Given a matrix Lie group $G$, a representation $\Pi$ of $G$ is a Lie group homomorphism $\Pi: G\longrightarrow\mathrm{GL}(V)$, where $V$ is a finite dimensional vector space and the general linear group $\mathrm{GL}(V)$ is the set of all linear isomorphisms of $V$. For each $g\in G$, $\Pi(g): V\longrightarrow V$ is a linear operator on $V$.

If $\mathfrak{g}$ is a Lie algebra, a representation of $\mathfrak{g}$ is a Lie algebra homomorphism $\pi: \mathfrak{g}\longrightarrow\mathrm{gl}(V)$, where $\mathrm{gl}(V)$ is the Lie algebra of $\mathrm{GL}(V)$.

If $\Pi$ or $\pi$ is a one-to-one homomorphism, then the representation is called faithful.

One may understand a representation as the action of a Lie group or a Lie algebra on the vector space $V$.

Example. [Trivial Representation] Let $G$ be a matrix Lie group. Define the trivial representation of $G$ by $$\Pi: G\longrightarrow\mathrm{GL}(1;\mathbb{C});\ A\longmapsto I.$$ This is an irreducible representation since $\mathbb C$ has no nontrivial subspace. If $\mathfrak{g}$ is a Lie algebra, the trivial representation of $\mathfrak{g}$, $\pi: \mathfrak{g}\longrightarrow\mathrm{gl}(1;\mathbb{C})$ is defined by $\pi(X)=0$ for all $X\in\mathfrak{g}$. This is also an irreducible representation.

Example. [The Adjoint Representation] Let $G$ be a matrix Lie group with Lie algebra $\mathfrak{g}$. The adjoint mapping $\mathrm{Ad}: G\longrightarrow\mathrm{GL}(\mathfrak{g})$ is defined by $$\mathrm{Ad}_A(X)=AXA^{-1}$$ for $A\in G$. We claim that $AXA^{-1}\in\mathfrak{g}$ for $A\in G$ and $X\in\mathfrak{g}$, so that $\mathrm{Ad}_A:\mathfrak{g}\longrightarrow\mathfrak{g}$. First note that for any invertible matrix $A$, $(AXA^{-1})^m=AX^mA^{-1}$. So, \begin{align*}
e^{AXA^{-1}}&=\sum_{m=0}^\infty\frac{(AXA^{-1})^m}{m!}\\
&=A\sum_{m=0}^\infty\frac{X^m}{m!}A^{-1}\\
&=Ae^XA^{-1}.
\end{align*}
Now for $A\in G$ and $X\in\mathfrak{g}$,
\begin{align*}
e^{tAXA^{-1}}&=e^{A\cdot tX\cdot A^{-1}}\\
&=Ae^{tX}A^{-1}\in G
\end{align*} and hence $AXA^{-1}\in\mathfrak{g}$. Note that $\mathrm{Ad}: G\longrightarrow\mathrm{GL}(\mathfrak{g})$ is a Lie group homomorphism. So $\mathrm{Ad}$ can be considered as a representation of $G$ acting on the Lie algebra $\mathfrak{g}$. We call $\mathrm{Ad}$ the adjoint representation of $G$. We can also define the adjoint representation of the Lie algebra $\mathfrak{g}$ as follows:
$$\mathrm{ad}:\mathfrak{g}\longrightarrow\mathrm{gl}(\mathfrak{g});\ \mathrm{ad}_X(Y)=[X,Y].$$ $\mathrm{ad}$ is a Lie algebra homomorphism.

Let $V$ be a finite dimensional real vector space. Then complexification of $V$, $V_{\mathbb{C}}$ is the space of formal linear combination $v_1+iv_2$ with $v_1,v_2\in V$. This is again a real vector space. If we define $$i(v_1+iv_2)=-v_2+iv_1,$$ then $V_{\mathbb{C}}$ becomes a complex vector space. For example, the complexification $\mathfrak{su}(2)_{\mathbb{C}}$ of the Lie algebra $\mathfrak{su}(2)$ is $\mathfrak{sl}(2;\mathbb{C})$.

Some Representations of $\mathrm{SU}(2)$

Let $V_m$ be the space of homogeneous polynomials in two variables with total degree $m\geq 0$
$$f(z_1,z_2)=a_0z_1^m+a_1z_1^{m-1}z_2+a_2z_1^{m-2}z_2^2+\cdots+a_mz_2^m.$$ Then $V_m$ is an $(m+1)$-dimensional complex vector space. Define $\Pi_m:\mathrm{SU}(2)\longrightarrow\mathrm{GL}(V_m)$ by $$[\Pi_m(U)f](z)=f(U^{-1}z).$$
Let us write $U^{-1}=\begin{pmatrix}
U_{11}^{-1} & U_{12}^{-1}\\
U_{21}^{-1} & U_{22}^{-1}
\end{pmatrix}$. Then $U^{-1}z=\begin{pmatrix}
U_{11}^{-1}z_1+U_{12}^{-1}z_2\\
U_{21}^{-1}z_1+U_{22}^{-1}z_2
\end{pmatrix}$, where $z=\begin{pmatrix}z_1\\z_2\end{pmatrix}\in\mathbb{C}^2$. So $[\Pi_m(U)f](z_1,z_2)$ is written as
$$[\Pi_m(U)f](z_1,z_2)=\sum_{k=0}^ma_k(U_{11}^{-1}z_1+U_{12}^{-1}z_2)^{m-k}(U_{21}^{-1}z_1+U_{22}^{-1}z_2)^k.$$
We now show that $\Pi_m$ is indeed a Lie group homomorphism: \begin{align*}
\Pi_m(U_1)[\Pi_m(U_2)f](z)&=\Pi_m(U_2f)(U_1^{-1}z)\\
&=f(U_2^{-1}U_1^{-1}z)\\
&=f((U_1U_2)^{-1}z)\\
&=[\Pi_m(U_1U_2)f](z).
\end{align*} Therefore, $\Pi_m$ is a finite dimensional complex representation of $\mathrm{SU}(2)$. Note that each $\Pi_m$ is irreducible and that every finite dimensional irreducible representation of $\mathrm{SU}(2)$ is equivalent to one and only one of the $\Pi_m$’s.

Now we compute the corresponding representation $\pi_m$ of the Lie algebra $\mathfrak{su}(2)$. $\pi_m$ can be computed as $$\pi_m(X)=\frac{d}{dt}\Pi_m(e^{tX})|_{t=0}.$$ So,
\begin{align*}
[\pi_m(X)f](z)&=\frac{d}{dt}[\Pi_m(e^{tX})f](z)|_{t=0}\\
&=\frac{d}{dt}f(e^{-tX}z)|_{t=0}.
\end{align*} Let $z(t)$ be a curve in $\mathbb{C}^2$ defined as $z(t)=e^{-tX}z$, so that $z(0)=z$. Write $z(t)=(z_1(t),z_2(t))$, where $z_i(t)\in\mathbb{C}$, $i=1,2$. By the chain rule, \begin{align*}
\pi_m(X)f&=\frac{\partial f}{\partial z_1}\frac{dz_1}{dt}|_{t=0}+\frac{\partial f}{\partial z_2}\frac{dz_2}{dt}|_{t=0}\\
&=-\frac{\partial f}{\partial z_1}(X_{11}z_1+X_{12}z_2)-\frac{\partial f}{\partial z_2}(X_{21}z_1+X_{22}z_2),\ \ \ \ \ \mbox{(1)}\end{align*}
since $\frac{dz}{dt}|_{t=0}=-Xz$.

Every finite dimensional complex representation of the Lie algebra $\mathfrak{su}(2)$ extends uniquely to a complex linear representation of the complexification of $\mathfrak{su}(2)$ and the complexification of $\mathfrak{su}(2)$ is isomorphic to $\mathfrak{sl}(2;\mathbb{C})$. Thus, the representation $\pi_m$ of $\mathfrak{su}(2)$ extends to a representation of $\mathfrak{sl}(2;\mathbb{C})$. Note that the Lie algebra $\mathfrak{sl}(2;\mathbb{C})$ is the set of all $2\times 2$ trace-free complex matrices, i.e. matrices of the form $\begin{pmatrix}\alpha & \beta\\\gamma & -\alpha\end{pmatrix}$ where $\alpha$, $\beta$ and $\gamma$ are complex numbers. So any element in $\mathfrak{sl}(2;\mathbb{C})$ can be uniquely written as $\alpha H+\beta X+\gamma Y$, where $H=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}$, $X=\begin{pmatrix}
0 & 1\\
0 & 0
\end{pmatrix}$, $Y=\begin{pmatrix}
0 & 0\\
1 & 0
\end{pmatrix}$.
Let us calculate $\pi_m$ for the basis members $H$, $X$, and $Y$.
$$(\pi_m(H)f)(z)=-\frac{\partial f}{\partial z_1}z_1+\frac{\partial f}{\partial z_2}z_2$$ so
$$\pi_m(H)=-z_1\frac{\partial}{\partial z_1}+z_2\frac{\partial}{\partial z_2}.$$
Applying $\pi_m(H)$ to a basis element $z_1^kz_2^{m-k}$, we obtain
\begin{align*}
\pi_m(H)z_1^kz_2^{m-k}&=-kz_1^kz_2^{m-k}+(m-k)z_1^kz_2^{m-k}\\
&=(m-2k)z_1^kz_2^{m-k}.
\end{align*}
This means that $z_1^kz_2^{m-k}$ is an eigenvector for $\pi_m(H)$ with eigenvalue $m-2k$. In particular, $\pi_m(H)$ is diagonalizable. Using (1) again we also obtain
$$\pi_m(X)=-z_2\frac{\partial}{\partial z_1},\ \pi_m(Y)=-z_1\frac{\partial}{\partial z_2}$$
and
\begin{align*}
\pi_m(X)z_1^kz_2^{m-k}&=-kz_1^{k-1}z_2^{m-k+1},\ \ \ \ \ \mbox{(2)}\\
\pi_m(Y)z_1^kz_2^{m-k}&=(k-m)z_1^{k+1}z_2^{m-k-1}.\ \ \ \ \ \mbox{(3)}
\end{align*}

Proposition. The representation $\pi_m$ is an irreducible representation of $\mathfrak{sl}(2;\mathbb{C})$.

Proof. Suppose that $W$ is a nonzero invariant subspace of $V_m$. We claim that $W=V_m$. Since $W\ne \{0\}$, there exists $w\in W$ with $w\ne 0$. $w$ can be uniquely written as
$$w=a_0z_1^m+a_1z_1^{m-1}z_2+a_2z_1^{m-2}z_2^2+\cdots+a_mz_2^m$$ with at least one of the $a_k$’s nonzero. Let $k_0$ be the smallest value of $k$ for which $a_k\ne 0$ and consider $\pi_m(X)^{m-k_0}w$. Since $\pi_m(X)$ lowers the power of $z_1$ by 1, $\pi_m(X)^{m-k_0}w$ will kill all the terms in $w$ except $a_{k_0}z_1^{m-k_0}z_2^{k_0}$. On the other hand,
$$\pi_m(X)^{m-k_0}(z_1^{m-k_0}z_2^{k_0})=(-1)^{m-k_0}(m-k_0)!z_2^m.$$ Since $\pi_m(X)^{m-k_0}(z_1^{m-k_0}z_2^{k_0})$ is a multiple of $z_2^m$ and $W$ is invariant, $W$ must contain $z_2^m$. It follows from (2) that $\pi_m(Y)^kz_2^m$ is a nonzero multiple of $z_1^kz_2^{m-k}$ for $0< k\leq m$. Hence, $W$ must contain $z_1^kz_2^{m-k}$, $0< k\leq m$. Since $W$ contains all the basis members of $V_m$, $z_1^kz_2^{m-k}$, $0\leq k\leq m$, then $W=V_m$.