First I want to give you some background how the question arised, before actually asking it.

Recently, in the context of quantum mechanics, I thought about the group $SO(3)$ and its Lie Algebra $so(3)$. Wherever I looked, I could only find a construction of $so(3)$ very concretely in terms of matrices, as being the tangent space of $SO(3)$ at the identity. When considering $SO(3)$ represented as $3\times3$ matrices, it follows directly (chain rule etc.), that elements in $so(3)$ are anti-symmetric, real $3\times3$ matrices and form a three dimensional vector space. Therefore (with the knowledge that they can be represented as matrices) we can take the commutator as being the Lie Bracket in $so(3)$, although in the Lie Algebra $so(3)$ a priori there is no product defined, which even rises the question why the commutator lies again in $so(3)$. Matrix multiplication in $SO(3)$ gives addition in $so(3)$ (chain rule).

It seems as if the Lie Bracket, as being the commutator, gets its definition from the fact, that we know elements in $so(3)$ can be represented as matrices.

Therefore my questions are the following:

What is $SO(3)$ as abstract group? How can we get hold of it and present it without matrices? Especially with regard to the second question:

How to get the Lie Algebra out of $SO(3)$ in an algebraically satisfying way, i.e. without the explicit construction of matrices?

Looking forward for interesting ideas!

Cheers, Niki

ps.: Although I am sure, if we take the ill of matrices, there should be at least a way of getting the commutator in $so(3)$ without multiplying matrices, but I only found the following argument for $R_i(t) \in SO(3) \ \forall t$ and $R_i(0) = id$: \begin{eqnarray} [\Omega_1,\Omega_2] = \frac{d}{dt} R_1(t)\Omega_2 R_1(t)^{-1} \mid_{t=0} \newline \mbox{where } \frac{d}{dt}R_i(t) \mid_{t=0} = \Omega_i \newline \mbox{with } R\Omega_i R^{-1} = \frac{d}{dt}R R_i(t) R^{-1} \mid_{t=0} \end{eqnarray}