Tangent vectors on SO(2) Parametrization of group SO(2) can be given as the following;
$$
        \begin{pmatrix}
        \cos( \theta) & \sin(\theta)\\
        - \sin(\theta) & \cos(\theta) \\
        \end{pmatrix}
$$
Expanding this curve near the identity element, we have
$$ X_{e} = \begin{pmatrix}
 0 & 1 \\
-1 & 0 \\
\end{pmatrix}
$$
Now, i do understand that this is the generator of left-invariant vector field along this curve. So this is supposed to be an tangent vector at the identity element. But i don't understand how i can relate this matrix element to basis of tangent vectors at the identity.
Dimension of a tangent space at a given point in the manifold should be equal to the dimension of the manifold, but how can i know the dimension of the manifold, espacially for this particular example?
Thanks.
 A: $SO(2)$ is homeomorphic to $S^1$, simply by the map that sends 
$$\left(\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)\mapsto e^{i\theta}.$$
Now $S^1$ is a 1- manifold, with a cover by charts given by $(\sigma_N,S^1- \{(0,1)\}$ and $(\sigma_S,S^1-\{(0,-1)\})$ where $\sigma_N$ means stereographic projection from the north pole, similarly for $\sigma_S$. It will now follow that $\text{SO}(2)$ is also a $1$ - manifold, i.e. the tangent space is one dimensional.
A: This vector is (the only vector of) the basis of the tangent space $T_e \mathrm{SO}(2)$ embedded into the tangent space $T_{I_2}M_2(\mathbb R)$.
In fact, $$\exp(tX_e) = 1 + tX_e + \frac{t^2 X_e^2}{2} + \frac{t^3 X_e^3}{6} + \ldots$$
$$= 1 + tX_e - \frac{t^2}{2} - \frac{t^3}{6} X_e + \ldots$$
$$= \cos(t) \cdot 1 + \sin(t) X_e,$$
which is to be expected from the general theory. Thus, the flow along the left-invariant vector field generated by $X_e$ gives us a covering map $\exp(-X_e): \mathbb{R} \to \mathrm{SO}(2)$, which is in fact a Lie group morphism.
Note that under the standard isomorphism $f: \mathrm{SO}(2) \to \mathrm{U}(1)$ we have $f(X_e) = i$, and the formula above becomes $e^{it} = \cos(t) + i \sin(t).$
It's simple yet nice stuff :)
