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.

If a smooth $n$-manifold $M \subset \mathbb{R}^m$ is globally defined as the preimage of a regular value of a smooth function $f:\mathbb{R}^n \to \mathbb{R}^{m-n}$, then we have the desired setup
$$TM=\{(x,v) \in \mathbb{R}^{m} \times \mathbb{R}^{m} \mid f(x)=0 \text{ and } df_{x}(v)=0\}.$$

If you read his answer more carefully, @jef808 was saying that we could try to use the defining equations to find linearly independent global sections. For example, one obvious solution to the equation $ux+vy=0$ is $(u,v)=(-y,x)$. Then the map $S^1\to TS^1$ given by $(x,y) \mapsto ((x,y),(-y,x))$ defines a nonvanishing vector field. We can generalize this construction to $S^{2n-1}$ and its tangent bundle
\begin{align*}
TS^{2n-1} &= \bigg \{((x_1,y_1,\ldots,x_n,y_n),(u_1,v_1,\ldots,u_n,v_n)) \in \mathbb{R}^{2n} \times \mathbb{R}^{2n} \mid \\
& \qquad \qquad \sum_{k=1}^n x_k^2 +\sum_{k=1}^n y_k^2=1 \text{ and } \sum_{k=1}^n u_k x_k + \sum_{k=1}^n v_k y_k =0 \bigg \}
\end{align*}
by defining a map
$$(x_1,y_1,\ldots,x_n,y_n)\mapsto ((x_1,y_1,\ldots,x_n,y_n),(-y_1,x_1,\ldots,-y_n,x_n)).$$
However, for $n >1$, we need *several* linearly independent nonvanishing sections. As we found out circa 1960, this is only possible for $S^1$, $S^3$, and $S^7$; try googling "parallelizability of spheres". So even when the defining equations are very simple, it may be practically impossible to see whether or not you can construct enough linearly independent global sections of the tangent bundle.
Similarly, if there was another way to use the defining equations to demonstrate triviality of a bundle, then Milnor, Bott, et cetera probably would have used it in the case of spheres.

One *can* ask if there are special cases where the defining equations shed light on the issue. I don't know if there are interesting families of special cases, but there are certainly ad hoc methods on a case-by-case basis:

**Example (Torus).** Even though we know that the torus has a trivial tangent bundle, it's not so apparent from a particular embedding $M \subset \mathbb{R}^3$, say
$$M=\{(x,y,z) \in \mathbb{R}^3 \mid f(x,y,z)=(x^2 +y^2+z^2+3)^2-16(x^2+y^2)=0\}.$$
We know that points $((x,y,z),(u,v,w))$ in the tangent space satisfy
\begin{align*}
df_{(x,y,z)}(u,v,w)&=\begin{bmatrix} -32x +4x(3+x^2+y^2+z^2) \\ -32y +4y(3+x^2+y^2+z^2) \\ 4z(3+x^2+y^2+z^2)
\end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} \\
&=-16(2x u+2yv)+2(3+x^2+y^2+z^2)(2xu+2yv+2zw) \\
&=0.\end{align*}
Similar to the $S^1$ case, we have an obvious solution of $(u,v,w)=(-y,x,0)$. Call this first vector field $\nu$. To find a second (linearly independent) vector field $\eta$, we could start by assuming $\eta=(u,v,w)$ is perpendicular to $\nu=(-y,x,0)$, i.e. $\eta=c_1(x,y,0)+c_2(0,0,1)$ for some $c_1,c_2$ (that are functions of $x,y,z$). Plugging $u=c_1 x$, $v=c_1 y$, $w=c_2$ into the above equation, we must have
$$-16(2x^2 c_1+2y^2 c_1)+2(3+x^2+y^2+z^2)(2x^2 c_1+2y^2c_1+2z c_2) =0,$$
equivalently written as
\begin{align*}2x^2 c_1+2y^2c_1+2z c_2 &=\frac{16(2x^2 c_1+2y^2 c_1)}{2(3+x^2+y^2+z^2)} \\
c_2 &=\frac{c_1}{z} \left(\frac{8(x^2 +y^2 )}{3+x^2+y^2+z^2} -x^2 +y^2\right).
\end{align*}
If we set $c_1=z$, this gives us a reasonably simple expression for $c_2$. One can easily check that $\nu(x,y,z)=(-y,x,0)$ and $\eta(x,y,z)=(x z, y z, c_2(x,y,z))$ define two linearly independent nonvanishing vector fields on $M$. Thus $M$ has a trivial tangent bundle.

## Best Answer

In general it's pretty difficult to visualise tangent bundles given that they have dimension twice the dimension of the base manifold; for example, $TS^2$ is $4$-dimensional. However, one can prove indirectly that this bundle is non-trivial (i.e. not a product). If it

werea product, then it would have a nowhere-zero section. In other words, one could construct a continuous vector field on the sphere which doesn't vanish anywhere (explicitly, if $TS^2 \cong S^2 \times \mathbb{R}^2$ then we could take our vector field to be constant $(1, 0)$ in the second factor). And this is impossible, by the hairy ball theorem.An example of a non-trivial normal bundle is easier. Here's one way. If $M$ is the total space of some vector bundle, and $N$ is the zero section, then the normal bundle of $N$ in $M$ is simply the bundle $M$ itself. So if $M$ is non-trivial (say $M$ is the Mobius bundle, i.e. a cylinder with a 'half twist'; this is non-trivial for the 'no nowhere-zero sections' reason above) then $N$ has non-trivial normal bundle in $M$.

As for why the tangent and normal bundles need not be the same, in general they don't even have the same dimension. If $N$ is an $n$-dimensional submanifold of an $m$-manifold $M$ then $TN$ is $2n$-dimensional whilst the normal bundle is $m$-dimensional.