Isomorphism between $\mathfrak{so}(3)$ and $\mathbb{R}^3$ Good morning. I'm studying Lie Groups and this question came out, I just can not find a way to answer it (probably because my understanding of Lie groups is pretty poor).

A lie group is a group wich is also a manifold and such that both multiplication and inverse maps are smooth.


The tangent space of a manifold $M$ at a point $p \in M$ is the $\mathbb{R}$-vector space $T_p(M)$ of all derivations at $p$, i.e. $T_p(M) = \{D \in \textrm{Lin}(C^\infty(M), \mathbb{R}) \mid D(fg) = D(f)g(p) + D(g)f(p)\}$.


The special orthogonal group of dimension $3$ is the group $SO(3)$ given by all $3 \times 3$ matrices $M$ such that $\det M = 1$ and $M^{-1} = M^T$ und the operation of matrix product. It is a lie group, since product and taking inverses of matrices are smooth functions from $\mathbb{R}^9$ to itself, and $SO(3)$ is an open set of $\mathbb{R}^9$.

My first question is: why is $\{L_1, L_2, L_3\}$ (given below) a basis for $T_1(SO(3))$?
$$L_1 = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{bmatrix}$$
$$L_2 = \begin{bmatrix}
0 & 0 & 1\\
0 & 0 & 0\\
-1 & 0 & 0
\end{bmatrix}$$
$$L_3 = \begin{bmatrix}
0 & -1 & 0\\
1 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}$$
I think it can be answered in parts:

*

*What is the dimension of $C^\infty(SO(3))$?

*What is the dimension of $\textrm{Lin}(C^\infty(SO(3)),\mathbb{R})$? It is the dimension of $C^\infty(SO(3))$, right? It is the dual space.

*(probably alredy answered but) what is a basis for $C^\infty(SO(3))$? I would like to know this so I can write $L_1,L_2$ and $L_3$ as linear transformations just for me to see that they're really derivations.

I'll try to finish the proof once I get to understand this better, but probably my next question would be:

*

*What are the left-invariant vector fields of $L_1, L_2$ and $L_3$?

Once i get this, the proof is finished, right? I just have to construct the isomorfism that sends $L_i$ to $e_i$ and check that $L_1,L_2$ and $L_3$ satisfy the same bracket relations as $e_1,e_2$ and $e_3$ with the cross product.
 A: The space $C^\infty (SO(3))$ of smooth functions on $SO(3)$ is infinite dimensional, see the proof here.
The space $\operatorname{Lin} (C^\infty (SO(3)), \mathbb R)$ is the dual space of $C^\infty (SO(3))$ and is also infinite dimensional. See What can be said about the dual space of an infinite-dimensional real vector space? for more information.
So asking the basis or the dimension of $C^\infty (SO(3))$ does not help you find the basis of $T_I SO(3)$. In general, the tangent space $T_pM$ of a manifold $M$ is defined as a linear subspace of $\operatorname{Lin} (C^\infty (M), \mathbb R)$. It turns out that it is finite dimensional and the dimension equal $\dim (M)$. This is proved in most introductory textbooks in smooth manifold, for example John M. Lee's Introduction to Smooth Manifold.
Since $SO(3)$ is defined as a submanifold of $\mathbb R^9$, one can define $T_I SO(3)$ as the set of all $A'(0)$, where $A : I \to \mathbb R^9$ is a smooth curve such that $A(s)\in SO(3)$ for all $s\in I$ and $A(0) =I$. This is by definition a subset of $\mathbb R^9$.
For any such curve $A$, $A(s)^t A(s) = I$. Differentiate and set $s=0$, using that $A(0) = I$, gives
$$ A'(0)^t + A'(0) = 0.$$
On the other hand, for any skew symmetric matrix $X$, let
$$A(t) = e^{tX} = I + tX + \cdots + \frac{1}{n!} (tX)^n + \cdots.$$
Then $A(0) = I, A'(0) = X$ and $A(t)$ is in $SO(3)$: that $A\in O(3)$ is proved here and $\det A = 1$ then follows from a continuity argument.
Thus $T_ISO(3)$ is identified as the space of all skew-symmetric $3\times 3$ matrices. It is then an easy exercise to check that $L_1, L_2, L_3$ forms a basis.
Recall that a vector field $X$ on a Lie group $G$ is left-invariant if $X = (\ell_g)_* X$ for all $g\in G$. Here $\ell_g : G\to G$ is the left multiplication. In the case for matrix group, $\ell_A X = AX$ is linear, thus  $$(\ell_A)_*X = AX.$$
In particular, the left invariant vector fields corresponding to $L_i$ are
$$ A\in SO(3) \mapsto AL_i \in T_A SO(3).$$
