Definition of Lie Group and Lie Algebra

Exponential Map

Intuitively, we can simply view Lie groups $G$ as the groups of elements $g(\alpha)\in G$ that are parameterized continuously by a set of real numbers $\alpha$. We denote their representations by $\cal{D}(\alpha)$, which can be constructed in the following steps:

1. We choose $\alpha$ such that $g(0)=e$. As a result, ${\cal{D}}(0)=\cal{I}$ is the identity operator.
2. For small parameters $\epsilon$, we are able to Taylor expand to the first order of each component  $\epsilon^a$: \begin{equation}{\cal{D}}(\epsilon)={\cal{I}}+i\sum_{a=1}^N\epsilon^a \cal{T}_a\,\end{equation} where ${\cal{T}}_1,\cdots , {\cal{T}}_N$ are operators called generators.
3. For finite parameters $\alpha$, the group property allows to make $k$ sequential transformations, each with small parameters $\alpha/k$. As a result, \begin{equation}{\cal{D}}(\alpha)=\lim_{k\rightarrow\infty}{\cal{D}}^k(\alpha/k)=\lim_{k\rightarrow\infty}\left({\cal{I}}+\frac{i}{k}\sum_{a=1}^N\alpha^a {\cal{T}}_a\right)^k=\exp\left[i\sum_{a=1}^N\alpha^a {\cal{T}}_a\right]\,,\end{equation} For simplicity, we will use Einstein summation convention and write the above exponential map as \begin{equation} {\cal{D}}(\alpha)= e^{i\,\alpha^a {\cal{T}}_a}\,,\tag{1}\end{equation} where an index appearing in both subscript and superscript implies summation.

Remarks:

1. Note that we have defined the addition, scaling and multiplication operations for operators. So for any operator $\cal{T}$, we are able to compute $e^{\cal{T}}\equiv \sum_{k=0}^{\infty}\frac{{\cal{T}}^k}{k!}$ with ${\cal{T}}^0\equiv \cal{I}$.
2. In terms of $\cal{D}(\alpha)$, the generator ${\cal{T}}_a$ in Eq. (1) is \begin{equation}{\cal{T}}_a=-i\left.\frac{\partial}{\partial \alpha^a}{\cal{D}}(\alpha)\right|_{\alpha=0}\,.\end{equation} So ${\cal{T}}_1,\cdots , {\cal{T}}_N$ are a basis that spans a vector space.
3. For matrix Lie groups, i.e. $G \subseteq \text{GL}(n, \mathbb{C})$, besides the representation (1), we can also write the group element itself in exponential map \begin{equation}\mathbf{g}(\alpha)=e^{i\,\alpha^a \mathbf{X}_a}\,,\tag{2}\end{equation} where the generators $\mathbf{X}_1,\cdots, \mathbf{X}_N$ are matrices. 
4. For Lie groups beyond matrix groups, Eq. (2) does not make sense. In this case, we requires a more rigorous definition of Lie groups from the view of manifold. In this view, generators are a basis of the tangent space $\text{T}_eG$ and the exponential map is defined by one-parameter subgroup. We won't go to details on this since we are mainly interested in matrix Lie groups in physics. Interested readers can check the book "Geometry, Topology and Physics" for details.

Lie Algebra

For the group multiplication $g(\alpha)g(\beta)=g(\gamma)$, we have \begin{equation}i\,\gamma^a {\cal{T}}_a=\log \left(e^{i\,\alpha^a \cal{T}_a}e^{i\,\beta^b \cal{T}_b}\right)\tag{3}\end{equation} for representation (1). Note that $\gamma$ is a function of $\alpha$ and $\beta$, we can Taylor expand both sides of the above equation in terms of the components $\alpha^a$ and $\beta^a$. The left side of Eq. (3) is  \begin{equation}\text{LHS}\equiv  A^{a}{\cal{T}}_a + B^{a}_{b}{\cal{T}}_a \alpha^b + C^{a}_{b}{\cal{T}}_a \beta^b + D^{a}_{bc}{\cal{T}}_a \alpha^b\alpha^c + E^{a}_{bc}{\cal{T}}_a \beta^b\beta^c + F^{a}_{bc}{\cal{T}}_a \alpha^b\beta^c +\cdots \tag{4}\end{equation} for some constants to be determined, while the right side of Eq. (3) turns out to be  \begin{equation}\text{RHS} = i\,\alpha^a {\cal{T}}_a+i\,\beta^a {\cal{T}}_a-\frac{1}{2}\alpha^a\beta^b\Big[\cal{T}_a\,,\cal{T}_b\Big]+\cdots \tag{5}\end{equation} where $\Big[\,,\,\Big]$ is the commutator between two operators as defined by \begin{equation}\Big[\cal{A}\,,\cal{B}\Big]\equiv\cal{A}\cal{B}-\cal{B}\cal{A}\,.\tag{6}\end{equation} Eq. (3) requires that the corresponding coefficients of each order in Eq. (4) and (5) must be equal:\begin{eqnarray} \text{zero order: } && A^a = 0\\  \text{first order: } && B^{a}_{b}=C^{a}_{b}=i\,\delta^a_{b}\\ \text{second order: } && D^{a}_{bc}=E^{a}_{bc}=0\,,\quad F^{c}_{ab}{\cal{T}}_c=-\frac{1}{2}\alpha^a\beta^b\Big[\cal{T}_a\,,\cal{T}_b\Big]\\ \text{higher order: }&&\cdots \end{eqnarray} The key point here is not to solve all the coefficients, but to find the constraint on the generators. The equality of the second order requires that $\Big[\cal{T}_a\,,\cal{T}_b\Big]$ must be a linear combination of generators denoted by \begin{equation}\Big[\cal{T}_a\,,\cal{T}_b\Big]=i\,f_{ab}^{\,\,\,\,c}\,\cal{T}_c\tag{7}\end{equation} for some constants $f_{ab}^{\,\,\,\,c}$.

Therefore, the closure requirement of Lie group imposes the commutation relation (7). Furthermore,  Eq. (7) is the only requirement on generators: By the Baker–Campbell–Hausdorff formula \begin{equation} e^{{X}}e^{{Y}}=\exp\left\{{{X}}+{{Y}}+\frac{1}{2}\Big[{{X}}\,,{{Y}}\Big] +\frac{1}{12}\Big[{{X}}\,, \Big[{{X}}\,,{{Y}}\Big]\Big] +\frac{1}{12}\Big[{{Y}}\,, \Big[{{Y}}\,,{{X}}\Big]\Big]+\cdots \right\}\,,\end{equation} once Eq. (7) is met, all higher orders automatically become a linear combination of generators and thus provides no additional constraints.

Remarks:

1. Since the commutation relation (7) among generators is the result of the closure property of Lie group multiplications, it is the same for all representations, i.e., for a different representation, say ${\cal{A}}(\alpha)\equiv e^{i\alpha^a \cal{A}_a}$, there is the same commutation relation: \begin{equation}\Big[\cal{A}_a\,,\cal{A}_b\Big]=i\,f_{ab}^{\,\,\,\,c}\,\cal{A}_c\,,\end{equation} where $\,f_{ab}^{\,\,\,\,c}$ is the same as that in Eq. (7).
2. With the relation (7), generators span a Lie algebra $\mathfrak{g}$. In general, a Lie algebra $\mathfrak{g}$ is a vector space equipped with a binary operation $[\,,\,]$ called Lie bracket that satisfies the properties:
• Bilinear: $\left[Z\,,c_1X+c_2Y\right]=c_1\left[Z\,,X\right]+c_2\left[Z\,,Y\right]$
• Skew: $\left[X\,,Y\right]=-\left[Y\,,X\right]$
• Jacobi identity: $\left[X\,,\left[Y\,,Z\right]\right]+\left[Y\,,\left[Z\,,X\right]\right]+\left[Z\,,\left[X\,,Y\right]\right]=0$.
3. The commutator (6) is a Lie bracket. It is straightforward to verify that it satisfies the above three properties of Lie bracket.

Adjoint Representation

Definition of adjoint representation of Lie group

Recall that the adjoint representation for finite groups is defined as $\cal{A}(g)|h\rangle = |ghg^{-1}\rangle$ where the group elements form a basis of the representation space.

Now for Lie groups, we define their adjoint representations as \begin{equation}{\cal{A}}(\alpha)\left|\cal{T}_b\right\rangle = \left| e^{i\alpha^a \cal{T}_a}\,\cal{T}_b\,e^{-i\alpha^c \cal{T}_c}\right\rangle\,,\tag{8}\end{equation}where their generators form a basis $|{\cal{T}}_1\rangle,\cdots, |{\cal{T}}_N\rangle$ of the representation space.

Similar to the finite group case, it is easy to verify that $\cal{A}(\alpha)$ as defined in Eq. (8) is a representation, i.e., if $e^{i\alpha^a \cal{T}_a}e^{i\beta^b \cal{T}_b}=e^{i\gamma^c \cal{T}_c}$, then $\cal{A}(\alpha)\cal{A}(\beta)=\cal{A}(\gamma)$.

Definition of adjoint representation of Lie algebra

We define the adjoint representation of Lie algebra as the generators of the adjoint representation of Lie group. That is, when we write Lie group's adjoint representation (8) in the exponential form as \begin{equation}{\cal{A}}(\alpha)\equiv e^{i\alpha^a\cal{A}_a}\,,\end{equation} $\cal{A}_a$ is the adjoint representation of Lie algebra. 

Remarks:

• The commutator relation (7) holds for $\cal{A}_a$ as well because they follow the same Lie group multiplication.
• Adjoint representation plays a key role in the classification of semisimple Lie algebra.

Note that generators span a representation space in which the vector addition and scaling operations are defined by \begin{equation} \left|\alpha {\cal{T}}_a+ \beta {\cal{T}}_b\right\rangle \equiv \alpha|{\cal{T}_a}\rangle+\beta |{\cal{T}_b}\rangle\,,\end{equation} we can then reduce Eq. (8) to \begin{equation}\color{red}{\cal{A}_a\,\left|\cal{T}_b\right\rangle=\left|\,\left[\cal{T}_a,\,\cal{T}_b\right]\, \right\rangle}\,.\tag{9}\end{equation} by Taylor expanding both sides of Eq. (8) to the first order of $\alpha$.

[Exercise 1] Some literature directly lists Eq. (9) as the definition of  the adjoint representation of Lie algebra. If so, prove that $\cal{A}_a$ defined in Eq. (9) preserves the commutator relation (7).

[Solution] Note that ${\cal{A}}_a{\cal{A}}_b\left| {\cal{T}_c}\right\rangle={\cal{A}}_a\left| \left[{\cal{T}_b}\,,{\cal{T}_c}\right]\right\rangle=\left| \left[{\cal{T}_a}\,,\left[{\cal{T}_b}\,,{\cal{T}_c}\right]\right]\right\rangle$, we can compute \begin{eqnarray}\Big[{\cal{A}}_a\,,{\cal{A}}_b\Big]\left| {\cal{T}_d} \right\rangle&=&{\cal{A}}_a{\cal{A}}_b\left| {\cal{T}_d}\right\rangle-{\cal{A}}_b{\cal{A}}_a\left| {\cal{T}_d} \right\rangle= \left| \left[{\cal{T}_a}\,,\left[{\cal{T}_b}\,,{\cal{T}_d}\right]\right]\right\rangle-\left| \left[{\cal{T}_b}\,,\left[{\cal{T}_a}\,,{\cal{T}_d}\right]\right]\right\rangle= \left| \left[\left[{\cal{T}_a}\,,{\cal{T}_b}\right]\,,{\cal{T}_d}\right]\right\rangle\\&=&i\,f_{ab}^{\,\,\,\,c}\left| \left[{\cal{T}_c}\,,{\cal{T}_d}\right]\right\rangle= i\,f_{ab}^{\,\,\,\,c} {\cal{A}}_c\left| {\cal{T}_d}\right\rangle\end{eqnarray} for any $\left|{\cal{T}_d}\right\rangle$. As a result, $\Big[{\cal{A}}_a\,,{\cal{A}}_b\Big]\left| {\cal{T}_d} \right\rangle=i\,f_{ab}^{\,\,\,\,c} {\cal{A}}_c$.

Transformation of adjoint representation of Lie algebra

The generators of a Lie group are not unique, since the exponential map (1) only depends on the linear combination of generators. We can express the same representation (1) in terms of a different set of generators ${\cal{T}}'_a$ and the parameters $\beta$ as \begin{equation} {\cal{D}}(\beta)= e^{i\,\beta^a {\cal{T}}'_a}\,.\end{equation} Consequently, the commutation relations becomes \begin{equation}\Big[\cal{T}'_a\,,\cal{T}'_b\Big]=i\,{f'}_{ab}^{\,\,\,\,c}\,\cal{T}'_c\end{equation} for different constants ${f'}_{ab}^{\,\,\,\,c}$, and the adjoint representation for the generator ${\cal{T}}'_a$ now becomes $\cal{A}'_a$ as defined by \begin{equation}\cal{A}'_a\,\left|\cal{T}'_b\right\rangle=\left|\,\left[\cal{T}'_a,\,\cal{T}'_b\right] \right\rangle\,.\tag{10}\end{equation} 

To find the relation between $\cal{A}_a$ in Eq. (9) and $\cal{A}'_a$ in Eq. (10), we introduce a general linear transformation $\cal{L}$ such that  \begin{equation}\left|{\cal{T}}'_a\right\rangle \equiv {\cal{L}}\left|{\cal{T}}_a\right\rangle =\sum_{b=1}^N\left|{\cal{T}}_b\right\rangle\mathbf{L}_{ba}\,.\tag{11}\end{equation} As a result, we have ${\cal{A}}'_a\,\left|{\cal{T}}'_b\right\rangle=\sum_{c,d=1}^N\left|\,\left[\cal{T}_c,\,\cal{T}_d\right] \right\rangle\mathbf{L}_{ca}\mathbf{L}_{db}=\sum_{c,d=1}^N{\cal{A}}_c\left|{\cal{T}}_d \right\rangle\mathbf{L}_{ca}\mathbf{L}_{db}=\sum_{c=1}^N{\cal{A}}_c\left|{\cal{T}}'_b \right\rangle\mathbf{L}_{ca}$ for any $\left|{\cal{T}}'_b\right\rangle$, and thus \begin{equation}{\cal{A}}'_a =\sum_{b=1}^N {\cal{A}}_b\mathbf{L}_{ba}\,.\tag{12}\end{equation} Remarks:

• Recall that ordinary operators remain the same when changing the basis of the representation space. Only their matrix form changes by a similar transformation. In contrast, here the adjoint representation changes as in Eq. (12) when changing the basis as in Eq. (11).
• The transformation matrix $\mathbf{L}$ applies on the subscript index $a$ in the adjoint representation.
• As seen in Eq. (11) and (12), the adjoint representation and its basis changes in the same way. It is expected since they corresponds to the same underlying generator.

Matrix form of adjoint representation of Lie algebra

We want to prove that the matrix element of $\cal{A}_a$ is \begin{equation}\color{red}{[\mathbf{A}_a]_{cb}=i\,f_{ab}^{\,\,\,\,c}}\,.\tag{13}\end{equation}

The proof is in two steps:

1. If $|{\cal{T}}_1\rangle,\cdots, |{\cal{T}}_N\rangle$ is an orthonormal basis, i.e.,  $\langle \cal{T}_a|\cal{T}_b\rangle=\delta_{ab}$, then by submitting Eq. (7) into Eq. (9), we have the matrix form of $\cal{A}_a$ as  \begin{equation}[\mathbf{A}_a]_{cb}=\left\langle\cal{T}_c\right| \cal{A}_a\left|\cal{T}_b\right\rangle =i\,f_{ab}^{\,\,\,\,c}\,.\tag{14}\end{equation}
2. Now in a general basis $|{\cal{T}}'_1\rangle,\cdots, |{\cal{T}}'_N\rangle$, the matrix form of the adjoint representation is \begin{equation} {\cal{A}}'_a\left|\cal{T}'_b\right\rangle =\sum_{c=1}^N \left|\cal{T}'_c\right\rangle\,\left[\mathbf{A}'_a\right]_{cb}\,.\tag{15}\end{equation} By submitting Eq. (11), (12) and (14) into Eq. (15), we obtain the relation \begin{equation}\mathbf{A}'_a=\sum_{b=1}^N\mathbf{L}^{-1}\mathbf{A}_b\mathbf{L}\,\mathbf{L}_{ba}\,.\tag{16}\end{equation} Meanwhile, since $\Big[{\cal{T}'_a}\,,{\cal{T}'_b}\Big]=\sum_{c,d=1}^N\Big[{\cal{T}}_c\,,{\cal{T}}_d\Big]\mathbf{L}_{ca}\mathbf{L}_{db}=i\sum_{c,d=1}^N f_{cd}^{\,\,\,\,e}\mathbf{L}_{ca}\mathbf{L}_{db}\,{\cal{T}}_e=i\sum_{c,d,g=1}^N f_{cd}^{\,\,\,\,e}\mathbf{L}_{ca}\mathbf{L}_{db}\mathbf{L}^{-1}_{ge}\,{\cal{T}}'_g$, we have \begin{equation} {f'}_{ab}^{\,\,\,\,c} = \sum_{d,m,n=1}^N \mathbf{L}^{-1}_{cm}\, f_{dn}^{\,\,\,\,m}\,\mathbf{L}_{nb}\,\mathbf{L}_{da}\,.\tag{17}\end{equation} Using Eq. (14), (16) and (17), we prove that \begin{equation}[\mathbf{A}'_a]_{cb}=i\,{f'}_{ab}^{\,\,\,\,c}\,\end{equation} holds in a general basis as well.

Remarks: 

• $[\mathbf{A}_a]_{cb}=\left\langle\cal{T}_c\right| \cal{A}_a\left|\cal{T}_b\right\rangle$ only holds in an orthonormal basis. But $[\mathbf{A}_a]_{cb}=i\,f_{ab}^{\,\,\,\,c}$ holds in any basis, as we proved above.

[Exercise 2] Some literature directly lists Eq. (13) as the definition of  the adjoint representation of Lie algebra. If so, prove that $\cal{A}_a$ defined in Eq. (13) preserves the commutator relation (7).

[Solution] Note that $\left\langle {\cal{T}_m} \right| {\cal{A}_a}{\cal{A}_b}\left|{\cal{T}_n}\right\rangle=\sum_{c=1}^N \left\langle {\cal{T}_m} \right| {\cal{A}_a}\left| {\cal{T}_c}\right\rangle\left\langle {\cal{T}_c}\right|{\cal{A}_b}\left|{\cal{T}_n}\right\rangle=-f_{ac}^{\,\,\,\,m}f_{bn}^{\,\,\,\,c}$, we can compute \begin{eqnarray} \left\langle {\cal{T}_m} \right| \Big[{\cal{A}_a}\,,{\cal{A}_b}\Big]\left|{\cal{T}_n}\right\rangle &=& \left\langle {\cal{T}_m} \right| {\cal{A}_a}{\cal{A}_b}\left|{\cal{T}_n}\right\rangle-\left\langle {\cal{T}_m} \right| {\cal{A}_b}{\cal{A}_a}\left|{\cal{T}_n}\right\rangle = -f_{ac}^{\,\,\,\,m}f_{bn}^{\,\,\,\,c} +f_{bc}^{\,\,\,\,m}f_{an}^{\,\,\,\,c}\\&=&-f_{ab}^{\,\,\,\,c}f_{cn}^{\,\,\,\,m}=i\,f_{ab}^{\,\,\,\,c}\left\langle {\cal{T}_m} \right| {\cal{A}_c}\left|{\cal{T}_n}\right\rangle\end{eqnarray} for any $\left\langle{\cal{T}_m}\right|$ and $\left|{\cal{T}_n}\right\rangle$, and thus $\Big[{\cal{A}}_a\,,{\cal{A}}_b\Big]\left| {\cal{T}_d} \right\rangle=i\,f_{ab}^{\,\,\,\,c} {\cal{A}}_c$. In third equality above, we have applied the relation \begin{equation}f_{ac}^{\,\,\,\,m}f_{bn}^{\,\,\,\,c}+f_{bc}^{\,\,\,\,m}f_{na}^{\,\,\,\,c}+f_{nc}^{\,\,\,\,m}f_{ab}^{\,\,\,\,c}=0\end{equation} which is obtained by submitting the commutator relation (7) into the Jacobi identity.