5.1 Lie algebras

对照

If a Lie group $G$ is defined as a differentiable manifold with a group law, one can consider the tangent space at the identity, and that will be the Lie algebra of $G$. We are however interested mainly in cases where $G$ is a matrix group, and in such cases the Lie algebra can be defined more concretely:

Definition · Lie algebra

Definition (Lie algebra). For $G$ a Lie group of $n$ by $n$ invertible matrices, the Lie algebra of $G$ (written $Lie(G)$ or $\mathfrak{g}$) is the space of $n$ by $n$ matrices $X$ such that $e^{tX}\in G$ for $t\in \mathbf{R}$.

Here the exponential of a matrix is given by usual power series formula for the exponential

$$e^A=1+A+\frac{1}{2}{A}^2+\cdots +\frac{1}{n!}{A}^n+\dots$$

which can be shown to converge (like the usual exponential), for any matrix $A$. While this definition is more concrete than defining a Lie algebra as a tangent space, it does not make obvious some general properties of a Lie algebra, in particular that a Lie algebra is a real vector space (see theorem 3.20 of [42]). Our main interest will be in using it to recognize certain specific Lie algebras corresponding to specific Lie groups.

Notice that while the group $G$ determines the Lie algebra $\mathfrak{g}$, the Lie algebra does not determine the group. For example, $O(n)$ and $SO(n)$ have the same tangent space at the identity, and thus the same Lie algebra, but elements in $O(n)$ not in the component of the identity (i.e., with determinant $-1$) can’t be written in the form $e^{tX}$ (since then you could make a path of matrices connecting such an element to the identity by shrinking $t$ to zero).

Note also that, for a given $X$, different values of $t$ may give the same group element, and this may happen in different ways for different groups sharing the same Lie algebra. For example, consider $G=U(1)$ and $G=\mathbf{R}$ , which both have the same Lie algebra $\mathfrak{g}=\mathbf{R}$ . In the first case an infinity of values of $t$ give the same group element, in the second, only one does. In chapter 6 we’ll see a more subtle example of this: $SU(2)$ and $SO(3)$ are different groups with the same Lie algebra.

We have $G\subset GL(n,\mathbf{C})$, and $X\in M(n,\mathbf{C})$, the space of $n$ by $n$ complex matrices. For all $t\in \mathbf{R}$, the exponential $e^{tX}$ is an invertible matrix (with inverse $e^{-tX}$), so in $GL(n,\mathbf{C})$. For each $X$, we thus have a path of elements of $GL(n,\mathbf{C})$ going through the identity matrix at $t=0$, with velocity vector

$$\frac{d}{dt}{e}^{tX}=X{e}^{tX}$$

which takes the value $X$ at $t=0:$

$$\frac{d}{dt}(e^{tX}{)}_{|t=0}=X$$

To calculate this derivative, use the power series expansion for the exponential, and differentiate term-by-term.

For the case $G=GL(n,\mathbf{C})$, we have $\mathfrak{gl}(n,\mathbf{C})=M(n,\mathbf{C})$, which is a linear space of the right dimension to be the tangent space to $G$ at the identity, so this definition is consistent with our general motivation. For subgroups $G\subset GL(n,\mathbf{C})$ given by some condition (for example that of preserving an inner product), we will need to identify the corresponding condition on $X\in M(n,\mathbf{C})$ and check that this defines a linear space.

The existence of such a linear space $\mathfrak{g}\subset M(n,\mathbf{C})$ will provide us with a distinguished representation on a real vector space, called the “adjoint representation”:

Definition · Adjoint representation

Definition (Adjoint representation). The adjoint representation $(Ad,\mathfrak{g})$ is given by the homomorphism

$$Ad:g\in G\to Ad(g)\in GL(\mathfrak{g})$$

where $Ad(g)$ acts on $X\in \mathfrak{g}$ by

$$(Ad(g))(X)=gX{g}^{-1}$$

To show that this is well-defined, one needs to check that $gX{g}^{-1}\in \mathfrak{g}$ when $X\in \mathfrak{g}$ , but this can be shown using the identity

$$e^{tgX{g}^{-1}}=g{e}^{tX}{g}^{-1}$$

which implies that $e^{tgX{g}^{-1}}\in G\mathrm{if}{e}^{tX}\in G$ . To check this identity, expand the exponential and use

$$(gX{g}^{-1}{)}^k=(gX{g}^{-1})(gX{g}^{-1})\dots (gX{g}^{-1})=g{X}^k{g}^{-1}$$

It is also easy to check that this is a homomorphism, with

$$Ad(g_1)Ad(g_2)=Ad(g_1{g}_2)$$

A Lie algebra $\mathfrak{g}$ is not just a real vector space, but comes with an extra structure on the vector space:

Definition · Lie bracket

Definition (Lie bracket). The Lie bracket operation on $\mathfrak{g}$ is the bilinear antisymmetric map given by the commutator of matrices

$$[\cdot ,\cdot ]:(X,Y)\in \mathfrak{g}\times \mathfrak{g}\to [X,Y]=XY-YX\in \mathfrak{g}$$

We need to check that this is well-defined, i.e., that it takes values in $\mathfrak{g}$.

Theorem

Theorem. If $X,Y\in \mathfrak{g}$, $[X,Y]=XY-YX\in \mathfrak{g}$.

Proof

Proof. Since $X\in \mathfrak{g}$ , we have $e^{tX}\in G$ and we can act on $Y\in \mathfrak{g}$ by the adjoint representation

$$Ad(e^{tX})Y=e^{tX}Y{e}^{-tX}\in \mathfrak{g}$$

As $t$ varies this gives us a parametrized curve in $\mathfrak{g}$. Its velocity vector will also be in $\mathfrak{g}$, so

$$\frac{d}{dt}(e^{tX}Y{e}^{-tX})\in \mathfrak{g}$$

One has (by the product rule, which can easily be shown to apply in this case)

$$\begin{array}{rl}\frac{d}{dt}(e^{tX}Y{e}^{-tX})& =\left(\frac{d}{dt}(e^{tX})\right)Y{e}^{-tX}+e^{tX}Y\left(\frac{d}{dt}{e}^{-tX}\right)\\ & =X{e}^{tX}Y{e}^{-tX}-e^{tX}YX{e}^{-tX}\end{array}$$

Evaluating this at $t=0$ gives

$$XY-YX$$

which is thus, from the definition, shown to be in $\mathfrak{g}$.

The relation

$$\frac{d}{dt}(e^{tX}Y{e}^{-tX}{)}_{|t=0}=[X,Y]\text{\hspace{1em}(5.1)}$$

used in this proof will be continually useful in relating Lie groups and Lie algebras.

To do calculations with a Lie algebra, one can choose a basis $X_1,X_2,\dots ,X_n$ for the vector space $\mathfrak{g}$, and use the fact that the Lie bracket can be written in terms of this basis as

$$[X_j,X_k]=\sum _{l=1}^n{c}_{jkl}{X}_l\text{\hspace{1em}(5.2)}$$

where $c_{jkl}$ is a set of constants known as the “structure constants” of the Lie algebra. For example, in the case of $\mathfrak{su}(2)$, the Lie algebra of $SU(2)$ has a basis $X_1,X_2,X_3$ satisfying

$$[X_j,X_k]=\sum _{l=1}^3{ϵ}_{jkl}{X}_l$$

(see equation 3.5) so the structure constants of $\mathfrak{su}(2)$ are the totally antisymmetric ${ϵ}_{jkl}$.

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上一节：5 Lie Algebras and Lie Algebra Representations · 下一节：5.2 Lie algebras of the orthogonal and unitary groups

附录：约定与记号

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正文：英文 · 原著转录

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原书 PDF · 印刷页 48、49、50

来源版本：2025-10-20

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