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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:

  1. 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:

  2. 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}

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closed as off topic by Qiaochu Yuan, Bruce Westbury, Mariano Suárez-Alvarez, Alain Valette, Leonid Positselski Nov 18 '11 at 15:27

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This is not quite appropriate for MO, but short answers: $\text{SO}(3)$ is abstractly the automorphism group of a real $3$-dimensional oriented inner product space $V$. On such a space $V$ one can define the adjoint of an operator in $\text{End}(V)$ and $\mathfrak{so}(3)$ is the subspace of antisymmetric operators with respect to the adjoint. – Qiaochu Yuan Nov 18 '11 at 13:43
warner, foundations of differentiable manifolds and lie groups – Yosemite Sam Nov 18 '11 at 14:55

Abstractly you may think of $SO\left(3\right)$ as the group of rotations in Euclidean $3$-space, that is the group of linear transformations of $\mathbb{R}^3$ which preserve the norm $\left|\left(x,y,z\right)\right|^2=x^2+y^2+z^2$. Alternatively, you may think of elements of $SO\left(3\right)$ as orthonormal bases, $\hat{x},\hat{y},\hat{z}$, which corresponds to the rotation which sends the standard basis $\hat{i},\hat{j},\hat{k}$ to $\hat{x},\hat{y},\hat{z}$.

You're already familiar with the matrix definition of $so\left(3\right)$, which is skew-symmetric matrices with Lie bracket given by the commutator of matrices. Any such matrix looks like $$\left(\begin{array}{ccc}0 & -\omega_{z} & \omega_{y}\\\ \omega_{z} & 0 & -\omega_{x}\\\ -\omega_{y} & \omega_{x} & 0\end{array}\right)$$This corresponds to an "angular velocity" vector $\left(\omega_x,\omega_y,\omega_z\right)$ and the commutator of two skew-symmetric matrices corresponds to the cross product of their corresponding angular velocities. So you see that $so\left(3\right)$ is isomorphic to $\mathbb{R}^3$ with the Lie bracket given by the cross product.

In general for a Lie group $G$, one may define the Lie bracket on its Lie algebra $g=T_e G$ as follows. For $a\in G$, let $C_a:G\to G$, the commutator by $a$, be $C_a\left(b\right)=aba^{-1}$.The derivative at $e$, $\left(dC_a\right)_e:g\to g$, is the "adjoint representation" of $G$ on its Lie algebra, $Ad:G\to GL\left(g\right)$, $Ad\left(a\right)=\left(dC_a\right)_e$. Taking the derivative again gives a map $ad:g\to gl\left(g\right)$ and the Lie bracket on $g$ is $$\left[x,y\right]=ad\left(x\right)\left(y\right)$$So you see, you don't need to use matrix multiplication to abstractly define the Lie bracket of a Lie algebra, thought of as the tangent space to a Lie group.

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Ah that's nice. Of course, the derivative of the commutator map gives one on the corresponing tangent spaces. But then the usual notation of your $(dC_a)_e$ map, i.e. $(dC_a)_e (g) = aga^{-1}$ is a purely formal expression and does a priori not correspond to actual products. But concerning $SO(3)$, I am very well aware that $SO(3)$ is the group of rotations preserving euclidian norm and orientation, but I wondered if there is a nice presentation of this group. It certainly is formally ok to let it be the group of automorphisms of a real 3-d oriented inner prod space, but a bit unsatisfying – Niki Nov 18 '11 at 15:28
In the phrase "rotations preserving euclidean norm and orientation", the sub-phrase "preserving euclidean norm and orientation" is redundant. – José Figueroa-O'Farrill Nov 18 '11 at 20:36

You can define $SO(n)$ as follows: Let $V$ be a real $n$-dimensional vector space with an inner product. Define $SO(n)$ to be the group of linear transformations $A: V \rightarrow V$ that preserve the inner product. In other words, $Av\cdot Aw = v\cdot w$, for any $v, w \in V$. It is straightforward to use this definition and directional differentiation to derive an abstract definition of the Lie algebra $so(n)$.

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Small nitpick: you have defined $O(n)$; $SO(n)$ is the identity component, i.e., orientation-preserving isometries. – José Figueroa-O'Farrill Nov 18 '11 at 20:35
José, you're absolutely right. Thanks! – Deane Yang Nov 19 '11 at 4:02

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