Basics of Group Representation

The representation of a group $G$ is a mapping from any $g\in G$ to a linear operator $\cal{D}(g)$ that preserves the group multiplication.

Note:

1. Recall that operators are linear transformations that map one vector to another vector. So the multiplication between two operators $\cal{D}(g_1)\cal{D}(g_2)$ is defined as two sequential transformations \begin{equation}\cal{D}(g_1)\cal{D}(g_2)\,\,|\psi\rangle:= \cal{D}(g_1)\,\Big(\cal{D}(g_2)|\psi\rangle\Big)\end{equation} when acting on any vector $|\psi\rangle$.
2. The requirement of preserving the group multiplication means \begin{equation}\cal{D}(g_1)\cal{D}(g_2)=\cal{D}(g_1g_2)\,\tag{1}\end{equation} for any $g_1,g_2\in G$.

[Exercise 1] Prove that the representation forms a group.

[Solution] 

1. Closure: By Eq. (1).
2. Associativity: \begin{eqnarray}\cal{D}(g_1)\Big(\cal{D}(g_2)\cal{D}(g_3)\Big)=\cal{D}(g_1)\cal{D}(g_2g_3)&=&\cal{D}(g_1(g_2g_3))\\&=&\cal{D}((g_1g_2)g_3)=\cal{D}(g_1g_2)\cal{D}(g_3)=\Big(\cal{D}(g_1)\cal{D}(g_2)\Big)\cal{D}(g_3)\end{eqnarray}
3. Identity element is $\cal{D}(e)$ because $\cal{D}(g)\cal{D}(e)=\cal{D}(ge)=\cal{D}(g)$ and $\cal{D}(e)\cal{D}(g)=\cal{D}(eg)=\cal{D}(g)$.
4. Inverse element for any $\cal{D}(g)$ is  $\cal{D}(g^{-1})$ because $\cal{D}(g^{-1})\cal{D}(g)=\cal{D}(g^{-1}g)=\cal{D}(e)$ and $\cal{D}(g)\cal{D}(g^{-1})=\cal{D}(gg^{-1})=\cal{D}(e)$.

Note: 

• From the above exercise, we can see that Eq. (1) already contains the requirements that \begin{equation}\cal{D}(e)=\cal{I}\,.\end{equation} and that all the operators in the representation must be invertible \begin{equation}\cal{D}^{-1}(g):=\cal{D}(g^{-1})\,.\end{equation}

As shown in this post, given a basis $|\alpha_1\rangle, \cdots, |\alpha_n\rangle$, there is one-to-one correspondence between an operator and its matrix form: \begin{equation}\cal{D}(g) \Big[|\alpha_1\rangle, \cdots, |\alpha_n\rangle\Big]=\Big[|\alpha_1\rangle, \cdots, |\alpha_n\rangle\Big]\,\mathbf{D}(g)\,.\end{equation} So in some textbook, the representation of a group $G$ is directly defined as a mapping from any $g\in G$ to a $n\times n$ invertible matrix of complex numbers $\mathbf{D}(g)$. 

Remarks:

1. All the invertible matrices form a group under matrix multiplication called general linear group $\text{GL}\left(n, \mathbb{C}\right)$.
2. There is no requirement for the dimension $n$. In other word, a group can has representations in any dimensions. 
3. The same operator $\cal{D}(g)$ has different but similar matrix forms under different basis \begin{equation}\mathbf{D'}(g)=\mathbf{S}^{-1}\,\mathbf{D}(g)\,\mathbf{S}\,.\end{equation} The central task of representation theory is to find a suitable basis within which the matrix form $\mathbf{D'}(g)$ becomes block diagonal if $\mathbf{D}(g)$ is not block diagonal.
4. In an orthonormal basis, $\mathbf{D}(e)=\mathbf{I}$. Then in any basis, $\mathbf{D'}(e)=\mathbf{I}$ by similar transformation.

Adjoint Representation

For any finite group $G$, its group elements can form a basis $|g_1\rangle, \cdots, |g_n\rangle$ of a representation space. There is a so-called adjoint representation $\cal{A}$ defined by \begin{equation}\cal{A}(g)\,|h\rangle :=| g hg^{-1}\tag{2}\rangle\end{equation} for any $g, h\in G$.

[Exercise 2] Verify that Eq. (2) forms a representation.

[Solution] We only need to check Eq. (2) preserves the group multiplication. Indeed, for any $g_1, g_2, h\in G$, 

\begin{equation}\cal{A}(g_1)\cal{A}(g_2)\, |h\rangle=\cal{A}(g_1) |g_2hg^{-1}_2\rangle= |g_1g_2hg^{-1}_2g^{-1}_1\rangle=|(g_1g_2)h(g_1g_2)^{-1}\rangle=\cal{A}(g_1g_2)|h\rangle\,,\end{equation} and thus $\cal{A}(g_1)\cal{A}(g_2)=\cal{A}(g_1g_2)$.

Note:

• If the basis $|g_1\rangle, \cdots, |g_n\rangle$ is orthonormal, i.e. $\langle g_i| g_j\rangle=\delta_{ij}$ for any $g_i,g_j\in G$, then the matrix element of $\cal{A}(g)$ is \begin{equation}\mathbf{A}_{ij}(g)=\langle g_i|\cal{A}(g)|g_j\rangle\,.\tag{3}\end{equation} 
• If the basis $|g_1\rangle, \cdots, |g_n\rangle$ is not orthonormal, the matrix form $\mathbf{A'}(g)$ differs from the above $\mathbf{A}(g)$ by a similar transformation.
• To find irreducible representation, one can start with any matrix form and diagonalize it. So for convenience, we usually use the matrix form (3) as the adjoint representation.