevant to the answer, but it’s a long while since I did sone serious topology and I remember very little. First, can any map $M \rightarrow S^{2n}$ be extended to $N$ (if it has degree $0$, it works)? Second, can any nonzero vector field on $N$ be part of a basis? (or in other words, can any map $N \rightarrow S^{2n}$ extend to some $N \rightarrow O(2n+1)$? $\endgroup$
    – Aphelli
    Commented Mar 31, 2022 at 17:44

2 Answers 2


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.

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    $\begingroup$ Given that the OP mentions this was a contest oroblem, I wonder if there is a way to do it without referencing a paper. But $+1$ from me either way. $\endgroup$ Commented Apr 1, 2022 at 23:57

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.

  • $\begingroup$ Can you elaborate on why there is a self map of $K(\mathbb{Z}, 2n)$ which makes the diagram commute? As an alternative, one can use the Serre spectral sequence associated to the fibration $BSO(2n)\rightarrow BSO(2n+1)$ to see that with rational coefficients, the Euler class is mapped to a generator of $H^{2n}(S^{2n})$. $\endgroup$ Commented Apr 3, 2022 at 0:56
  • $\begingroup$ @JasonDeVito I don't think I'm saying anything too profound about the square in the second diagram. The map along the bottom left $S^{2n} \to BSO(2n) \to K(\mathbb{Z}, 2n)$ is some class $d \in H^{2n} S^{2n}$, i.e., $d$ times a generator represented by the top map $S^{2n} \to K(\mathbb{Z}, 2n)$. I just decided to typeset it as a square since there isn't a lot of good options for drawing diagrams on MSE. $\endgroup$
    – JHF
    Commented Apr 4, 2022 at 12:41
  • $\begingroup$ That makes perfect sense. I don't remember what time I asked that question, but I'm going to assume that it was either really late or pre-caffeine in the morning ;-). Sorry to bother! $\endgroup$ Commented Apr 4, 2022 at 18:19
  • $\begingroup$ @JasonDeVito You can see that the map $S^{2n} \rightarrow BSO(2n)$ classifies the tangent bundle by letting $M=S^{2n}$ which gives the relation $T(S^{2n})=kT(S^{2n})$, which implies $k:S^{2n} \rightarrow S^{2n}$ is $1$. $\endgroup$ Commented Apr 5, 2022 at 19:59
  • $\begingroup$ @ConnorMalin: Very nice. I wonder if this answer and these comments are converging on the kind of answer which was intended for a contest? $\endgroup$ Commented Apr 6, 2022 at 19:15

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