Manifold embedded in euclidean space with nontrivial normal bundle 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.
 A: 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.
