5.4 Lie algebra representations

对照

We have defined a group representation as a homomorphism (a map of groups preserving group multiplication)

$$\pi :G\to GL(n,\mathbf{C})$$

We can similarly define a Lie algebra representation as a map of Lie algebras preserving the Lie bracket:

Definition · Lie algebra representation

Definition (Lie algebra representation). A (complex) Lie algebra representation $(\varphi ,V)$ of a Lie algebra $\mathfrak{g}$ on an $n$ dimensional complex vector space $V$ is given by a real-linear map

$$\varphi :X\in \mathfrak{g}\to \varphi (X)\in \mathfrak{gl}(n,\mathbf{C})=M(n,\mathbf{C})$$

satisfying

$$\varphi ([X,Y])=[\varphi (X),\varphi (Y)]$$

Such a representation is called unitary if its image is in $\mathfrak{u}(n)$, i.e., if it satisfies

$$\varphi (X{)}^{†}=-\varphi (X)$$

More concretely, given a basis $X_1,X_2,\dots ,X_d$ of a Lie algebra $\mathfrak{g}$ of dimension $d$ with structure constants $c_{jkl}$, a representation is given by a choice of complex $n$ dimensional matrices $\varphi (X_j)$ satisfying the commutation relations

$$[\varphi (X_j),\varphi (X_k)]=\sum _{l=1}^d{c}_{jkl}\varphi (X_l)$$

The representation is unitary when the matrices are skew-adjoint.

The notion of a Lie algebra representation is motivated by the fact that the homomorphism property causes the map $\pi$ to be largely determined by its behavior infinitesimally near the identity, and thus by the derivative ${\pi }^{\prime }$ . One way to define the derivative of such a map is in terms of velocity vectors of paths, and this sort of definition in this case associates to a representation $\pi :G\to GL(n,\mathbf{C})$ a linear map

$${\pi }^{\prime }:\mathfrak{g}\to M(n,\mathbf{C})$$

where

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

Figure 5.1: Derivative of a representation $\pi :G\to GL(n,\mathbf{C})$ , illustrated in terms of “velocity” vectors along paths.

For the case of $U(1)$ we classified in theorem 2.3 all irreducible representations (homomorphisms $U(1)\to GL(1,\mathbf{C})={\mathbf{C}}^{\ast }$) by looking at the derivative of the map at the identity. For general Lie groups $G$, something similar can be done, showing that a representation $\pi$ of $G$ gives a representation of the Lie algebra (by taking the derivative at the identity), and then trying to classify Lie algebra representations.

Theorem

Theorem. If $\pi :G\to GL(n,\mathbf{C})$ is a group homomorphism, then

$${\pi }^{\prime }:X\in \mathfrak{g}\to {\pi }^{\prime }(X)=\frac{d}{dt}(\pi (e^{tX}){)}_{|t=0}\in \mathfrak{gl}(n,\mathbf{C})=M(n,\mathbf{C})$$

satisfies

$$\pi (e^{tX})=e^{t{\pi }^{\prime }(X)}$$

2. For $g\in G$

$${\pi }^{\prime }(gX{g}^{-1})=\pi (g){\pi }^{\prime }(X)(\pi (g){)}^{-1}$$

3. ${\pi }^{\prime }$ is a Lie algebra homomorphism:

$${\pi }^{\prime }([X,Y])=[{\pi }^{\prime }(X),{\pi }^{\prime }(Y)]$$

Proof

Proof. 1. We have

$$\begin{array}{rl}\frac{d}{dt}\pi (e^{tX})& ={\frac{d}{ds}\pi (e^{(t+s)X})|}_{s=0}\\ & ={\frac{d}{ds}\pi (e^{tX}{e}^{sX})|}_{s=0}\\ & ={\pi (e^{tX})\frac{d}{ds}\pi (e^{sX})|}_{s=0}\\ & =\pi (e^{tX}){\pi }^{\prime }(X)\end{array}$$

So $f(t)=\pi (e^{tX})$ satisfies the differential equation $\frac{d}{dt}f=f{\pi }^{\prime }(X)$ with initial condition $f(0)=1$. This has the unique solution $f(t)=e^{t{\pi }^{\prime }(X)}$.

2. We have

$$\begin{array}{rl}{e}^{t{\pi }^{\prime }(gX{g}^{-1})}& =\pi (e^{tgX{g}^{-1}})\\ & =\pi (g{e}^{tX}{g}^{-1})\\ & =\pi (g)\pi (e^{tX})\pi (g{)}^{-1}\\ & =\pi (g)e^{t{\pi }^{\prime }(X)}\pi (g{)}^{-1}\end{array}$$

Diferentiating with respect to $t$ at $t=0$ gives

$${\pi }^{\prime }(gX{g}^{-1})=\pi (g){\pi }^{\prime }(X)(\pi (g){)}^{-1}$$

3. Recall equation 5.1:

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

so

$$\begin{array}{rl}{\pi }^{\prime }([X,Y])& ={\pi }^{\prime }\left({\frac{d}{dt}(e^{tX}Y{e}^{-tX})|}_{t=0}\right)\\ & ={\frac{d}{dt}{\pi }^{\prime }(e^{tX}Y{e}^{-tX})|}_{t=0}\text{(by linearity)}\\ & ={\frac{d}{dt}(\pi (e^{tX}){\pi }^{\prime }(Y)\pi (e^{-tX}))|}_{t=0}\text{(by 2.)}\\ & ={\frac{d}{dt}(e^{t{\pi }^{\prime }(X)}{\pi }^{\prime }(Y)e^{-t{\pi }^{\prime }(X)})|}_{t=0}\text{(by 1.)}\\ & =[{\pi }^{\prime }(X),{\pi }^{\prime }(Y)]\end{array}$$

This theorem shows that we can study Lie group representations $(\pi ,V)$ by studying the corresponding Lie algebra representation $({\pi }^{\prime },V)$. This will generally be much easier since the ${\pi }^{\prime }$ are linear maps. Unlike the non-linear maps $\pi$, the map ${\pi }^{\prime }$ is determined by its value on basis elements $X_j$ of $\mathfrak{g}$. The ${\pi }^{\prime }(X_j)$ will satisfy the same bracket relations as the $X_j$ (see equation 5.2). We will proceed in this manner in chapter 8 when we construct and classify all $SU(2)$ and $SO(3)$ representations, finding that the corresponding Lie algebra representations are much simpler to analyze. Note though that representations of the Lie algebra $\mathfrak{g}$ do not necessarily correspond to representations of the group $G$ (when they do they are called “integrable”). For a simple example, looking at the proof of theorem 2.3, one gets unitary representations of the Lie algebra of $U(1)$ for any value of the constant $k$, but these are only representations of the group $U(1)$ when $k$ is integral.

For any Lie group $G,$ we have seen that there is a distinguished representation, the adjoint representation $(Ad,\mathfrak{g})$ . The corresponding Lie algebra representation is also called the adjoint representation, but written as $\left(A{d}^{\prime },\mathfrak{g}\right)=\left(ad,\mathfrak{g}\right)$ From the fact that

$$Ad(e^{tX})(Y)=e^{tX}Y{e}^{-tX}$$

we can differentiate with respect to $t$ and use equation 5.1 to get the Lie algebra representation

$$ad(X)(Y)=\frac{d}{dt}(e^{tX}Y{e}^{-tX}{)}_{|t=0}=[X,Y]\text{\hspace{1em}(5.4)}$$

This leads to the definition:

Definition · Adjoint Lie algebra representation

Definition (Adjoint Lie algebra representation). $(ad,\mathfrak{g})$ is the Lie algebra representation given by

$$X\in \mathfrak{g}\to ad(X)$$

where $ad(X)$ is defined as the linear map from $\mathfrak{g}$ to itself given by

$$Y\to [X,Y]$$

Note that this linear map $ad(X)$ , which can be written as $[X,\cdot ]$ , can be thought of as the infinitesimal version of the conjugation action

$$(\cdot )\to e^{tX}(\cdot )e^{-tX}$$

The Lie algebra homomorphism property of $ad$ says that

$$ad([X,Y])=ad(X)\circ ad(Y)-ad(Y)\circ ad(X)$$

where these are linear maps on $\mathfrak{g}$, with ◦ composition of linear maps, so operating on $Z\in \mathfrak{g}$ we have

$$ad([X,Y])(Z)=(ad(X)\circ ad(Y))(Z)-(ad(Y)\circ ad(X))(Z)$$

Using our expression for $ad$ as a commutator, we find

$$[[X,Y],Z]=[X,[Y,Z]]-[Y,[X,Z]]$$

This is called the Jacobi identity. It could have been more simply derived as an identity about matrix multiplication, but here we see that it is true for a more abstract reason, reflecting the existence of the adjoint representation. It can be written in other forms, rearranging terms using antisymmetry of the commutator, with one example the sum of cyclic permutations

$$[[X,Y],Z]+[[Z,X],Y]+[[Y,Z],X]=0$$

Lie algebras can be defined much more abstractly as follows:

Definition · Abstract Lie algebra

Definition (Abstract Lie algebra). An abstract Lie algebra over a field $k$ is a vector space $A$ over $k$, with a bilinear operation

$$[\cdot ,\cdot ]:(X,Y)\in A\times A\to [X,Y]\in A$$

satisfying

1. Antisymmetry:

$$[X,Y]=-[Y,X]$$

2. Jacobi identity:

$$[[X,Y],Z]+[[Z,X],Y]+[[Y,Z],X]=0$$

Such Lie algebras do not need to be defined as matrices, and their Lie bracket operation does not need to be defined in terms of a matrix commutator (although the same notation continues to be used). Later on we will encounter important examples of Lie algebras that are defined in this more abstract way.

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上一节：5.3 A summary · 下一节：5.5 Complexification

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