I'm trying to find the left-invariant vector fields on the circle $S^1$.
If I understand correctly, $S^1$ is given the group structure of the multiplicative group of complex numbers on the unit circle in the complex plane, and $S^1$ is made into a $C^\infty$ manifold by forming the maximal atlas from the four projection maps from open semi-circles of $S^1$ onto the $x$- and $y$-axes.
Given a point $p\in S^1$, if $p$ is on the upper or lower open semicircles, then a basis for $T_pS^1$ is given by $\{\partial/\partial \bar x \rvert_p\}$, where $\bar x$ is the projection onto the $x$-axis, while if $p$ is on the left or right open semicircles, then a basis for $T_pS^1$ is given by $\{\partial/\partial \bar y \rvert_p\}$.
Given a vector field $X$ on $S^1$, $X$ is left-invariant provided that for all points $g,h\in S^1$, $$ (\ell_g)_{*,h}(X_h) = X_{gh} \text, $$ where $\ell_g\colon S^1 \to S^1$ is left-multiplication by $g$.
To obtain an arbitrary left-invariant vector field $X$ on $S^1$, I should be able to just pick an arbitrary tangent vector $A \in T_1S^1$ and see what it generates. Let $A = a(\partial/\partial \bar y\rvert_1) \in T_1S^1$. Then given a point $p\in S^1$, I can define $$ X_p = (\ell_p)_{*,1}(A). $$ However, I don't see how this is getting me any closer to concretely writing down what $X$ looks like. Any suggestions?
If you suppose that the circle $S^1$ is embedded in the complex plane $C$ you can represent an element of $S^1$ by $e^{it}, t\in R$. The tangent space $T_1S^1$ is the vertical line $x=0$. Let $(0,u)\in T_1S^1$, and $e^{it}\in S^1$, we denote by $l_{e^{it}}$ the left multiplication defined by $e^{it}$, thus you have $e^{it}(e^{iu})=e^{it}e^{iu}$. Remark that this map may be extended to $C$ in a linear map (the rotation of angle $t$), thus is equal to its differential, this implies that $l_{e^{it}}(0,u)=e^{it}(0,u)=(-u\sin(t),u\cos(t))$.