Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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}

share|improve this question

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

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

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
add comment

2 Answers 2

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.

share|improve this answer
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
add comment

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 $so(n)$.

share|improve this answer
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
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.