Let $U$ be the set $U$ of quaternions of unit length. I know that $U\times S^1$ is compact, connected and is a $2n$ manifold in a $2n+1$ dimensional vector space $V$.
How can I construct a differentiable tangent vector field on $U\times S^1$ that has no zeroes?
What I know:
A tangent vector field is a $C^k$ map $F:U\times S^1\rightarrow \mathbb{R}^7$ s.t. $F(x)\in T_x(U\times S^1)$ for all $x\in U\times S^1$. But how can I construct one?
Firstly the dimension of $U$ is $3$ thus the dimension of $U\times S^1$ is 4.
Consider $u$ a vector orthogonal to $(1,0,0,0)$, you can define the vector field $X_u$ on $U$ by $X_u(x)=xu$. Remark that since the map $L_x:U\rightarrow U$ defined by $L_x(y)=xy$ is a diffeomorphism, $X_u(x)=d{L_x}_{(1,0,0,0)}(u)$. Now take any vector $V$, on $S^1$ which does not have zeroes (for example if $S^1=R/\{x\rightarrow x+1\}$ the vector ${\partial\over{\partial x}}$ is invariant by $x\rightarrow x+1$ so induces a vector on $S^1$) and defined $X(x,y)=(X_u(x),V(y))$.