6.1 The rotation group in three dimensions

6.1 The rotation group in three dimensions

Concept links · terms present in this machine draft; source roles are unverified: vector space · orthogonal group · Lie algebra · Lie algebra representation · adjoint representation · Lie group · Lie bracket

Rotations in ${\mathbf{R}}^2$ about the origin are given by elements of $SO(2)$ , with a counterclockwise rotation by an angle $\theta$ given by the matrix

$$R(\theta )=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$$

This can be written as an exponential, $R(\theta )=e^{\theta L}=\cos \theta 1+L\sin \theta$ for

$$L=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$$

Here $SO(2)$ is a commutative Lie group with Lie algebra $\mathfrak{so}(2)=\mathbf{R}$ . Note that we have a representation on $V={\mathbf{R}}^2$ here, but it is a real representation, not one of the complex ones we have when we have a representation on a quantum mechanical state space.

In three dimensions the group $SO(3)$ is three dimensional and non-commutative. Choosing a unit vector $w$ and angle $\theta$, one gets an element $R(\theta ,\mathbf{w})$ of $SO(3)$ , rotation by an angle $\theta$ about the $w$ axis. Using standard basis vectors ${\mathbf{e}}_j$ , rotations about the coordinate axes are given by

$$R(\theta ,{\mathbf{e}}_1)=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos \theta & -\sin \theta \\ 0& \sin \theta & \cos \theta \end{array}\right),R(\theta ,{\mathbf{e}}_2)=\left(\begin{array}{ccc}\cos \theta & 0& \sin \theta \\ 0& 1& 0\\ -\sin \theta & 0& \cos \theta \end{array}\right)$$ $$R(\theta ,{\mathbf{e}}_3)=\left(\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right)$$

A standard parametrization for elements of $SO(3)$ is in terms of 3 “Euler angles” $\varphi ,\theta ,$ $\psi$ with a general rotation given by

$$R(\varphi ,\theta ,\psi )=R(\psi ,{\mathbf{e}}_3)R(\theta ,{\mathbf{e}}_1)R(\varphi ,{\mathbf{e}}_3)\text{\hspace{1em}(6.1)}$$

i.e., first a rotation about the z-axis by an angle $\varphi ,$ then a rotation by an angle $\theta$ about the new x-axis, followed by a rotation by $\psi$ about the new $z-$ axis. Multiplying out the matrices gives a rather complicated expression for a rotation in terms of the three angles, and one needs to figure out what range to choose for the angles to avoid multiple counting.

The infinitesimal picture near the identity of the group, given by the Lie algebra structure on $\mathfrak{so}(3)$ , is much easier to understand. Recall that for orthogonal groups the Lie algebra can be identified with the space of antisymmetric matrices, so in this case there is a basis

$$l_1=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)l_2=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ -1& 0& 0\end{array}\right)l_3=\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right)$$

which satisfy the commutation relations

$$[l_1,l_2]=l_3,[l_2,l_3]=l_1,[l_3,l_1]=l_2$$

Note that these are exactly the same commutation relations (equation 3.5) satisfied by the basis vectors $X_1,X_2,X_3$ of the Lie algebra $\mathfrak{su}(2)$ , so $\mathfrak{so}(3)$ and $\mathfrak{su}(2)$ are isomorphic Lie algebras. They both are the vector space ${\mathbf{R}}^3$ with the same Lie bracket operation on pairs of vectors. This operation is familiar in yet another context, that of the cross-product of standard basis vectors ${\mathbf{e}}_j$ in ${\mathbf{R}}^3$ :

$${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3,{\mathbf{e}}_2\times {\mathbf{e}}_3={\mathbf{e}}_1,{\mathbf{e}}_3\times {\mathbf{e}}_1={\mathbf{e}}_2$$

We see that the Lie bracket operation

$$(X,Y)\in {\mathbf{R}}^3\times {\mathbf{R}}^3\to [X,Y]\in {\mathbf{R}}^3$$

that makes ${\mathbf{R}}^3$ a Lie algebra $\mathfrak{so}(3)$ is the cross-product on vectors in ${\mathbf{R}}^3$

So far we have three diferent isomorphic ways of putting a Lie bracket on ${\mathbf{R}}^3$ , making it into a Lie algebra:

1. Identify ${\mathbf{R}}^3$ with antisymmetric real 3 by 3 matrices and take the matrix commutator as Lie bracket.

2. Identify ${\mathbf{R}}^3$ with skew-adjoint, traceless, complex 2 by 2 matrices and take the matrix commutator as Lie bracket.

3. Use the vector cross-product on ${\mathbf{R}}^3$ to get a Lie bracket, $i$.e., define

$$[\mathbf{v},\mathbf{w}]=\mathbf{v}\times \mathbf{w}$$

Something very special that happens for orthogonal groups only in dimension $n=3$ is that the vector representation (the defining representation of $SO(n)$ matrices on ${\mathbf{R}}^n)$ is isomorphic to the adjoint representation. Recall that any Lie group $G$ has a representation $(Ad,\mathfrak{g})$ on its Lie algebra $g$. $\mathfrak{so}(n)$ can be identified with the antisymmetric n by $n$ matrices, so is of (real) dimension $\frac{n^2-n}{2}$ . Only for $n=3$ is this equal to $n_{:}$ the dimension of the representation on vectors in ${\mathbf{R}}^n$ This corresponds to the geometrical fact that only in 3 dimensions is a plane (in all dimensions rotations are built out of rotations in various planes) determined uniquely by a vector (the vector perpendicular to the plane). Equivalently, only in 3 dimensions is there a cross-product $\text{ v }\times \text{ w }$ which takes two vectors determining a plane to a unique vector perpendicular to the plane.

The isomorphism between the vector representation $({\pi }_{vector},{\mathbf{R}}^3)$ on column vectors and the adjoint representation $(Ad,\mathfrak{so}(3))$ on antisymmetric matrices is given by

$$\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right)\leftrightarrow v_1{l}_1+v_2{l}_2+v_3{l}_3=\left(\begin{array}{ccc}0& -v_3& v_2\\ v_3& 0& -v_1\\ -v_2& v_1& 0\end{array}\right)$$

or in terms of bases by

$${\mathbf{e}}_j\leftrightarrow l_j$$

For the vector representation on column vectors, ${\pi }_{vector}(g)=g$ and ${\pi }_{vector}^{\prime }(X)=$ $X$ , where $X$ is an antisymmetric 3 by 3 matrix, and $g=e^X$ is an orthogonal 3 by 3 matrix. Both act on column vectors by the usual multiplication.

For the adjoint representation on antisymmetric matrices

$$Ad(g)\left(\begin{array}{ccc}0& -v_3& v_2\\ v_3& 0& -v_1\\ -v_2& v_1& 0\end{array}\right)=g\left(\begin{array}{ccc}0& -v_3& v_2\\ v_3& 0& -v_1\\ -v_2& v_1& 0\end{array}\right)g^{-1}$$

The corresponding Lie algebra representation is given by

$$ad(X)\left(\begin{array}{ccc}0& -v_3& v_2\\ v_3& 0& -v_1\\ -v_2& v_1& 0\end{array}\right)=[X,\left(\begin{array}{ccc}0& -v_3& v_2\\ v_3& 0& -v_1\\ -v_2& v_1& 0\end{array}\right)]$$

where $X$ is ${}_{\text{ a 3 }}$ by 3 antisymmetric matrix.

One can explicitly check that these representations are isomorphic, for instance by calculating how basis elements $l_j\in \mathfrak{so}(3)$ act. On vectors, these $l_j$ act by matrix multiplication, giving for instance, for $j=1$

$$l_1{\mathbf{e}}_1=0,l_1{\mathbf{e}}_2={\mathbf{e}}_3,l_1{\mathbf{e}}_3=-{\mathbf{e}}_2$$

On antisymmetric matrices one has instead the isomorphic relations

$$(ad(l_1))(l_1)=0,(ad(l_1))(l_2)=l_3,(ad(l_1))(l_3)=-l_2$$

Chapter contents · Next section: 6.2 Spin groups in three and four dimensions · 中文版本

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