6.5 Lie algebras and Lie groups

中文

An important class of non-associative algebras is due to the Norwegian mathematician Sophus Lie (1842–1899). Lie’s work on transformations of surfaces (known as contact transformations) gave rise to a class of continuous groups now known as Lie groups. These encompass essentially all the important groups that appear in mathematical physics, such as the orthogonal, unitary and symplectic groups. This subject is primarily a branch of differential geometry and a detailed discussion will appear in Chapter 19. Lie’s principal discovery was that Lie groups were related to a class of non-associative algebras that in turn are considerably easier to classify. These algebras have come to be known as Lie algebras. A more complete discussion of Lie algebra theory and, in particular, the Cartan–Dynkin classification for semisimple Lie algebras can be found in [3–5].

Lie algebras

A Lie algebra is a real or complex vector space with a law of composition or bracket product $[X,Y]$ satisfying

(LA1) $[X,Y]=-[Y,X](\mathrm{antisymmetry})$.

(LA2) $[X,aY+bZ]=a[X,Y]+b[X,Z]$ (distributive law).

(LA3) $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$ (Jacobi identity).

By (LA1) the bracket product is also distributive on the first argument,

$$[aX+bY,Z]=-[Z,aX+bY]=-a[Z,X]-b[Z,Y]=a[X,Z]+b[Y,Z].$$

Lie algebras are therefore algebras in the general sense, since Eq. (6.1) holds for the bracket product. The Jacobi identity replaces the associative law.

Example 6.7

Any associative algebra, such as the algebra ${\mathcal{M}}_n$ of $n\times n$ matrices discussed in Example 6.2, can be converted to a Lie algebra by defining the bracket product to be the commutator of two elements

$$[X,Y]=XY-YX.$$

Conditions (LA1) and (LA2) are trivial to verify, while the Jacobi identity (LA3) is straightforward:

$$\begin{array}{l}[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]\\ =X(YZ-ZY)-(YZ-ZY)X+Y(ZX-XZ)\\ -(ZX-XZ)Y+Z(XY-YX)-(XY-YX)Z\\ =XYZ-XZY-YZX+ZYX+YZX-YXZ\\ -ZXY+XZY+ZXY-ZYX-XYZ+YXZ\\ =0.\end{array}$$

The connection between brackets and commutators motivates the terminology that if a Lie algebra $\mathcal{L}$ has all bracket products vanishing, $[X,Y]=0$ for all $X$, $Y\in \mathcal{L}$ , it is said to be abelian.

Given a basis $X_1,\dots ,X_n$ of $\mathcal{L}$, let $C_{ij}^k=-C_{ji}^k$ be the structure constants with respect to this basis,

$$[X_i,X_j]=C_{ij}^k{X}_k.\text{\hspace{1em}(6.21)}$$

Given the structure constants, it is possible to calculate the bracket product of any pair of vectors $A=a^i{X}_i$ and $B=b^j{X}_j$ :

$$[A,B]=a^i{b}^j[X_i,X_j]=a^i{b}^j{C}_{ij}^k{X}_k.\text{\hspace{1em}(6.22)}$$

A Lie algebra is therefore abelian if and only if all its structure constants vanish. It is important to note that structure constants depend on the choice of basis and are generally different in another basis $X_i^{\prime }={A^{\prime }}_i^j{X}_j$

Example 6.8

Consider the set ${\mathcal{T}}_2$ of real upper triangular $2\times 2$ matrices, having the form

$$\mathsf{A}=\left(\begin{array}{cc}a& b\\ 0& c\end{array}\right).$$

Since

$$\left[\left(\begin{array}{cc}a& b\\ 0& c\end{array}\right),\left(\begin{array}{cc}d& e\\ 0& f\end{array}\right)\right]=\left(\begin{array}{cc}0& ae+bf-bd-ce\\ 0& 0\end{array}\right),\text{\hspace{1em}(6.23)}$$

these matrices form a Lie algebra with respect to the commutator product. The following three matrices form a basis of this Lie algebra:

$${\mathsf{X}}_1=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),{\mathsf{X}}_2=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),{\mathsf{X}}_3=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right),$$

having the commutator relations

$$[{\mathrm{X}}_1,{\mathrm{X}}_2]=-{\mathrm{X}}_1,[{\mathrm{X}}_1,{\mathrm{X}}_3]={\mathrm{X}}_1,[{\mathrm{X}}_2,{\mathrm{X}}_3]=\mathrm{O}.$$

The corresponding structure constants are

$$C_{12}^1=-C_{21}^1=-1,C_{13}^1=-C_{31}^1=1,\text{ all other }{C}_{jk}^i=0.$$

Consider a change of basis to

$${\mathbf{X}}_1^{\prime }={\mathbf{X}}_2+{\mathbf{X}}_3=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),$$ $${\mathbf{X}}_2^{\prime }={\mathbf{X}}_1+{\mathbf{X}}_3=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right),$$ $${\mathsf{X}}_3^{\prime }={\mathsf{X}}_1+{\mathsf{X}}_2=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right).$$

The commutation relations are

$$[{\mathrm{X}}_1^{\prime },{\mathrm{X}}_2^{\prime }]=\mathrm{O},[{\mathrm{X}}_1^{\prime },{\mathrm{X}}_3^{\prime }]=\mathrm{O},[{\mathrm{X}}_2^{\prime },{\mathrm{X}}_3^{\prime }]=-2{\mathrm{X}}_1={\mathrm{X}}_1^{\prime }-{\mathrm{X}}_2^{\prime }-{\mathrm{X}}_3^{\prime },$$

with corresponding structure constants

$$C_{23}^{\prime 1}=-C_{23}^{\prime 2}=-C_{23}^{\prime 3}=1,\text{ all other }{C}_{jk}^{\prime i}=0.$$

As before, an ideal of a Lie algebra $\mathcal{L}$ is a subset $\mathcal{I}$ such that $[X,Y]\in \mathcal{I}$ for all $X\in \mathcal{L}$ and all $Y\in \mathcal{I}$ , a condition written more briefly as $[\mathcal{L},\mathcal{I}]\subseteq \mathcal{I}$ . From (LA1) it is clear tha any right or left ideal must be two-sided. If $\mathcal{I}$ is an ideal of $\mathcal{L}$ it is possible to form a factor Lie algebra on the space of cosets $X+\mathcal{I}$ , with bracket product defined by

$$[X+\mathcal{I},Y+\mathcal{I}]=[X,Y]+\mathcal{I}.$$

This product is independent of the choice of representative from the cosets $X+\mathcal{I}$ and $Y+\mathcal{I}$ , since

$$[X+\mathcal{I},Y+\mathcal{I}]=[X,Y]+[X,\mathcal{I}]+[\mathcal{I},Y]+[\mathcal{I},\mathcal{I}]\subseteq [X,Y]+\mathcal{I}.$$

Example 6.9

In Example 6.8 let $\mathcal{B}$ be the subset of matrices of the form

$$\left(\begin{array}{cc}0& x\\ 0& 0\end{array}\right).$$

From the product rule Eq. (6.23) it follows that $\mathcal{B}$ forms an ideal of ${\mathcal{T}}_2$ . Every coset $\mathsf{X}+\mathbf{B}$ clearly has a diagonal representative

$$\mathsf{X}=\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right),$$

and since diagonal matrices always commute with each other, the factor algebra is abelian,

$$[\mathsf{X}+\mathcal{B},\mathsf{Y}+\mathcal{B}]=\mathsf{O}+\mathcal{B}.$$

The linear map $\phi :{\mathcal{T}}_2\to {\mathcal{T}}_2/\mathcal{B}$ defined by

$$\phi :\left(\begin{array}{cc}a& x\\ 0& b\end{array}\right)⟼\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)+\mathcal{B}$$

is a Lie algebra homomorphism, since by Eq. (6.23) it follows that $[\mathsf{X},\mathsf{Y}]\in B$ for any pair of upper triangular matrices $X$ and $Y$, and

$$\phi ([\mathsf{X},\mathsf{Y}])=\mathsf{O}+\mathcal{B}=[\phi (\mathsf{X}),\phi (\mathsf{Y})].$$

The kernel of the homomorphism $\phi$ consists of those matrices having zero diagonal elements,

$$\ker \phi =\mathcal{B}.$$

This example is an illustration of Theorem 6.1.

Matrix Lie groups

In Chapter 19 we will give a rigorous definition of a Lie group, but for the present purpose we may think of a Lie group as a group whose elements depend continuously on $n$ real parameters ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ . For simplicity we will assume the group to be a matrix group, whose elements can typically be written as

$$\mathsf{A}=\Gamma ({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n).$$

The identity is taken to be the element corresponding to the origin ${\lambda }_1={\lambda }_2=\cdot \cdot \cdot ={\lambda }_n=0$

$$\mathsf{I}=\Gamma (0,0,\dots ,0).$$

Example 6.10

The general member of the rotation group is an orthogonal $3\times 3$ matrix $A$ that can be written in terms of three angles $\psi ,\theta ,\varphi$.

$$\mathsf{A}=\left(\begin{array}{ccc}\cos \varphi \cos \psi & \cos \varphi \sin \psi & -\sin \varphi \\ \sin \theta \sin \varphi \cos \psi -\cos \theta \sin \psi & \sin \theta \sin \varphi \sin \psi +\cos \theta \cos \psi & \sin \theta \cos \varphi \\ \cos \theta \sin \varphi \cos \psi +\sin \theta \sin \psi & \cos \theta \sin \varphi \sin \psi -\sin \theta \cos \psi & \cos \theta \cos \varphi \end{array}\right).$$

As required, the identity element $I$ corresponds to $\theta =\varphi =\psi =0$ . These angles are similar but not identical to the standard Euler angles of classical mechanics, which have an unfortunate degeneracy at the identity element. Group elements near the identity have the form $\mathsf{A}=\mathsf{I}+ϵ\mathsf{X}\left(ϵ\ll 1\right)$ , where

$$\mathsf{I}={\mathsf{AA}}^T=(\mathsf{I}+ϵ\mathsf{X})(\mathsf{I}+ϵ{\mathsf{X}}^T)=\mathsf{I}+ϵ(\mathsf{X}+{\mathsf{X}}^T)+O({ϵ}^2).$$

If we only keep terms to first order in this equation then $X$ must be antisymmetric; $\mathsf{X}=-{\mathsf{X}}^T$

Although the product of two antisymmetric matrices is not in general antisymmetric, the set of $n\times n$ antisymmetric matrices is closed with respect to commutator products and forms a Lie algebra:

$$[\mathrm{X},\mathrm{Y}{]}^T=(\mathrm{XY}-\mathrm{YX}{)}^T={\mathrm{Y}}^T{\mathrm{X}}^T-{\mathrm{X}}^T{\mathrm{Y}}^T=(-\mathrm{Y})(-\mathrm{X})+\mathrm{X}(-\mathrm{Y})=\mathrm{YX}-\mathrm{XY}=-[\mathrm{X},\mathrm{Y}].$$

The Lie algebra of $3\times 3$ antisymmetric matrices may be thought of as representing ‘infinitesimal rotations’, or orthogonal matrices ‘near the identity’. Every $3\times 3$ antisymmetric matrix $X$ can be written in the form

$$\mathsf{X}=\left(\begin{array}{ccc}0& x^3& -x^2\\ -x^3& 0& x^1\\ x^2& -x^1& 0\end{array}\right)=\sum _{i=1}^3{x}^i{\mathsf{X}}_i$$

where

$${\mathsf{X}}_1=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right),{\mathsf{X}}_2=\left(\begin{array}{ccc}0& 0& -1\\ 0& 0& 0\\ 1& 0& 0\end{array}\right),{\mathsf{X}}_3=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 0\\ 0& 0& 0\end{array}\right).\text{\hspace{1em}(6.24)}$$

The basis elements ${\mathsf{X}}_i$ are called infinitesimal generators of the group and satisfy the following commutation relations:

$$[{\mathsf{X}}_1,{\mathsf{X}}_2]=-{\mathsf{X}}_3,[{\mathsf{X}}_2,{\mathsf{X}}_3]=-{\mathsf{X}}_1,[{\mathsf{X}}_3,{\mathsf{X}}_1]=-{\mathsf{X}}_2.\text{\hspace{1em}(6.25)}$$

This example is typical of the procedure for creating a Lie algebra from the group elements ‘near the identity’. More generally, if $G$ is a matrix Lie group whose elements depend on $n$ continuous parameters

$$\mathsf{A}=\Gamma ({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)\text{ with }\mathsf{I}=\Gamma (0,0,\dots ,0),$$

define the infinitesimal generators by

$${\mathbf{X}}_i={\frac{\partial \Gamma }{\partial {\lambda }_i}|}_{\lambda =0}(\lambda \equiv ({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n)),\text{\hspace{1em}(6.26)}$$

so that elements near the identity can be written

$$\mathsf{A}=\mathsf{I}+\sum _{i=1}^nϵa^i{\mathsf{X}}_i+O({ϵ}^2).$$

The group structure of $G$ implies the commutators of the ${\mathsf{X}}_i$ are always linear combinations of the ${\mathsf{X}}_i$ , satisfying Eq. (6.21) for some structure constants $C_{ji}^k=-C_{ij}^k$ . The proof will be given in Chapter 19.

One-parameter subgroups

A one-parameter subgroup of a Lie group $G$ is the image $\phi (\mathbb{R})$ of a homomorphism $\phi :\mathbb{R}\to G$ of the additive group of real numbers into $G$. Writing the elements of a one parameter subgroup of a matrix Lie group simply as $\begin{array}{r}\mathbf{A}(t)=\Gamma (a^1(t),a^2(t),\dots ,a^n(t))\end{array}$ , the homomorphism property requires that

$$\mathsf{A}(t+s)=\mathsf{A}(t)\mathsf{A}(s).\text{\hspace{1em}(6.27)}$$

It can be shown that through every element $g$ in a neighbourhood of the identity of a Lie group there exists a one-parameter subgroup $\phi$ such that $g=\phi (1)$ .

Applying the operation ${\frac{\mathrm{d}}{\mathrm{d}s}|}_{s=0}$ to Eq. (6.27) results in

$${\frac{\mathrm{d}}{\mathrm{d}s}\mathsf{A}(t+s)|}_{s=0}={\frac{\mathrm{d}}{\mathrm{d}t}\mathsf{A}(t+s)|}_{s=0}=\mathsf{A}(t)\frac{\mathrm{d}\mathsf{A}(s)}{\mathrm{d}s}{|}_{s=0}.$$

Hence

$$\frac{\mathrm{d}}{\mathrm{d}t}\mathsf{A}(t)=\mathsf{A}(t)\mathsf{X}\text{\hspace{1em}(6.28)}$$

where

$$\begin{array}{rl}\mathsf{X}& ={\frac{\mathrm{d}A(s)}{\mathrm{d}s}|}_{s=0}={\sum }_i{\frac{\partial \Gamma (\lambda )}{\partial {\lambda }^i}|}_{\lambda =0}{\frac{\mathrm{d}{a}^i(s)}{\mathrm{d}s}|}_{s=0}\\ & ={\sum }_i{\frac{\mathrm{d}{a}^i}{\mathrm{d}s}|}_{s=0}{X}_i.\end{array}$$

The unique solution of the differential equation Eq. (6.28) that satisfies the boundary condition $\mathsf{A}(0)=\mathsf{I}$ is $\mathsf{A}(t)={\mathrm{e}}^{t\mathsf{X}}$ , where the exponential of the matrix $t\mathsf{X}$ is defined by the power series

$${\mathrm{e}}^{t\mathsf{X}}=\mathsf{I}+t\mathsf{X}+\frac{1}{2!}{t}^2{\mathsf{X}}^2+\frac{1}{3!}{t}^3{\mathsf{X}}^3+\dots$$

The group property ${\mathrm{e}}^{(t+s)\mathsf{X}}={\mathrm{e}}^{t\mathsf{X}}{\mathrm{e}}^{s\mathsf{X}}$ follows from the fact that if $A$ and $B$ are any pair of commuting matrices $\mathsf{AB}=\mathsf{BA}$ then ${\mathrm{e}}^{\mathsf{A}}{\mathrm{e}}^{\mathsf{B}}={\mathrm{e}}^{\mathsf{A}+\mathsf{B}}$

In a neighbourhood of the identity consisting of group elements all connected to the identity by one-parameter subgroups, it follows that any group element ${\mathsf{A}}_1$ can be written as the exponential of a Lie algebra element

$${\mathrm{A}}_1=\mathrm{A}(1)={\mathrm{e}}^{\mathrm{X}}=1+\mathrm{X}+\frac{1}{2!}{\mathrm{X}}^2+\frac{1}{3!}{\mathrm{X}}^3+\dots \text{ where }\mathrm{X}\in \mathcal{G}.$$

Given a Lie algebra, say by specifying its structure constants, it is possible to reverse this process and construct the connected neighbourhood of the identity of a unique Lie group. Since the structure constants are a finite set of numbers, as opposed to the complicated set of functions needed to specify the group products, it is generally much easier to classify Lie groups by their Lie algebras than by their group products.

Example 6.11

In Example 6.10 the one-parameter group ${\mathrm{e}}^{t{\mathsf{X}}_1}$ generated by the infinitesimal generator ${\mathsf{X}}_1$ is found by calculating the first few powers

$${\mathsf{X}}_1=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& -1& 0\end{array}\right),{\mathsf{X}}_1^2=\left(\begin{array}{ccc}0& 0& 0\\ 0& -1& 0\\ 0& 0& -1\end{array}\right),{\mathsf{X}}_1^3=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right)=-{\mathsf{X}}_1,$$

as all higher powers follow a simple rule

$${\mathsf{X}}_1^4=-{\mathsf{X}}_1^2,{\mathsf{X}}_1^5={\mathsf{X}}_1,{\mathsf{X}}_1^6={\mathsf{X}}_1^2,\text{etc.}$$

From the exponential expansion

$${\mathrm{e}}^{t{\mathsf{X}}_1}=\mathsf{I}+t{\mathsf{X}}_1+\frac{1}{2!}{t}^2{\mathsf{X}}_1^2+\frac{1}{3!}{t}^3{\mathsf{X}}_1^3+\dots$$

it is possible to calculate all components

$${\left({\mathrm{e}}^{t{\mathsf{X}}_1}\right)}_{11}=1$$ $${\left({\mathrm{e}}^{t{\mathrm{X}}_1}\right)}_{22}=1-\frac{t^2}{2!}+\frac{t^4}{4!}-\cdots =\cos t$$ $${\left({\mathrm{e}}^{t{X}_1}\right)}_{23}=0+t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots =\sin t,\text{ etc. }$$

Hence

$${\mathrm{e}}^{t{X}_1}=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos t& \sin t\\ 0& -\sin t& \cos t\end{array}\right),$$

which represents a rotation by the angle $t$ about the $x$-axis. It is straightforward to verify the one-parameter group law

$${\mathrm{e}}^{t{\mathsf{X}}_1}{\mathrm{e}}^{s{\mathsf{X}}_1}={\mathrm{e}}^{(t+s){\mathsf{X}}_1}.$$

Exercise

Show that ${\mathsf{e}}^{t{\mathsf{X}}_2}$ and ${\mathrm{e}}^{t{\mathsf{X}}_3}$ represent rotations by angle $t$ about the $y$-axis and $z$-axis respectively.

Complex Lie algebras

While most of the above discussion assumes real Lie algebras, it can apply equally to complex Lie algebras. As seen in Section 6.2, it is always possible to regard a complex vector space $\mathcal{G}$ as being a real space ${\mathcal{G}}^R$ of twice the number of dimensions, by simply restricting the field of scalars to the real numbers. In this way any complex Lie algebra of dimension $n$ can also be considered as being a real Lie algebra of dimension $2n$ . It is important to be aware of whether it is the real or complex version of a given Lie algebra that is in question.

Example 6.12

In Example 2.15 of Chapter 2 it was seen that the $2\times 2$ unitary matrices form a group $SU(2)$ . For unitary matrices near the identity, $\mathsf{U}=\mathsf{I}+ϵ\mathsf{A}$

$$\mathsf{I}={\mathsf{UU}}^{†}=\mathsf{I}+ϵ(\mathsf{A}+{\mathsf{A}}^{†})+O({ϵ}^2).$$

Hence $A$ must be anti-hermitian,

$$\mathsf{A}+{\mathsf{A}}^{†}=\mathsf{O}.$$

Special unitary matrices are required to have the further restriction that their determinant is 1,

$$\det \mathsf{U}=\left|\begin{array}{cc}1+ϵa_{11}& ϵa_{12}\\ ϵa_{12}& 1+ϵa_{22}\end{array}\right|=1+ϵ(a_{11}+a_{22})+O({ϵ}^2),$$

and the matrix $A$ must be trace-free as well as being anti-hermitian,

$$\mathsf{A}=\left(\begin{array}{cc}ic& b+ia\\ -b+ia& -ic\end{array}\right)(a,b,c\in \mathbb{R}).$$

Such matrices form a real Lie algebra, as they constitute a real vector space and are closed with respect to commutator product,

$$[\mathsf{A},{\mathsf{A}}^{\prime }]=\mathsf{A}{\mathsf{A}}^{\prime }-{\mathsf{A}}^{\prime }\mathsf{A}=\left(\begin{array}{cc}2i(b{a}^{\prime }-a{b}^{\prime })& 2(a{c}^{\prime }-c{a}^{\prime })+2i(c{b}^{\prime }-b{c}^{\prime })\\ -2(a{c}^{\prime }-c{a}^{\prime })+2i(c{b}^{\prime }-b{c}^{\prime })& -2i(b{a}^{\prime }-a{b}^{\prime })\end{array}\right).$$

Any trace-free anti-hermitian matrix may be cast in the form

$$\mathsf{A}=ia{\sigma }_1+ib{\sigma }_2+ic{\sigma }_3$$

where ${\sigma }_i$ are the Pauli matrices,

$${\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\text{\hspace{1em}(6.29)}$$

whose commutation relations are easily calculated,

$$[{\sigma }_1,{\sigma }_2]=2i{\sigma }_3,[{\sigma }_2,{\sigma }_3]=2i{\sigma }_1,[{\sigma }_3,{\sigma }_1]=2i{\sigma }_2.\text{\hspace{1em}(6.30)}$$

Although this Lie algebra consists of complex matrices, note that it is not a complex Lie algebra since multiplying an anti-hermitian matrix by a complex number does not in general result in an anti-hermitian matrix. However multiplying by real scalars does retain the anti-hermitian property. A basis for this Lie algebra is

$${\mathsf{X}}_1=\frac{1}{2}i{\sigma }_1,{\mathsf{X}}_2=\frac{1}{2}i{\sigma }_2,{\mathsf{X}}_3=\frac{1}{2}i{\sigma }_3,$$

and the general Lie algebra element $A$ has the form

$$\mathsf{A}=2a{\mathsf{X}}_1+2b{\mathsf{X}}_2+2c{\mathsf{X}}_3(a,b,c\in \mathbb{R}).$$

By Eq. (6.30) the commutation relations between the ${\mathsf{X}}_k$ are

$$[{\mathsf{X}}_1,{\mathsf{X}}_2]=-{\mathsf{X}}_3,[{\mathsf{X}}_2,{\mathsf{X}}_3]=-{\mathsf{X}}_1,[{\mathsf{X}}_3,{\mathsf{X}}_1]=-{\mathsf{X}}_2,$$

which shows that this Lie algebra is in fact isomorphic to the Lie algebra of the group of $3\times 3$ orthogonal matrices given in Example 6.10. Denoting these real Lie algebras by $SU(2)$ and $s\mathcal{O}(3)$ respectively, we have

$$\mathcal{S}\mathcal{U}(2)\cong \mathcal{S}\mathcal{O}(3).$$

However, the underlying groups are not isomorphic in this case, although there does exist a homomorphism $\phi :\mathrm{SU}(2)\to \mathrm{SO}(3)$ whose kernel consists of just the two elements $\pm \mathsf{I}$. This is the so-called spinor representation of the rotation group. Strictly speaking it is not a representation of the rotation group – rather, it asserts that there is a representation of $SU(2)$ as the rotation group in ${\mathbb{R}}^3$

Example 6.13

A genuinely complex Lie algebra is $\mathcal{SL}(2,\mathbb{C})$ , the Lie algebra of the group of $2\times 2$ complex unimodular matrices. As in the preceding example the condition of unimodularity, or having determinant 1, implies that the infinitesimal generators are tracefree,

$$\det (\mathsf{I}+ϵ\mathsf{A})=1⟹\mathrm{tr}\mathsf{A}=a_{11}+a_{22}=0.$$

The set of complex trace-free matrices form a complex Lie algebra since (a) it forms a complex vector space, and (b) it is closed under commutator products by Eq. (2.15),

$$\mathrm{tr}[\mathsf{A},\mathsf{B}]=\mathrm{tr}(\mathsf{AB})-\mathrm{tr}(\mathsf{BA})=0.$$

This complex Lie algebra is spanned by

$${{\rm Y}}_1=\frac{1}{2}i{\sigma }_1=\frac{1}{2}\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right),{{\rm Y}}_2=\frac{1}{2}i{\sigma }_2=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),{{\rm Y}}_3=\frac{1}{2}i{\sigma }_3=\frac{1}{2}\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right),$$

for if $\mathsf{A}=[A_{ij}]$ is trace-free then

$$\mathsf{A}=\alpha {\mathsf{Y}}_1+\beta {\mathsf{Y}}_2+\gamma {\mathsf{Y}}_3$$

where

$$\alpha =-i(A_{12}+A_{21}),\beta =A_{12}-A_{21},\gamma =-2i{A}_{11}.$$

The Lie algebra $\mathcal{SL}(2,\mathbb{C})$ is isomorphic as a complex Lie algebra to the Lie algebra $S\mathcal{O}(3,\mathbb{C})$ of infinitesimal complex orthogonal transformations. The latter Lie algebra is spanned, as a complex vector space, by the same matrices ${\mathsf{X}}_i$ defined in Eq. (6.26) to form a basis of the real Lie algebra $s\mathcal{O}(3)$ . Since, by Eq. (6.30), the commutation relations of the ${\mathsf{Y}}_i$ are

$$[{{\rm Y}}_1,{{\rm Y}}_2]=-{{\rm Y}}_3,[{{\rm Y}}_2,{{\rm Y}}_3]=-{{\rm Y}}_1,[{{\rm Y}}_3,{{\rm Y}}_1]=-{{\rm Y}}_2,$$

comparison with Eq. (6.25) shows that the linear map $\phi :\mathcal{SL}(2,\mathbb{C})\to \mathcal{S}\mathcal{O}(3,\mathbb{C})$ defined by $\phi ({\mathsf{Y}}_i)={\mathsf{X}}_i$ is a Lie algebra isomorphism.

However, as a real Lie algebra the story is quite different since the matrices ${\mathsf{Y}}_1,{\mathsf{Y}}_2$ and ${\mathsf{Y}}_3$ defined above are not sufficient to span $\mathcal{SL}(2,\mathbb{C}{)}^R$ . If we supplement them with the matrices

$${\mathsf{Z}}_1=\frac{1}{2}{\sigma }_1=\frac{1}{2}\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\mathsf{Z}}_2=\frac{1}{2}{\sigma }_2=\frac{1}{2}\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\mathsf{Z}}_3=\frac{1}{2}{\sigma }_3=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),$$

then every member of $\mathcal{SL}(2,\mathbb{C})$ can be written uniquely in the form

$$\mathsf{A}=a_1{\mathsf{Y}}_1+a_2{\mathsf{Y}}_2+a_3{\mathsf{Y}}_3+b_1{\mathsf{Z}}_1+b_2{\mathsf{Z}}_2+b_3{\mathsf{Z}}_3(a_i,b_i\in \mathbb{R})$$

where

$$\begin{array}{l}{b}_3=A_{11}+\overline{A_{11}},\\ a_3=-i(A_{11}-\overline{A_{11}}),\\ b_1=A_{12}+A_{21}+\overline{A_{12}+A_{21}},\text{ etc. }\end{array}$$

Hence the ${\mathsf{Y}}_i$ and $Z_i$ span $\mathcal{SL}(2,\mathbb{C})$ as a real vector space, which is a real Lie algebra determined by the commutation relations

$$[{{\rm Y}}_1,{{\rm Y}}_2]=-{{\rm Y}}_3[{{\rm Y}}_2,{{\rm Y}}_3]=-{{\rm Y}}_1[{{\rm Y}}_3,{{\rm Y}}_1]=-{{\rm Y}}_2,\text{\hspace{1em}(6.31)}$$

$$[{\mathsf{Z}}_1,{\mathsf{Z}}_2]={\mathsf{Y}}_3[{\mathsf{Z}}_2,{\mathsf{Z}}_3]={\mathsf{Y}}_1[{\mathsf{Z}}_3,{\mathsf{Z}}_1]={\mathsf{Y}}_2,\text{\hspace{1em}(6.32)}$$

$$[{\mathsf{Y}}_1,{\mathsf{Z}}_2]=-{\mathsf{Z}}_3[{\mathsf{Y}}_2,{\mathsf{Z}}_3]=-{\mathsf{Z}}_1[{\mathsf{Y}}_3,{\mathsf{Z}}_1]=-{\mathsf{Z}}_2,\text{\hspace{1em}(6.33)}$$

$$[{\mathsf{Y}}_1,{\mathsf{Z}}_3]={\mathsf{Z}}_3[{\mathsf{Y}}_2,{\mathsf{Z}}_1]={\mathsf{Z}}_3[{\mathsf{Y}}_3,{\mathsf{Z}}_2]={\mathsf{Z}}_1,\text{\hspace{1em}(6.34)}$$

$$[{\mathsf{Y}}_1,{\mathsf{Z}}_1]=0[{\mathsf{Y}}_2,{\mathsf{Z}}_2]=0[{\mathsf{Y}}_3,{\mathsf{Z}}_3]=0.\text{\hspace{1em}(6.35)}$$

Example 6.14

Lorentz transformations are defined in Section 2.7 by

$${\mathbf{x}}^{\prime }=\mathsf{Lx},\mathsf{G}={\mathsf{L}}^T\mathsf{GL}$$

where

$$\mathsf{G}=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& -1\end{array}\right).$$

Hence infinitesimal Lorentz transformations $\mathsf{L}=\mathsf{I}+ϵ\mathsf{A}$ satisfy the equation

$${\mathsf{A}}^T\mathsf{G}+\mathsf{GA}=\mathsf{O},$$

which reads in components

$$A_{ij}+A_{ji}=0,A_{4i}-A_{i4}=0,A_{44}=0$$

where indices $i,j$ range from 1 to 3. It follows that the Lie algebra of the Lorentz group is spanned by six matrices

$${{\rm Y}}_1=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& -1& 0& 0\\ 0& 0& 0& 0\end{array}\right),{{\rm Y}}_2=\left(\begin{array}{cccc}0& 0& -1& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),{{\rm Y}}_3=\left(\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),$$ $${\mathsf{Z}}_1=\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\end{array}\right),{\mathsf{Z}}_2=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right),{\mathsf{Z}}_3=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right).$$

These turn out to have exactly the same commutation relations Eq. (6.31)–Eq. (6.35) as the generators of $\mathcal{SL}(2,\mathbb{C})$ in the previous example. Hence the real Lie algebra $\mathcal{SL}(2,\mathbb{C})$ is isomorphic to the Lie algebra of the Lorentz group $\mathrm{SO}(3,1)$. Since the complex Lie algebras $\mathcal{SL}(2,\mathbb{C})$

and $\mathcal{S}\mathcal{O}(3,\mathbb{C})$ were shown to be isomorphic in Example 6.13, their real versions must also be isomorphic. We thus have the interesting sequence of isomorphisms of real Lie algebras,

$$\mathcal{S}\mathcal{O}(3,1)\cong \mathcal{SL}(2,\mathbb{C})\cong \mathcal{S}\mathcal{O}(3,\mathbb{C}).$$

Problems

Problem 6.12

As in Example 6.12, $n\times n$ unitary matrices satisfy $\mathsf{U}{\mathsf{U}}^{†}=\mathsf{I}$ and those near the identity have the form

$$\mathsf{U}=\mathsf{I}+ϵ\mathsf{A}(ϵ\ll 1)$$

where $A$ is anti-hermitian, $\mathsf{A}=-{\mathsf{A}}^{†}$

(a) Show that the set of anti-hermitian matrices form a Lie algebra with respect to the commutator $[\mathsf{A},\mathsf{B}]=\mathsf{AB}-\mathsf{BA}$ as bracket product.

(b) The four Pauli matrices ${\sigma }_{\mu }(\mu =0,1,2,3)$ are defined by

$${\sigma }_0=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),{\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right).$$

Show tha $\begin{array}{r}{\mathsf{Y}}_{\mu }=\frac{1}{2}i{\sigma }_{\mu }\end{array}$ form a basis of the Lie algebra of $U(2)$ and calculate the structure constants.

(c) Show that the one-parameter subgroup generated by ${\mathsf{Y}}_1$ consists of matrices of the form

$${\mathrm{e}}^{t{{\rm Y}}_1}=\left(\begin{array}{cc}\cos \frac{1}{2}t& i\sin \frac{1}{2}t\\ i\sin \frac{1}{2}t& \cos \frac{1}{2}t\end{array}\right).$$

Calculate the one-parameter subgroups generated by ${\mathsf{Y}}_2,{\mathsf{Y}}_3$ and ${\mathsf{Y}}_0$.

Problem 6.13

Let $u$ be an $n\times 1$ column vector. A non-singular matrix $A$ is said to stretch $u$ if it is an eigenvector of $A$,

$$\mathbf{Au}=\lambda \mathbf{u}.$$

Show that the set of all non-singular matrices that stretch $u$ forms a group with respect to matrix multiplication, called the stretch group $of\mathbf{u}$.

(a) Show that the $2\times 2$ matrices of the form

$$\left(\begin{array}{cc}a& a+c\\ b+c& b\end{array}\right)(c\ne 0,a+b+c\ne 0)$$

form the stretch group of the $2\times 1$ column vector $\mathbf{u}=\left(\genfrac{}{}{0ex}{}{1}{-1}\right)$

(b) Show that the Lie algebra of this group is spanned by the matrices

$${\mathsf{X}}_1=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right),{\mathsf{X}}_2=\left(\begin{array}{cc}0& 0\\ 1& 1\end{array}\right),{\mathsf{X}}_3=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right).$$

Calculate the structure constants for this basis.

(c) Write down the matrices that form the one-parameter subgroups ${\mathrm{e}}^{t{\mathsf{X}}_1}$ and ${\mathrm{e}}^{t{\mathsf{X}}_3}$

Problem 6.14

Show that $2\times 2$ trace-free matrices, having $\mathrm{tr}\mathsf{A}=A_{11}+A_{22}=0$ , form a Lie algebra with respect to bracket product $[\mathsf{A},\mathsf{B}]=\mathsf{AB}-\mathsf{BA}$

(a) Show that the following matrices form a basis of this Lie algebra:

$${\mathsf{X}}_1=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),{\mathsf{X}}_2=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),{\mathsf{X}}_3=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$$

and compute the structure constants for this basis.

(b) Compute the one-parameter subgroups ${\mathsf{e}}^{t{\mathsf{X}}_1},{\mathsf{e}}^{t{\mathsf{X}}_2}$ and ${\mathrm{e}}^{t{\mathsf{X}}_3}$ .

Problem 6.15

Let $\mathcal{L}$ be the Lie algebra spanned by the three matrices

$${\mathsf{X}}_1=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),{\mathsf{X}}_2=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right),{\mathsf{X}}_3=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right).$$

Write out the structure constants for this basis, with respect to the usual matrix commutator bracket product.

Write out the three one-parameter subgroups ${\mathrm{e}}^{t{\mathsf{X}}_i}$ generated by these basis elements, and verify in each case that they do in fact form a one-parameter group of matrices.