5.2 Lie algebras of the orthogonal and unitary groups

对照

The groups we are most interested in are the groups of linear transformations preserving an inner product: the orthogonal and unitary groups. We have seen that these are subgroups of $\mathcal{G}\mathcal{L}(n,\mathbf{R})$ or $GL(n,\mathbf{C})$ , consisting of those elements $\Omega$ satisfying the condition

$$\Omega {\Omega }^{†}=1$$

In order to see what this condition becomes on the Lie algebra, write $\Omega =e^{tX}$ for some parameter $t$, and $X$ a matrix in the Lie algebra. Since the transpose of a product of matrices is the product (order-reversed) of the transposed matrices, i.e.,

$$(XY{)}^T=Y^T{X}^T$$

and the complex conjugate of a product of matrices is the product of the complex conjugates of the matrices, one has

$$(e^{tX}{)}^{†}=e^{t{X}^{†}}$$

The condition

$$\Omega {\Omega }^{†}=1$$

thus becomes

$$e^{tX}(e^{tX}{)}^{†}=e^{tX}{e}^{t{X}^{†}}=1$$

Taking the derivative of this equation gives

$$e^{tX}{X}^{†}{e}^{t{X}^{†}}+X{e}^{tX}{e}^{t{X}^{†}}=0$$

Evaluating this at $t=0$ gives

$$X+X^{†}=0$$

so the matrices we want to exponentiate must be skew-adjoint (it can be shown that this is also a sufficient condition), satisfying

$$X^{†}=-X$$

Note that physicists often choose to define the Lie algebra in these cases as self-adjoint matrices, then multiplying by $i$ before exponentiating to get a group element. We will not use this definition, with one reason that we want to think of the Lie algebra as a real vector space, so want to avoid an unnecessary introduction of complex numbers at this point.

5.2.1 Lie algebra of the orthogonal group

Recall that the orthogonal group $O(n)$ is the subgroup of $GL(n,\mathbf{R})$ of matrices $\Omega$ satisfying ${\Omega }^T={\Omega }^{-1}$. We will restrict attention to the subgroup $SO(n)$ of matrices with determinant $1$, which is the component of the group containing the identity, with elements that can be written as

$$\Omega =e^{tX}$$

These give a path connecting $\Omega$ to the identity (taking $e^{sX},s\in [0,t]$). We saw above that the condition ${\Omega }^T={\Omega }^{-1}$ corresponds to skew-symmetry of the matrix $X$

$$X^T=-X$$

So in the case of $G=SO(n)$, we see that the Lie algebra $\mathfrak{so}(n)$ is the space of skew-symmetric $(X^T=-X)$ $n$ by $n$ real matrices, together with the bilinear, antisymmetric product given by the commutator:

$$(X,Y)\in \mathfrak{so}(n)\times \mathfrak{so}(n)\to [X,Y]\in \mathfrak{so}(n)$$

The dimension of the space of such matrices will be

$$1+2+\cdots +(n-1)=\frac{n^2-n}{2}$$

and a basis will be given by the matrices ${ϵ}_{jk}$ , with $j,k=1,\dots ,n,j<k$ defined as

$$({ϵ}_{jk}{)}_{lm}=\{\begin{array}{ll}-1& \text{if }j=l,k=m\\ +1& \text{if }j=m,k=l\\ 0& \text{otherwise}\end{array}\text{\hspace{1em}(5.3)}$$

In chapter 6 we will examine in detail the $n=3$ case, where the Lie algebra $\mathfrak{so}(3)$ is ${\mathbf{R}}^3$ , realized as the space of antisymmetric real 3 by 3 matrices, with a basis the three matrices ${ϵ}_{12},{ϵ}_{13},{ϵ}_{23}$

5.2.2 Lie algebra of the unitary group

For the case of the group $U(n)$ the unitarity condition implies that $X$ is skewadjoint (also called skew-Hermitian), satisfying

$$X^{†}=-X$$

So the Lie algebra $\mathfrak{u}(n)$ is the space of skew-adjoint $n$ by $n$ complex matrices, together with the bilinear, antisymmetric product given by the commutator:

$$(X,Y)\in \mathfrak{u}(n)\times \mathfrak{u}(n)\to [X,Y]\in \mathfrak{u}(n)$$

Note that these matrices form a subspace of ${\mathbf{C}}^{n^2}$ of half the dimension, so of real dimension $n^2$. $\mathfrak{u}(n)$ is a real vector space of dimension $n^2$, but it is NOT a space of real $n$ by $n$ matrices. It is the space of skew-Hermitian matrices, which in general are complex. While the matrices are complex, only real linear combinations of skew-Hermitian matrices are skew-Hermitian (recall that multiplication by $i$ changes a skew-Hermitian matrix into a Hermitian matrix). Within this space of skew-Hermitian complex matrices, if one looks at the subspace of real matrices one gets the sub-Lie algebra $\mathfrak{so}(n)$ of antisymmetric matrices (the Lie algebra of $SO(n)\subset U(n)$).

Any complex matrix $Z\in M(n,\mathbf{C})$ can be written as a sum of

$$Z=\frac{1}{2}(Z+Z^{†})+\frac{1}{2}(Z-Z^{†})$$

where the first term is self-adjoint, the second skew-Hermitian. This second term can also be written as $i$ times a self-adjoint matrix

$$\frac{1}{2}(Z-Z^{†})=i\left(\frac{1}{2i}(Z-Z^{†})\right)$$

so we see that we can get all of $M(n,\mathbf{C})$ by taking all complex linear combinations of self-adjoint matrices.

There is an identity relating the determinant and the trace of a matrix

$$\det (e^X)=e^{\mathrm{trace}(X)}$$

which can be proved by conjugating the matrix to upper-triangular form and using the fact that the trace and the determinant of a matrix are conjugation invariant. Since the determinant of an $SU(n)$ matrix is 1, this shows that the Lie algebra $\mathfrak{su}(n)$ of $SU(n)$ will consist of matrices that are not only skew-Hermitian, but also of trace zero. So in this case $\mathfrak{su}(n)$ is again a real vector space, with the trace zero condition a single linear condition giving a vector space of real dimension $n^2-1$

One can show that $U(n)$ and $\mathfrak{u}(n)$ matrices can be diagonalized by conjugation by a unitary matrix and thus show that any $U(n)$ matrix can be written as an exponential of something in the Lie algebra. The corresponding theorem is also true for $SO(n)$ but requires looking at diagonalization into 2 by 2 blocks. It is not true for $O(n)$ (you can’t reach the disconnected component of the identity by exponentiation). It also turns out to not be true for the groups $SL(n,\mathbf{R})$ and $SL(n,\mathbf{C})$ for $n\ge 2$ (while the groups are connected, they have elements that are not exponentials of any matrix in $\mathfrak{sl}(n,\mathbf{R})$ or $\mathfrak{sl}(2,\mathbf{C})$ respectively).

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上一节：5.1 Lie algebras · 下一节：5.3 A summary

附录：约定与记号

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