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Let $X$ be a differentiable $n$-manifold embedded in some $\mathbb{R}^{n+1}$.

I have two questions.

I have read that if $X$ is compact and orientable, then the normal bundle of the embedding is trivial. What could be the idea of a proof of this?

Is there an imaginable compact $n$-manifold $X\subseteq \mathbb{R}^{n+1}$ such that the normal bundle of the embedding is nontrivial? Maybe this is a silly question but I have the impression that every embedding $X\subseteq \mathbb{R}^{n+k}$ has a trivial normal bundle.

André
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  • @user10676 You seem to be assuming that the manifold $X$ is parallelizable, i.e., that its tangent bundle is trivial. – yasmar Feb 12 '12 at 17:02

1 Answers1

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1) Yes, it is true that any closed submanifold $X_n \subset \mathbb R^{n+1}$ is orientable , even if $X$ is not compact.
Once you have orientability, the normal bundle is necessarily trivial.
Indeed, there exists an oriented frame $X_1, X_2,...,X_n\; (X_i\in \Gamma (X,TX))$ for $X$.
For every $x\in X$ there are two vectors in $ T_x(\mathbb R^{n+1})$ orthogonal to $T_xX$ and of length $1$.
By selecting $n(x)$, the one such that the basis $X_1(x), X_2(x),...,X_n(x),n(x)$ of $T_x(\mathbb R^{n+1})$ is direct, you obtain a nowhere zero section $n\in \Gamma (X,N)$ trivializing $N$.

2) No, this is false: some manifolds have no embeddings with trivial normal bundle in any $\mathbb R^{n+k}$. Here is why:

From the exact sequence of vector bundles on $X$

$$ 0\to TX\to T\mathbb R^{n+k}|X \to N\to 0 $$ you deduce the equality of Stiefel-Whitney classes $w_1(TX)=w_1(N)$.
So if the normal bundle $N$ were trivial, you would conclude that $w_1(TX)=0$.
But this is false for all even dimensional real projective spaces $\mathbb P^{2r}(\mathbb R) $ .
So in any embedding $\mathbb P^{2r}(\mathbb R)\hookrightarrow \mathbb R^{n+k}$ these projective spaces have non-trivial normal bundles.

Georges Elencwajg
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  • Interesting. This is most probably again stupid, but does this mean that I can add any trivial bundle to the tangent bundle of $S^2$ and get a trivial bundle? This is somehow the question if $\oplus$ is well defined on isomorphism classes of vector bundles. – André Feb 13 '12 at 20:06
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    Dear @André: Yes, if you add any trival bundle $\theta^k$ to $TS^2$, you will get a trivial bundle. This follows from a formal calculation, namely: $TS^2\oplus \theta^k=(TS^2\oplus \theta)\oplus \theta^{k-1}=\theta^3 \oplus \theta^{k-1}=\theta^{k+2}$ – Georges Elencwajg Feb 13 '12 at 20:39
  • Sorry for reviving a (very) old answer: In your 1) you say that any closed submanifold $X_n \subset R^{n+1}$ is orientable. Maybe I am missing something elementary, but how does this work for the Klein bottle? Are you implicitly assuming an immersion, rather than an embedding? – ThePuix Nov 24 '20 at 19:34
  • @ThePuix: the Klein bottle cannot be embedded into $\mathbb R^3$. – Georges Elencwajg Nov 24 '20 at 22:52
  • But it can be immersed, hence giving rise to a closed subset of $\mathbb{R}^3$? – ThePuix Nov 24 '20 at 23:12
  • @ThePuix your last comment about immersions and closed subsets is irrelevant to my answer. You don't seem to realize that being an immersion is weaker than being an embedding. – Georges Elencwajg Nov 25 '20 at 13:22
  • Yes, my first comment was confusing: I meant to switch immersion and embedding. I realize that an immersion is a weaker notion than an embedding. My misunderstanding might be that the immersed Klein bottle is a closed set, but is not a manifold under this immersion and is such irrelevant to your answer. Sorry for the useless comments! – ThePuix Nov 25 '20 at 14:16