Tangent vector field of a Lie group as a manifold

292 Views Asked by At

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?

2

There are 2 best solutions below

0
On

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 $X(x,y)=(X_u(x),V(y))$.

0
On

Both $S^1$ and $U$ are Lie groups. Lie groups are known to be parallelizable manifolds, so in particular it is possible to find a nowhere vanishing vector field $X$ on $U$ and a nowhere vanishing one $Y$ on $S^1$. The vector field $U \times S^1 \ni (u,s) \mapsto (X_u, Y_s) \in T_u U \times T_s S^1 \simeq T_{(u,s)} (U \times S^1)$ is therefore a field as desired.


A more constructive approach is to pick a $0 \ne v \in \mathfrak g$ (the Lie algebra of $G = U \times S^1$) and consider "translating" it: if $L_g : G \to G$ is the translation by $g \in G$, then $x \mapsto (\Bbb d _e L_x) (v) \in T_x G$ (where $e \in G$ is the identity element of $G$) is a nowhere vanishing vector field (because $\Bbb d _e L_x : T_e G \to T_x G$ is an isomorphism).