6.2 Spin groups in three and four dimensions

6.2 Spin groups in three and four dimensions

Concept links · terms present in this machine draft; source roles are unverified: orthogonal group · Lie algebra · Lie group · group homomorphism

A subtle and remarkable property of the orthogonal groups $SO(n)$ is that they come with an associated group, called $Spin(n)$ , with every element of $SO(n)$ corresponding to two distinct elements of $Spin(n)$ . There is a surjective group homomorphism

$$\Phi :Spin(n)\to SO(n)$$

with the inverse image of each element of $SO(n)$ given by two distinct elements of $Spin(n)$

Digression. The topological reason for this is that, $(forn>2)$ the fundamental group of $SO(n)$ is non-trivial, with ${\pi }_1(SO(n))={\mathbf{Z}}_2$ (in particular there is a non-contractible loop in $SO(n)$ , contractible if you go around it twice). $Spin(n)$ is topologically the simply-connected double cover of $SO(n)$ , and the covering map $\Phi :Spin(n)\to SO(n)$ can be chosen to be a group homomorphism.

$Spin(n)$ is a Lie group of the same dimension as $SO(n)$ , with an isomorphic tangent space at the identity, so the Lie algebras of the two groups are isomorphic: $\mathfrak{so}(n)\simeq \mathfrak{spin}(n)$

In chapter 29 we will explicitly construct the groups $Spin(n)$ for any $n_{:}$ but here we will only do this for $n=3$ and $n=4$ , using methods specific to these two cases. In the cases $n=5$ (where $Spin(5)=Sp(2)$ , the 2 by 2 norm-preserving quaternionic matrices) and $n=6$ (where $Spin(6)=SU(4))$ special methods can be used to identify $Spin(n)$ with other matrix groups. For $n>6$ the group $Spin(n)$ will be a matrix group, but distinct from other classes of such groups.

Given such a construction of $Spin(n)$ , we also need to explicitly construct the homomorphism $\Phi$, and show that its derivative ${\Phi }^{\prime }$ is an isomorphism of Lie algebras. We will see that the simplest construction of the spin groups here uses the group $Sp(1)$ of unit-length quaternions, with $Spin(3)=Sp(1)$ and $Spin(4)=Sp(1)\times Sp(1)$ . By identifying quaternions and pairs of complex numbers, we can show that $Sp(1)=SU(2)$ and thus work with these spin groups as either 2 by 2 complex matrices (for $Spin(3))$ ), or pairs of such matrices (for $Spin(4))$ .

6.2.1 Quaternions

Concept links · terms present in this machine draft; source roles are unverified: vector space · Lie group

The quaternions are a number system (denoted by $H$) generalizing the complex number system, with elements $q\in \mathbf{H}$ that can be written as

$$q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k},q_j\in \mathbf{R}$$

with $i$, $j$, $k$ $\eta \in \mathbf{H}$ satisfying

$${\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2=-1,\mathbf{ij}=-\mathbf{ji}=\mathbf{k},\mathbf{ki}=-\mathbf{ik}=\mathbf{j},\mathbf{jk}=-\mathbf{kj}=\mathbf{i}$$

and a conjugation operation that takes

$$q\to \bar{q}=q_0-q_1\mathbf{i}-q_2\mathbf{j}-q_3\mathbf{k}$$

This operation satisfies (for $u,v\in \mathbf{H})$

$$\overline{uv}=\bar{v}\bar{u}$$

As a vector space over $R$, $H$ is isomorphic with ${\mathbf{R}}^4$ . The length-squared function on this ${\mathbf{R}}^4$ can be written in terms of quaternions as

$$|q{|}^2=q\bar{q}=q_0^2+q_1^2+q_2^2+q_3^2$$

and is multiplicative since

$$|uv{|}^2=uv\overline{uv}=uv\bar{v}\bar{u}=|u{|}^2|v{|}^2$$

Using

$$\frac{q\bar{q}}{|q{|}^2}=1$$

one has a formula for the inverse of a quaternion

$$q^{-1}=\frac{\bar{q}}{|q{|}^2}$$

The length one quaternions thus form a group under multiplication, called $Sp(1)$ . There are also Lie groups called $Sp(n)$ for larger values of $n,$ consisting of invertible matrices with quaternionic entries that act on quaternionic vectors preserving the quaternionic length-squared, but these play no significant role in quantum mechanics so we won’t study them further. $Sp(1)$ can be identified with the three dimensional sphere since the length one condition on $q$ is

$$q_0^2+q_1^2+q_2^2+q_3^2=1$$

the equation of the unit sphere $S^3\subset {\mathbf{R}}^4$

6.2.2 Rotations and spin groups in four dimensions

Concept links · terms present in this machine draft; source roles are unverified: linear map · Lie group

Pairs $(u,v)$ of unit quaternions give the product group $Sp(1)\times Sp(1)$ . An element $(u,v)$ of this group acts on $q\in \mathbf{H}={\mathbf{R}}^4$ by left and right quaternionic multiplication

$$q\to uq{v}^{-1}$$

This action preserves lengths of vectors and is linear in $q,$ so it must correspond to an element of the group $SO(4)$ . One can easily see that pairs $(u,v)$ and $\left(-u,-v\right)$ give the same linear transformation of ${\mathbf{R}}^4$ , so the same element of $SO(4)$ and show that $SO(4)$ is the group $Sp(1)\times Sp(1)$ , with the two elements $(u,v)$ and $\left(-u,-v\right)$ identified. The name $Spin(4)$ is given to the Lie group $Sp(1)\times Sp(1)$ that “double covers” $SO(4)$ in this manner, with the covering map

$$\Phi :(u,v)\in Sp(1)\times Sp(1)=Spin(4)\to \{q\to uq{v}^{-1}\}\in SO(4)$$

6.2.3 Rotations and spin groups in three dimensions

Concept links · terms present in this machine draft; source roles are unverified: vector space · linear map · Lie algebra · U(1) · Lie bracket

Later on we’ll encounter $Spin(4)$ and $SO(4)$ again, but for now we’re interested in the subgroup $Spin(3)$ that only acts non-trivially on 3 of the dimensions, and double covers not $SO(4)$ but $SO(3)$ . To find this, consider the subgroup of $Spin(4)$ consisting of pairs $(u,v)$ of the form $(u,u)$ (a subgroup isomorphic to $Sp(1)$ , since elements correspond to a single unit length quaternion $u$). This subgroup acts on quaternions by conjugation

$$q\to uq{u}^{-1}$$

an action which is trivial on the real quaternions (since $u(q_{0}1)u^{-1}=q_{0}1)$ . It preserves and acts nontrivially on the space of “pure imaginary” quaternions of the form

$$q=\vec{v}=v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k}$$

which can be identified with the vector space ${\mathbf{R}}^3$ . An element $u\in Sp(1)$ acts on $\vec{v}\in {\mathbf{R}}^3\subset \mathbf{H}$ as

$$\vec{v}\to u\vec{v}{u}^{-1}$$

This is a linear action, preserving the length |⃗v|, so it corresponds to an element of $SO(3)$ . We thus have a map (which can easily be checked to be a homomorphism)

$$\Phi :u\in Sp(1)\to \{\vec{v}\to u\vec{v}{u}^{-1}\}\in SO(3)$$

Figure 6.1: Double cover $Sp(1)\to SO(3)$

Both u and −$u$ act in the same way on $\vec{v},$ so we have two elements in $Sp(1)$ corresponding to the same element in SO(3). One can show that $\Phi$ is a surjective map (any element of SO(3) is $\Phi$ of something), so it is what is called a “covering” map, specifically a two-fold cover. It makes $Sp(1)$ a double cover of $SO(3)$ , and we give this group the name ${}^{\mathfrak{a}}Spin(3{)}^{\mathfrak{n}}$ . This also allows us to characterize more simply $SO(3)$ as a geometrical space. It is $S^3=Sp(1)=Spin(3)$ with opposite points on the three-sphere identified. This space is known as $\mathbf{R}{\dot{\mathbf{P}}}^3$ , real projective 3-space, which can also be thought of as the space of lines through the origin in ${\mathbf{R}}^4$ (each such line intersects $S^3$ in two opposite points).

Digression. The covering map $\Phi$ is an example of a topologically non-trivial cover. Topologically, it is not true that $S^3\simeq {\mathbf{RP}}^3\times (+1,-1)$. $S^3$ is a connected space, not two disconnected pieces. This topological non-triviality implies that globally there is no possible homomorphism going in the opposite direction from $\Phi$ $(i.e.,SO(3)\to Spin(3))$ . This can be done locally, picking a local patch in $SO(3)$ and taking the inverse of $\Phi$ to a local patch in $Spin(3)$ , but this won’t work if we try and extend it globally to all of $SO(3)$

The identification ${\mathbf{R}}^2=\mathbf{C}$ allowed us to represent elements of the unit circle group $U(1)$ as exponentials $e^{i\theta }$ , where $i\theta$ was in the Lie algebra $\mathfrak{u}(1)=i\mathbf{R}$ of U(1). $Sp(1)$ behaves in much the same way, with the Lie algebra sp(1) now the space of all pure imaginary quaternions, which can be identified with ${\mathbf{R}}^3$ by

$$\mathbf{w}=\left(\begin{array}{c}{w}_1\\ w_2\\ w_3\end{array}\right)\in {\mathbf{R}}^3\leftrightarrow \vec{w}=w_1\mathbf{i}+w_2\mathbf{j}+w_3\mathbf{k}\in \mathbf{H}$$

Unlike the $U(1)$ case, there’s a non-trivial Lie bracket, the commutator of quaternions.

Elements of the group $Sp(1)$ are given by exponentiating such Lie algebra elements, which we will write in the form

$$u(\theta ,\mathbf{w})=e^{\theta \vec{w}}$$

where $\theta \in \mathbf{R}$ and $\vec{w}$ is a purely imaginary quaternion of unit length. Since

$${\vec{w}}^2=(w_1\mathbf{i}+w_2\mathbf{j}+w_3\mathbf{k}{)}^2=-(w_1^2+w_2^2+w_3^2)=-1$$

the exponential can be expanded to show that

$$e^{\theta \vec{w}}=\cos \theta +\vec{w}\sin \theta$$

Taking $\theta$ as a parameter, the $u(\theta ,\mathbf{w})$ give paths in $Sp(1)$ going through the identity at $\theta =0$ , with velocity vector ⃗w since

$$\frac{d}{d\theta }u(\theta ,\mathbf{w}{)}_{|\theta =0}=(-\sin \theta +\vec{w}\cos \theta {)}_{|\theta =0}=\vec{w}$$

We can explicitly evaluate the homomorphism $\Phi$ on such elements $u(\mathbf{\theta },\mathbf{w})\in$ $Sp(1)$ , with the result that $\Phi$ takes $u(\theta ,\mathbf{w})$ to a rotation by an angle $2\theta$ around the axis w:

Theorem 6.1.

$$\Phi (u(\theta ,\mathbf{w}))=R(2\theta ,\mathbf{w})$$

Proof

First consider the special case $\mathbf{w}={\mathbf{e}}_3$ of rotations about the 3-axis.

$$u(\theta ,{\mathbf{e}}_3)=e^{\theta \mathbf{k}}=\cos \theta +\mathbf{k}\sin \theta$$

and

$$u(\theta ,{\mathbf{e}}_3{)}^{-1}=e^{-\theta \mathbf{k}}=\cos \theta -\mathbf{k}\sin \theta$$

so $\Phi (u(\theta ,{\mathbf{e}}_3))$ is the rotation that takes v (identified with the quaternion $\vec{v}=$ $v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k})$ to

$$\begin{array}{rl}u(\theta ,{\mathbf{e}}_3)\vec{v}u(\theta ,{\mathbf{e}}_3{)}^{-1}& =(\cos \theta +\mathbf{k}\sin \theta )(v_1\mathbf{i}+v_2\mathbf{j}+v_3\mathbf{k})(\cos \theta -\mathbf{k}\sin \theta )\\ & =(v_1({\cos }^2\theta -{\sin }^2\theta )-v_2(2\sin \theta \cos \theta ))\mathbf{i}\\ & +(2v_1\sin \theta \cos \theta +v_2({\cos }^2\theta -{\sin }^2\theta ))\mathbf{j}+v_3\mathbf{k}\\ & =(v_1\cos 2\theta -v_2\sin 2\theta )\mathbf{i}+(v_1\sin 2\theta +v_2\cos 2\theta )\mathbf{j}+v_3\mathbf{k}\end{array}$$

This is the orthogonal transformation of ${\mathbf{R}}^3$ given by

$$\mathbf{v}=\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right)\to \left(\begin{array}{ccc}\cos 2\theta & -\sin 2\theta & 0\\ \sin 2\theta & \cos 2\theta & 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{c}{v}_1\\ v_2\\ v_3\end{array}\right)\text{\hspace{1em}(6.2)}$$

The same calculation can readily be done for the case of ${\mathbf{e}}_1$ , then use the Euler angle parametrization of equation 6.1 to show that a general $u(\theta ,\mathbf{w})$ can be written as a product of the cases already worked out. □

Notice that as $\theta$ goes from 0 to $2\pi ,u(\theta ,\mathbf{w})$ traces out a circle in $Sp(1)$ . The homomorphism $\Phi$ takes this to a circle in $SO(3)$ , one that gets traced out twice as $\theta$ goes from 0 to $2\pi$ , explicitly showing the nature of the double covering above that particular circle in $SO(3)$ .

The derivative of the map $\Phi$ will be a Lie algebra homomorphism, a linear map

$${\Phi }^{\prime }:\mathfrak{sp}(1)\to \mathfrak{so}(3)$$

It takes the Lie algebra $\mathfrak{sp}(1)$ of pure imaginary quaternions to the Lie algebra so(3) of 3 by 3 antisymmetric real matrices. One can compute it easily on basis vectors, using for instance equation 6.2 above to find for the case $\vec{w}=$ k

$$\begin{array}{rl}{\Phi }^{\prime }(\mathbf{k})& =\frac{d}{d\theta }\Phi (\cos \theta +\mathbf{k}\sin \theta {)}_{|\theta =0}\\ & ={\left(\begin{array}{ccc}-2\sin 2\theta & -2\cos 2\theta & 0\\ 2\cos 2\theta & -2\sin 2\theta & 0\\ 0& 0& 0\end{array}\right)}_{|\theta =0}\\ & =\left(\begin{array}{ccc}0& -2& 0\\ 2& 0& 0\\ 0& 0& 0\end{array}\right)=2l_3\end{array}$$

Repeating this on other basis vectors one finds that

$${\Phi }^{\prime }(\mathbf{i})=2l_1,{\Phi }^{\prime }(\mathbf{j})=2l_2,{\Phi }^{\prime }(\mathbf{k})=2l_3$$

Thus ${\Phi }^{\prime }$ is an isomorphism of $\mathfrak{sp}(1)$ and so(3) identifying the bases

$$\frac{\mathbf{i}}{2},\frac{\mathbf{j}}{2},\frac{\mathbf{k}}{2}\mathrm{and}{l}_1,l_2,l_3$$

Note that it is the $\frac{\mathbf{i}}{2},\frac{\mathbf{j}}{2},\frac{\mathbf{k}}{2}$ that satisfy simple commutation relations

$$\left[\frac{\mathbf{i}}{2},\frac{\mathbf{j}}{2}\right]=\frac{\mathbf{k}}{2},\left[\frac{\mathbf{j}}{2},\frac{\mathbf{k}}{2}\right]=\frac{\mathbf{i}}{2},\left[\frac{\mathbf{k}}{2},\frac{\mathbf{i}}{2}\right]=\frac{\mathbf{j}}{2}$$

6.2.4 The spin group and SU(2)

Concept links · terms present in this machine draft; source roles are unverified: vector space · Lie algebra · Pauli matrices · Lie group · complexification

Instead of doing calculations using quaternions with their non-commutativity and special multiplication laws, it is more conventional to choose an isomorphism between quaternions $H$ and a space of 2 by 2 complex matrices, and work with matrix multiplication and complex numbers. The Pauli matrices can be used to give such an isomorphism, taking

$$1\to 1=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),\mathbf{i}\to -i{\sigma }_1=\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right),\mathbf{j}\to -i{\sigma }_2=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$$ $$\mathbf{k}\to -i{\sigma }_3=\left(\begin{array}{cc}-i& 0\\ 0& i\end{array}\right)$$

The correspondence between H and 2 by 2 complex matrices is then given by

$$q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}\leftrightarrow \left(\begin{array}{cc}{q}_0-i{q}_3& -q_2-i{q}_1\\ q_2-i{q}_1& q_0+i{q}_3\end{array}\right)$$

Since

$$det\left(\begin{array}{cc}{q}_0-i{q}_3& -q_2-i{q}_1\\ q_2-i{q}_1& q_0+i{q}_3\end{array}\right)=q_0^2+q_1^2+q_2^2+q_3^2$$

we see that the length-squared function on quaternions corresponds to the determinant function on 2 by 2 complex matrices. Taking $q\in Sp(1)$ , so of length one, the corresponding complex matrix is in $SU(2)$

Under this identification of H with 2 by 2 complex matrices, we have an identification of Lie algebras sp $(1)=\mathfrak{su}(2)$ between pure imaginary quaternions and skew-Hermitian trace-zero 2 by 2 complex matrices

$$\vec{w}=w_1\mathbf{i}+w_2\mathbf{j}+w_3\mathbf{k}\leftrightarrow \left(\begin{array}{cc}-i{w}_3& -w_2-i{w}_1\\ w_2-i{w}_1& i{w}_3\end{array}\right)=-i\mathbf{w}\cdot \mathbf{\sigma }$$

The basis $\frac{\mathbf{i}}{2},\frac{\mathbf{j}}{2},\frac{\mathbf{k}}{2}$ of $\mathfrak{sp}(1)$ gets identified with a basis for the Lie algebra $\mathfrak{su}(2)$ which written in terms of the Pauli matrices is

$$X_j=-i\frac{{\sigma }_j}{2}$$

with the $X_j$ satisfying the commutation relations

$$[X_1,X_2]=X_3,[X_2,X_3]=X_1,[X_3,X_1]=X_2$$

which are precisely the same commutation relations as for so(3)

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

We now have three isomorphic Lie algebras $\mathfrak{sp}(1)=\mathfrak{su}(2)=\mathfrak{so}(3)$ , with elements that get identified as follows

$$\left(w_1\frac{\mathbf{i}}{2}+w_2\frac{\mathbf{j}}{2}+w_3\frac{\mathbf{k}}{2}\right)\leftrightarrow -\frac{i}{2}\left(\begin{array}{cc}{w}_3& w_1-i{w}_2\\ w_1+i{w}_2& -w_3\end{array}\right)\leftrightarrow \left(\begin{array}{ccc}0& -w_3& w_2\\ w_3& 0& -w_1\\ -w_2& w_1& 0\end{array}\right)$$

This isomorphism identifies basis vectors by

$$\frac{\mathbf{i}}{2}\leftrightarrow -i\frac{{\sigma }_1}{2}\leftrightarrow l_1$$

etc. The first of these identifications comes from the way we chose to identify H with 2 by 2 complex matrices. The second identification is ${\Phi }^{\prime }$ , the derivative at the identity of the covering map $\Phi$.

On each of these isomorphic Lie algebras we have adjoint Lie group (Ad) and Lie algebra $(ad)$ representations. $Ad$ is given by conjugation with the corresponding group elements in $Sp(1),SU(2)$ and $SO(3)$ . ad is given by taking commutators in the respective Lie algebras of pure imaginary quaternions, skew-Hermitian trace-zero 2 by 2 complex matrices and 3 by 3 real antisymmetric matrices.

Note that these three Lie algebras are all three dimensional real vector spaces, so these are real representations. To get a complex representation, take complex linear combinations of elements. This is less confusing in the case of $\mathfrak{su}(2)$ than for $\mathfrak{sp}(1)$ since taking complex linear combinations of skew-Hermitian trace-zero 2 by 2 complex matrices gives all trace-zero 2 by 2 matrices (the Lie algebra $\mathfrak{sl}(2,\mathbf{C}))$ ).

In addition, recall that there is a fourth isomorphic version of this representation, the representation of $SO(3)$ on column vectors. This is also a real representation, but can straightforwardly be complexified. Since $\mathfrak{so}(3)$ and $\mathfrak{su}(2)$ are isomorphic Lie algebras, their complexifications so $(3{)}_{\mathbf{C}}$ and $\mathfrak{sl}(2,\mathbf{C})$ will also be isomorphic.

In terms of 2 by 2 complex matrices, Lie algebra elements can be exponentiated to get group elements in $SU(2)$ and define

$$\Omega (\theta ,\mathbf{w})=e^{\theta (w_1{X}_1+w_2{X}_2+w_3{X}_3)}=e^{-i\frac{\theta }{2}\mathbf{w}\cdot \mathbf{\sigma }}\text{\hspace{1em}(6.3)}$$

$$=1\cos \frac{\theta }{2}-i(\mathbf{w}\cdot \sigma )\sin \frac{\theta }{2}\text{\hspace{1em}(6.4)}$$

Transposing the argument of theorem 6.1 from H to complex matrices, one finds that, identifying

$$\mathbf{v}\leftrightarrow \mathbf{v}\cdot \mathbf{\sigma }=\left(\begin{array}{cc}{v}_3& v_1-i{v}_2\\ v_1+i{v}_2& -v_3\end{array}\right)$$

one has

$$\Phi (\Omega (\theta ,\mathbf{w}))=R(\theta ,\mathbf{w})$$

with $\Omega (\theta ,\mathbf{w})$ acting by conjugation, taking

$$\mathbf{v}\cdot \mathbf{\sigma }\to \Omega (\theta ,\mathbf{w})(\mathbf{v}\cdot \mathbf{\sigma })\Omega (\theta ,\mathbf{w}{)}^{-1}=(R(\theta ,\mathbf{w})\mathbf{v})\cdot \mathbf{\sigma }\text{\hspace{1em}(6.5)}$$

Note that in changing from the quaternionic to complex case, we are treating the factor of 2 diferently, since in the future we will want to use $\Omega (\theta ,\mathbf{w})$ to perform rotations by an angle $\theta$. In terms of the identification $SU(2)=Sp(1)$ , we have $\Omega (\theta ,\mathbf{w})=u(\frac{\theta }{2},\mathbf{w})$

Recall that any $SU(2)$ matrix can be written in the form

$$\begin{array}{c}\left(\begin{array}{cc}\alpha & \beta \\ -\overline{\beta }& \overline{\alpha }\end{array}\right)\\ \alpha =q_0-i{q}_3,\beta =-q_2-i{q}_1\end{array}$$

with $\alpha ,\beta \in \mathbf{C}$ arbitrary complex numbers satisfying $|\alpha {|}^2+|\beta {|}^2=1$ . A somewhat unenlightening formula for the map $\Phi :SU(2)\to SO(3)$ in terms of such explicit $SU(2)$ matrices is given by

$$\Phi \left(\begin{array}{cc}\alpha & \beta \\ -\overline{\beta }& \overline{\alpha }\end{array}\right)=\left(\begin{array}{ccc}\mathrm{Re}({\alpha }^2-{\beta }^2)& \mathrm{Im}({\alpha }^2+{\beta }^2)& -2\mathrm{Re}(\alpha \beta )\\ -\mathrm{Im}({\alpha }^2-{\beta }^2)& \mathrm{Re}({\alpha }^2+{\beta }^2)& 2\mathrm{Im}(\alpha \beta )\\ 2\mathrm{Re}(\alpha \overline{\beta })& 2\mathrm{Im}(\alpha \overline{\beta })& |\alpha {|}^2-|\beta {|}^2\end{array}\right)$$

See [83], page 123-4, for a derivation.

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