Show $X \times S^1$ is parallelizable if $X$ admits a non-vanishing normal vector field Let $X \subset \mathbb R^{n+1}$ be an embedded submanifold of codimension 1 such that there is a map $F:X \to \mathbb R^{n+1}$ such that for all $x \in X$, $F(x)$ is in the orthogonal complement of $T_xX$ in $\mathbb R^{n+1}$. Show that $X \times S^1$ admits a global frame.
I think we have 2 global normal vector fields on $X \times S^1$ namely $(N,0)$ and $(0,M)$ where $M,N$ are global normal fields of $X, S^1$ respectively. So I guess if $X$ was just embedded in $\mathbb R^2$ then their cross product would work. But I have no idea how to approach this in higher dimensions. Any ideas?
 A: At any point $x \in X$, The normal space $N_{x}(X)$ must be $1$-dimensional, so any vector orthogonal to a normal vector must be a tangent vector (since we can write any vector space space as the direct sum of a subspace and its orthogonal complement, and the dimension of $T_{x}(X)$ is the dimension of $X$). Let $N(X)$ be the non-vanishing normal vector field. Take $v^{j}(x,z)$ for $(x,z) \in X \times \mathbb{S}^1$ to be
$$v^{j}(x,z) = \big(e_{j} - \langle \frac{N(x)}{|N(x)|},e_{j} \rangle N(x),  \langle \frac{N(x)}{|N(x)|},e_{j} \rangle iz  \big)$$
where $e_j$ is the $j$th standard basis vector, and we embed $\mathbb{S}^1$ into $\mathbb{C}$ so that we can get a tangent vector to $z$ by multiplying by $i$. Note that 
$$e_{j} - \langle \frac{N(x)}{|N(x)|},e_{j} \rangle e_{j}$$
is a tangent vector because it is orthogonal to $N(x)$, and by our above discussion it must be in the tangent space. Thus $v^j$ is a vector field, so we need to show that $v^{1}, \cdots, v^{n}$ are all linearly independent at a point. Suppose that there exists $x$, and $c_j$ not all zero, for which
$$\sum c_{j} v^{j}(x,z) = 0$$
Then it must be the case that
$$\sum c_{j}e_{j} = \lambda N(x), \,\,\,\,\, \lambda \neq 0$$
since the only way the projection to the orthogonal complement could vanish would be if the vector itself was parallel to $N(x)$. But now, if we take the dot product of both sides of this equality with $\frac{N(x)}{|N(x)|}$, we get
$$\sum c_{j}  \langle \frac{N(x)}{|N(x)|},e_{j} \rangle = \lambda |N(x)|$$
where the left-hand side is zero by hypothesis, since it is the right coordinate of $\sum c_{j} v^{j}(x,z)$, which was the zero vector. But this implies that $\lambda |N(x)| = 0$, and since $\lambda \neq 0$, we must have that $N(x) = 0$, which is impossible since we assumed $N$ was a non-vanishing normal field.
