Lie Algebra

This article is one of Lie Group & Representation contents.

Generators

Let parameterize the identity element as $\alpha=0$. Thus assume that in some neighborhood of the identity, the group elements can be described by a function of $N$ real parameters, $\alpha_a$ for $a=1$ to $N$, such that $$g(\alpha)|_{\alpha=0}=e$$

Then if we find a representation of the group, the linear operators of the representation will be parameterized the same way, and $$D(\alpha)|_{\alpha=0}=1$$

Then in some neighborhood of the identity element, we can Taylor expand $D(\alpha)$, and if we are close enough, just keep the first term: $$D(d\alpha)=1+id\alpha_a X_a +\cdots$$ where we have called the parameter $d\alpha$ to remind you that it is infinitesimal. Also, $$X_a\equiv -i\frac{\partial}{\partial \alpha_a}D(\alpha)|_{\alpha=0}$$

The $X_a$ for $a=1$ to $N$ are called the generators of the group. If the representation is unitary, the $X_a$ will be hermitian operators.

We can write the group elements (at least in some neighborhood of $e$) in terms of the generators: $$D(\alpha)=\lim_{k\rightarrow \infty} (1+i\alpha_a X_a /k)^k=e^{i\alpha_a X_a}$$ This called exponential parameterization. 

Lie Algebras

What is group multiplication of exponential parameterization? $$e^{i\alpha_a X_a} e^{i\beta_b X_b}=e^{i\delta_a X_a}$$ for some $\delta$. Expanding this shows generators form an algebra under commutation.

$$i\delta_a X_a = \ln (1+e^{i\alpha_a X_a} e^{i\beta_b X_b} - 1)$$ Keep terms up to second order in the parameters $\alpha$ and $\beta$, using the Taylor expansion of $\ln(1+K)$ where $$ K=e^{i\alpha_a X_a} e^{i\beta_b X_b} - 1= (1+i\alpha_a X_a - \frac{1}{2} (\alpha_a X_a)^2 + \cdots ) (1+i\beta_b X_b - \frac{1}{2} (\beta_b X_b)^2 + \cdots)-1 = i\alpha_a X_a + i\beta_a X_a - \alpha_a X_a \beta_b X_b - \frac{1}{2}(\alpha_a X_a)^2 - \frac{1}{2}(\beta_a X_a)^2 + \cdots $$

This gives $$i\delta_a X_a = K - \frac{1}{2} K^2 + \cdots = i\alpha_a X_a + i\beta_aX_a - \alpha_a X_a\beta_b X_b - \frac{1}{2}(\alpha_aX_a)^2 - \frac{1}{2} (\beta_a X_a)^2 + \frac{1}{2} (\alpha_a X_a + \beta_a X_a)^2 + \cdots = i\alpha_a X_a + i\beta_a X_a - \frac{1}{2}[\alpha_a X_a,\beta_b X_b]+\cdots$$ 

Now we can calculate $\delta_a$, again in an expansion in $\alpha$ and $\beta$. $$[\alpha_a X_a, \beta_b X_b]=-2i(\delta_c - \alpha_c - \beta_c)X_c + \cdots \equiv i\gamma_c X_c$$ where the $i$ is put in to make $\gamma$ real and the $\cdots$ represent terms that have more than two factors of $\alpha$ and $\beta$. Since this must be true for all $\alpha$ and $\beta$, we must have $$\gamma_c = \alpha_a \beta_b f_{abc}$$ for some constants $f_{abc}$, thus $$[X_a, X_b]=if_{abc}X_c$$ where f_{abc}=-f{bac}$$ because $[A,B]=-[B,A]$. $f_{abc}$ called sturcture constants, and it determine Lie algebra.

If representation is unitary, structure constant is real. $$[X_a,X_b]^\dagger=-if_{abc}^*X_c=[X_b,X_a]=if_{bac}X_c=-if_{abc}X_c$$

The Jacobi Identity

$$[X_a,[X_b,X_c]]+[X_b,[X_c,X_a]]+[X_c,[X_a,X_b]]=0$$

The Adjoint Representation

The structure constants themselves generate a representation of thje algebra called the adjoint representation. If we use the Lie algebra, we can compute $$[X_a,[X_b,X_c]]=if_{bcd}[X_a,X_d]=-f_{bcd}f_{ade}X_e$$ so Jacobi identity implies $$f_{bcd}f_{ade}+f_{abd}f_{cde}+f_{cad}f_{bde}=0$$

Defining a set of matrices $T_a$ $$[T_a]_{bc}\equiv -if_{abc}$$ then Jacobi identity can be rewritten as $$[T_a,T_b]=if_{abc}T_c$$

Thus the structure constants themselves furish a representation of the algebra. This is called the adjoint representation. The dimension of a representation is the dimension of the linear space of which it acts (just as for a finite group). The dimension of the adjoint representation is just the number of independent generators, which is the number of real parameters required to describe a group element. Note that since the $f_{abc}$s are real, the generators of the adjoint representation are pure imaginary.

We would like to have a convenient scalar product on the linear space of generators in the adjoint representation, to turn it into a vector space. A good one is the trace in the adjoint representation $$\mbox{Tr}(T_aT_b)$$ This is a real symmetric matrix. We will next show that we can put it into a very simple canonical form. We can chage its form by making a linear transformation on the $X_a$, which in turn, induces a linear transformation on the structure constants. Suppose $$X_a\rightarrow X'_a=L_{ab}X_b$$ then $$[X'_a,X'_b]=iL_{ad}L_{be}f_{dec}X_c = iL{ad}L_{be}f_{deg}L_{gh}^{-1}L_{hc}X_c=iL_{ad}L_{be}f_{deg}L_{gc}^{-1}X'_c$$ so $$f_{abc}\rightarrow f'_{abc}=L_{ad}L_{be}f_{deg}L_{gc}^{-1}$$ 

If we then define a new $T_a$s with the transformed $f$s, $$[T_a]_{bc}\rightarrow [T'_a]_{bc}=L_{ad}L[T_d]L^{-1}$$

In other words, a linear transformation on the $X_a$s induces a linear transforamtion on the $T_a$s which involves both a similarity transformation and the same linear transformation on the $a$ index that lables the generator. But in the trace the similarity transformation doesn't matter, so $$\mbox{Tr}(T_aT_b)\rightarrow \mbox{Tr}(T'_aT'_b)=L_{ac}L_{bd}\mbox{Tr}(T_cT_d)$$

Thus we can diagonalize the trace by choosing an appropriate $L$ (here we only need an orthogonal matrix). Suppose we have done this (and dropped the primes), so the $$\mbox{Tr}(T_aT_b)=k^a\delta_{ab}\mbox{ no sum}$$

We still have the freedom to rescale the generators (by making a diagonal $L$ transformation), so for example, we could choose all the non-zero $k^a$s to have absolute value 1. Bu, we cannot change the sign of the $k^a$s (because $L$ appears squared int the transformation).

For now, we will assume that the $k^a$s are positive. This defines the class of algebras that we study in this book. They are called compact Lie algebras. We will come back briefly below to algebras in which some are zero. And we will take $$\mbox{Tr}(T_aT_b)=\lambda \delta_{ab}$$ for some convenient possitive $\lambda$. In this basis, the structure constants are completely antisymmetric, because we can write $$f_{abc}=-i\lambda^{-1}\mbox{Tr}([T_a,T_b]T_c)$$ which is completely antisymmetric because of the cyclic property of the trace. $$\mbox{Tr}([T_a,T_b]T_c)=(T_aT_bT_c-T_bT_aT_c)=(T_bT_cT_a-T_cT_bT_a)=\mbox{Tr}([T_b,T_c]T_a)$$ which implies $$f_{abc}=f_{bca}$$

This shows complete antisymmetry of $f_{abc}$ $$f_{abc}=f_{bca}=f_{cab}=-f_{bac}=-f_{acb}=-f_{cba}$$ In this basis, the adjoint representation is unitary, because the $T_a$ are imaginary and antisymmetric, and therefore hermitian.

Simple Algebras and Groups

An invariant subalgebra is some set of generators which goes into itself under commutation with any element of the algebra. That is, if $X$ is any generator in the invariant subalgebra and $Y$ is any generator in the whole algebra, $[Y,X]$ is a generator in the invariant subalgebra. When exponentiated, an invariant subalgebra generates an invariant subgroup. To see this note that $$h=e^{iX},\ g=e^{iY},\ g^{-1}hg=e^{iX'}$$ where $$X'=e^{-iY}Xe^{iY}=X-i[Y,X]-\frac{1}{2}[Y,[Y,x]]+\cdots$$

Note that the easy way to see this is to consider $$X'(\epsilon)=e^{-i\epsilon Y} X e^{i\epsilon Y}$$ then Taylor expand in $\epsilon$ and set $\epsilon=1$. Each derivative brings another commutator. Evidently, each of the terms in $X'$ is in the subalgebra, and thus $e^{iX'}$ is in the subgroup, which is therefore invariant.

The whole algebra and 0 are trivial invariant subalgebras. An algebra which has no nontrivial invariant subalgebra is called simple. A simple algebra generates a simple group.

The adjoint representation of a simple Lie algebra satistying $\mbox{Tr}(T_aT_b)=\lambda\delta_{ab}$ is irreducible. To see this assume the contrary. Then there is an invariant subspace in the adjoint representation. But the sates of the adjoint representation correspond to generators, so this means that we can find a basis in which the invariant subspace is spanned by some subset of the generators, $T_r$ for $r=1$ to $K$. Call the rest of the generators $T_x$ for $x=K+1$ to $N$. Then because the $r$s span an invariant subspace, we mus have $$[T_a]_{xr}=-if_{axr}=0$$ for all $a,x$ and $r$. Because of the complete antisymmtery of the structure constants, this means that all components of $f$ that have two $r$s and onre $x$ or two $x$s and one $r$ vanish. But that means that the nonzero structure constants involve eigther three $r$s or three $x$s, and thus the algebra falls into two nontrivial invariant subalgebras, and is not simple. Thus the adjoint representation of a simple Lie algebra satisfying $\mbox{Tr}(T_aT_b)=\lambda\delta_{ab}$ is irreducible.

We will often find it uesful to discuss special Abelian invariant subalgebras consisting of a single generator which commutes with all the generators of the group (or of some subgroup we are interested in). We will call such an algebra a $U(1)$ factor of the group. $U(1)$ is the group of pahse tranformations. $U(1)$ factors do not appear in the structure constants at all. These Abelian invariant subalgebras correspond to direction in the space of generators for which $k_a=0$ in $\mbox{Tr}(T_aT_b)=k^a\delta_{ab}\mbox{ no sum}$. If $X_a$ is a $U(1)$ generator, $f_{abc}=0$ for all $b$ and $c$. That also means that the corresponding $k^a$ is zero, so the trace scalar product does not give a norm on the space. The structure constants do not tell us anything about the $U(1)$ subalgebras.

Algebras without Abelian invariant subalgebras are called semisimple. They are bulit, as we will see, by putting simple algebras together. In these algebras, every generator has a non-zero commutator with some other generator. Because of the cyclic property of the structure constants, $f_{abc}=-i\lambda^{-1}\mbox{Tr}([T_a,T_b]T_c)$, this also implies that every generator is a linear combination of commutators of generators. In such a case, the structure constants carry a great deal of information. We will use them to determine the entire structure of the algebra and its representations. From here on, unless explicitly stated, we will discuss semisomple algebras, and we will deal with representations by unitary operators.

States and Operators

The generators of a representation can be thought of as either linear operators or matrices, just as representations of finite groups: $$X_a\left| i \right\rangle = \left| j \right\rangle \left\langle j \right| X_a \left| i \right\rangle = \left| j \right\rangle [X_a]_{ji}$$ with the sum on $j$ understood. The states form row vectors and the matrix representing a linear operator acts on the right.

In the Hilbert space on which the representation acts, the group elements can be thought of as transformations on the states. The group element $e^{i\alpha_aX_a}$ maps or transforms the kets as follows: $$\left| i \right\rangle \rightarrow \left| i'\right\rangle = e^{i\alpha_a X_a}\left| i \right\rangle$$ Taking the adjoint show that the corresponding bras transform as $$\left\langle i \right| \rightarrow \left\langle i'\right|=\left\langle i \right| e^{-i\alpha_a X_a}$$ The ket obtained by acting on $\left| i \right\rangle$ with an operator $O$ is a sum of kets, and therefore must also transform like $$O\left| i \right\rangle \rightarrow e^{i\alpha_aX_a}O\left| i \right\rangle=e^{i\alpha_aX_a}Oe^{-i\alpha_aX_a}e^{i\alpha_aX_a}\left| i \right\rangle = O'\left| i'\right\rangle$$ This implies that any operator $O$ transforms as follows: $$O\rightarrow O'=e^{i\alpha_a X_a}Oe^{-i\alpha_a X_a}$$ The transformation leaves all matrix elements invariant.

The action of the algebra on these onjects is related to the change in the state of operator under an infinitesimal transformation. $$-i\delta\left| i \right\rangle =-i((1+i\alpha_a X_a)\left| i \right\rangle - \left| i \right\rangle )=\alpha_a X_a \left| i\right\rangle,\ -i\delta \left\langle i \right| = -\left\langle i \right| \alpha_a X_a,\ -i\delta O = [\alpha_a X_a, O]$$

Thus, corresponding to the action of the generator $X_a$ on a ket $$X_a\left| i \right\rangle$$ is $-X_a$ acting on a bra $$-\left\langle i\right| X_a$$ and the commutator of $X_a$ with an iperator $$[X_a,O]$$. Then the invariance of a matrix element $\left\langle i \right| O \left| i \right\rangle$ is expressed by the fact, $$\left\langle i \right| O(X_a\left| i \right\rangle)+\left\langle i [X_a,O]\left| i \right\rangle - (\left\langle i \right| X_a \right\rangle O \left| i \right\rangle = 0$$

Reference

Howard Georgi - Lie Algebras in Particle Physics