I think I am stuck with the following statement that I read on the Encyclopedia of Mathematics website regarding Isotropy representations:
"If $G$ is a Lie group acting smoothly and transitively on $M$ then the tangent space $T_xM$ to a point of the manifold can be naturally identified with the space $\mathfrak{g} / \mathfrak{g}_x$ where $\mathfrak g$ and $\mathfrak g_x$ stand for the Lie algebra of $G$ respectively the isotropy subgroug $G_x$."
Now if I let $S^1 \cong SO(2) = \left\{\left(\begin{matrix}\cos \theta & - \sin \theta& 0 \\ \sin\theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{matrix}\right) \colon \theta \in \mathbb R\right\}$ act on the sphere $S^2$ via $(A,x) \mapsto Ax$ then the isotropy subgroup of the northpole $N = (0,0,1)$ is all of $S^1$, that is $\mathfrak{g}_N = \mathfrak{g}$ and so the above identification statement says that $$ T_NS^2 \cong \mathfrak{g} / \mathfrak{g} = \{e\} $$ But this cannot be true since $T_NS^2 \cong \mathbb{R}^2$ where the right hand side is a one - point set .. what am I missing ?
The hypothesis of the statement is that $G$ acts transitively on $M$, meaning that for any $p,q\in M$ there exists $g\in G$ such that $g\centerdot p=q$. In your example, $S^1$ does not act transitively on $S^2$.