# The Lorentz Group

Let $\mathbb{L}^4$ be $\mathbb{R}^4$ with the Lorentzian inner product $\langle\ ,\ \rangle$ defined by
$$\langle v,w\rangle=-v^0w^0+v^1w^1+v^2w^2+v^3w^3$$
for $v={}^t(v^0,v^1,v^2,v^3),w={}^t(w^0,w^1,w^2,w^3)\in\mathbb{R}^4$. In particular,
$$||v||^2=-(v^0)^2+(v^1)^2+(v^2)^2+(v^3)^2.$$
$\mathbb{L}^4$ is called the Minkowski $4$-spacetime or simply 4-spacetime. In physics, $\mathbb{L}^4$ is commonly denoted by $\mathbb{R}^{3+1}$. Another common notation for $\mathbb{L}^4$ in differential geometry is $\mathbb{R}^4_1$.

A linear transformation $A:\mathbb{L}^4\longrightarrow\mathbb{L}^4$ is called a Lorentz transformation if it is a Lorentzian isometry i.e. a Lorentzian inner product preserving map. Note that $\langle v,w\rangle$ can be written in matrix form as
$$\langle v,w\rangle={}^tv(g_{ij})w,$$
where
$$(g_{ij})=\begin{pmatrix} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}.$$
The matrix $(g_{ij})$ is called a Lorentzian metric tensor. The set of all Lorentz transformations forms a Lie group called Lorentz group and is denoted  by $\mathrm{O}(3,1)$:
$$\mathrm{O}(3,1)=\{A\in\mathrm{GL}(\mathbb{L}^4): {}^tA(g_{ij})A=(g_{ij})\}.$$
$\mathrm{O}(3,1)$ contains conventional rotations such as one in the $x^1-x^2$ plane
$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & \cos\theta & -\sin\theta & 0\\ 0 & \sin\theta & \cos\theta & 0\\ 0 & 0 & 0 & 1 \end{pmatrix},\ 0\leq\theta<2\pi$$
plus Lorentz boots which may be regarded as rotation between space and time directions. An example of boosts is
$$\begin{pmatrix} \cosh\phi & \sinh\phi & 0 & 0\\ \sinh\phi & \cosh\phi & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix},\ -\infty<\phi<\infty.$$
Lorentz transformations leave the origin (present) fixed due to linearity. The set of all isometries of $\mathbb{L}^4$ constains Lorentz transformations and translations. It is a Lie group called Poincaré group.

For $A\in\mathrm{O}(3,1)$, $\det A=\pm 1$. Those Lorentz transformations with determinant 1 are spatial orientation (parity) preserving transformations. They form a Lie subgroup of $\mathrm{O}(3,1)$ and is denoted by $\mathrm{SO}(3,1)$:
$$\mathrm{SO}(3,1)=\{A\in\mathrm{O}(3,1): \det A=1\}.$$
$\mathrm{SO}(3,1)$ has two connected components. ($\mathrm{O}(3,1)$ has four connected components.) The identiy componenent of $\mathrm{SO}(3,1)$ is denoted by $\mathrm{SO}^+(3,1)$. $\mathrm{SO}^+(3,1)$ is the group of both time orientation and parity preserving Lorentz transformations.

# Orthogonal Group $\mathrm{O}(n)$ and Symmetry

Denote by $\langle\ ,\ \rangle$ the standard Euclidean inner product in $\mathbb{R}^n$. Then for any $v,w\in\mathbb{R}^n$,
$$\langle v,w\rangle={}^tvw.$$

Definition. A bijective map $A: \mathbb{R}^n\longrightarrow\mathbb{R}^n$ is said to be an isometry if it preserves the inner product $\langle\ ,\ \rangle$ i.e. for any vectors $v,w\in\mathbb{R}^n$
$$\langle Av,Aw\rangle=\langle v,w\rangle.$$
Isometries are usually called symmetries in physics.

The set of all isometries of $\mathbb{R}^n$ forms a group with composition $\circ$. The group is denoted by $\mathrm{E}(n)$ and called the Euclidean motion group. The orthogonal group $\mathrm{O}(n)$ is a subgroup of $\mathrm{E}(n)$ and that it is a group of all isometries of $\mathbb{R}^n$ which leaves the origin fixed. In other words,

Theorem. $\mathrm{O}(n)$ is the group of all linear isometries of $\mathbb{R}^n$.

Proof. Let $A\in\mathrm{O}(n)$. Then
\begin{align*}
\langle Av,Aw\rangle&=\langle v,{}^tAAw\rangle\\
&=\langle v,w\rangle
\end{align*}
since ${}^tAA=I$. Hence, $A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry. Conversely, if $A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry, then for any $v,w\in\mathbb{R}^n$
\begin{align*}
\langle Av,Aw\rangle=\langle v,w\rangle&\Longrightarrow\langle{}^tAAv,w\rangle=\langle v,w\rangle\\
&\Longrightarrow\langle {}^tAAv-v,w\rangle=0.
\end{align*}
Since $w$ is arbitrary,
$$({}^tAA-I)v=0.$$
Since $v$ is also arbitrary,
$${}^tAA=I$$
i.e. $A$ is an orthogonal matrix.

Remark. For any $v,w\in\mathbb{R}^n$,
$$||v+w||^2=||v||^2+2\langle v,w\rangle+||w||^2$$
where $||v||=\sqrt{\langle v,v\rangle}$, the Euclidean norm of $v$. Plugging in $Av$ and $Aw$ for $v$ and $w$ respectively, we obtain
$$\langle Av,Aw\rangle=\frac{1}{2}(||A(v+w)||^2-||Av||^2-||Aw||^2).$$
This equation tells that a linear isomorphism $A:\mathbb{R}^n\longrightarrow\mathbb{R}^n$ is an isometry if and only if it preserves the norm  $||\cdot||$.

Let $A\in\mathrm{O}(n)$. Since $\det A=\det {}^tA$, $\det A=\pm 1$. This implies that $\mathrm{O}(n)$ has two connected components. One that contains orthogonal matrices whose determinant is $1$, i.e. $\mathrm{SO}(n)$, and the other that contains orthogonal matrices whose determinant is $-1$. The identity component $\mathrm{SO}(n)$ of $\mathrm{O}(n)$ is the group of all linear isometries that preserve orientation. Clearly $\mathrm{SO}(n)$ is a normal subgroup of $\mathrm{O}(n)$.

The Lie groups $\mathrm{O}(n)$ and $\mathrm{SO}(n)$ are compact.

# Quantum Angular Momentum in $\mathbb{R}^{2+2}$ and $\mathfrak{su}(1,1)$ Representation

It can be shown that quantum angular momentum
\begin{align*}
L_x&=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\
L_y&=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\
L_z&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}
can be obtained purely mathematically by $\mathfrak{su}(2)$ Lie algebra representation as discussed here. Since $\mathfrak{su}(2)$ representation contains information on the symmetry of $\mathbb{R}^3$, one can speculate that the symmetry of the background space plays a crucial role in quantum mechanics.

For fun, let us consider quantum mechanics in $\mathbb{R}^{2+2}$, a 4-space with 2 time dimensions. (I said for fun, so let us not worry about whether it is physically meaniful or not for now.) In this case, can we also derive quantum angular momentum (or something like it) by a Lie algebra representation? If so, what is a relevant Lie algebra? To answer this question, we need to understand the symmetry of $\mathbb{R}^{2+2}$.

The rotations (actually Euclidean rotation and Lorentz boosts), in particular orthochronous Lorentz transformations i.e. time-orientation and parity preserving Lorentz transformations in  Minkowski 3-space $\mathbb{R}^{2+1}$ form the special pseudo orthogonal group $\mathrm{SO}^+(2,1)$, the identity component of the Lorentz group $\mathrm{O}(2,1)$. The $2+1$ dimensional spacetime $\mathbb{R}^{2+1}$ can be identified with the set of $2\times 2$ matrices of the form
$$\underline{X}=\begin{pmatrix} \eta & x+iy\\ -(x-iy) & -\eta \end{pmatrix}$$
with the inner product $\langle\ ,\ \rangle$ defined by
$$\langle\underline{X},\underline{Y}\rangle=-\frac{1}{2}\mathrm{tr}\left[\underline{X}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \underline{Y}\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} \right]$$
In particular
$$|\underline{X}|^2=\det\underline{X}$$
Here the matrix $\underline{X}$ is identified with the 3-vector $X=(\eta,x,y)\in\mathbb{R}^{2+1}$. The identification is an isometry. The indefinte special unitary group $\mathrm{SU}(1,1)$ acts isometrically on $\mathbb{R}^{2+1}$ by the action
$$\mathrm{SU}(1,1)\times\mathbb{R}^{2+1}\longrightarrow\mathbb{R}^{2+1};\ (U,X)\longmapsto UXU^{-1}$$
For a fixed $U\in\mathrm{SU}(1,1)$, the map
$$\mathbb{R}^{2+1}\longrightarrow\mathbb{R}^{2+1};\ X\longmapsto UXU^{-1}$$
is an orthochronous Lorentz transformation of $\mathbb{R}^3$. Thus the Lie group action induces a Lie group representation $\rho:\mathrm{SU}(1,1)\longrightarrow\mathrm{SO}^+(2,1)$. Since both $U$ and $-U$ result the same isometry, the representation $\rho$ is a 2:1 map. The kernal of $\rho$ is $\mathbb{Z}_2=\{\pm I\}$, so we have $\mathrm{SU}(1,1)/\mathbb{Z}_2=\mathrm{SO}^+(2,1)$. The quotient group $\mathrm{SU}(1,1)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(1,1)$ is called the projective indefinite special unitary group. The double cover $\mathrm{SU}(1,1)$ of $\mathrm{SO}^+(2,1)$ is connected (but not simply connected) and there exists a short exact sequence
$$1\rightarrow\mathbb{Z}_2\rightarrow\mathrm{SU}(1,1)\rightarrow\mathrm{SO}^+(2,1)\rightarrow 1$$
So $\mathrm{SU}(1,1)$ may be regarded as the spin group $\mathrm{Spin}(2,1)$. Since the spin group $\mathrm{Spin}(p,q)$ of a split signature is required to be connected (but not necessarily simply connected), it may not uniquely exist unlike the spin group $\mathrm{Spin}(n)$. For instance, the linear special group $\mathrm{SL}(2,\mathbb{R})$ may also be considered as the spin group $\mathrm{Spin}(2,1)$.

Let $\mathcal{K}$ be the space of states $\psi$ as smooth functions on $\mathbb{R}^{2+1}$. It should be noted that $\mathcal{K}$ is not a Hilbert space but rather a Krein space. I will discuss about this some other time. Define a map $\Pi:\mathrm{SU(1,1)}\longrightarrow\mathrm{GL(\mathcal{K})}$ as follows: For each $U\in\mathrm{SU}(1,1)$, $\Pi(U):\mathcal{K}\longrightarrow\mathcal{K}$ is an isomorphism defined by
$$[\Pi(U)\psi](v)=\psi(\rho(U)^{-1}v),\ v\in\mathbb{R}^{2+1}$$
where $\rho$ is the covering map $\rho: \mathrm{SU}(1,1)\stackrel{2:1}{\longrightarrow}\mathrm{SO}^+(2,1)$. $\Pi$ is indeed a group homomorphism: For $U_1,U_2\in\mathrm{SU}(1,1)$,
\begin{align*}
\Pi(U_1)[\Pi(U_2)\psi](v)&=\Pi(U_2\psi)(\rho(U_1)^{-1}v)\\
&=\psi(\rho(U_2)^{-1}\rho(U_1)^{-1}v)\\
&=\psi((\rho(U_1)\rho(U_2))^{-1}v)\\
&=\psi(\rho(U_1U_2)^{-1}v)\\
&=[\Pi(U_1U_2)\psi](v)
\end{align*}
Hence, $\Pi$ is an infinite dimensional real representation of $\mathrm{SU}(1,1)$. The corresponding $\mathfrak{su}(1,1)$ Lie algebra representation $\pi$ can be computed as
$$\pi(X)=\frac{d}{dt}\Pi(e^{tX})|_{t=0}$$
So,
\begin{align*}
[\pi(X)\psi](v)&=\frac{d}{dt}[\Pi(e^{tX})\psi](v)|_{t=0}\\
&=\frac{d}{dt}\psi(\rho(e^{tX})^{-1}v)|_{t=0}
\end{align*}
The Lie algebra $\mathfrak{su}(1,1)$ has the canonical basis
\begin{align*}
X_1&=\frac{1}{2}\sigma_1=\frac{1}{2}\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}\\
X_2&=\frac{1}{2}\sigma_2=\frac{1}{2}\begin{pmatrix}
0 & i\\
-i & 0
\end{pmatrix}\\
X_3&=\frac{i}{2}\sigma_3=\frac{1}{2}\begin{pmatrix}
i & 0\\
0 & -i
\end{pmatrix}
\end{align*}
Let us calculate $\pi$ for the basis member $X_1$. $e^{\phi X_1}=\begin{pmatrix} \cosh\frac{\phi}{2} & \sinh\frac{\phi}{2}\\ \sinh\frac{\phi}{2} & \cosh\frac{\phi}{2} \end{pmatrix}$ and $\rho(e^{\phi X_1})=R_\phi^y$ where $R_\phi^y=\begin{pmatrix} \cosh\phi & -\sinh\phi & 0\\ -\sinh\phi & \cosh\phi & 0\\ 0 & 0 & 1\end{pmatrix}$ is rotation in $\mathbb{R}^{2+1}$ about the $y$-axis by a hyperbolic angle $\phi$. (Although I conveniently call this a rotation, note that this is not a Euclidean rotation but a rotation in spacetime. It is called a Lorentz boost in physics.) Let $v(\phi)$ be a curve in $\mathbb{R}^{2+1}$ defined by
$$v(\phi)=\rho(e^{\phi X_1})^{-1}v=(R_\phi^y)^{-1}v$$ so that $v(0)=v$. Write $v(\phi)=(\eta(\phi),x(\phi),y(\phi))$ and $v=(\eta,x,y)$. Then by the chain rule,
\begin{align*}
[\pi(X_1)\psi](v)&=\frac{\partial\psi}{\partial \eta}\frac{dx}{d\phi}|_{\phi=0}+\frac{\partial\psi}{\partial x}\frac{dx}{d\phi}|_{\phi=0}+\frac{\partial\psi}{\partial y}\frac{dy}{d\phi}|_{\phi=0}\\
&=x\frac{\partial\psi}{\partial \eta}+\eta\frac{\partial\psi}{\partial x}
\end{align*}
Hence,
$$\pi(X_1)=x\frac{\partial}{\partial \eta}+\eta\frac{\partial}{\partial x}$$
Using
\begin{align*}
e^{\phi X_2}&=\begin{pmatrix}
\cosh\frac{\phi}{2} & i\sinh\frac{\phi}{2}\\
-i\sinh\frac{\phi}{2} & \cosh\frac{\phi}{2}
\end{pmatrix},\ \rho(e^{\phi X_2})=R_\phi^x=\begin{pmatrix}
\cosh\phi & 0 & -\sinh\phi\\
0 & 1 & 0\\
-\sinh\phi & 0 & \cosh\phi
\end{pmatrix}\\
e^{\theta X_3}&=\begin{pmatrix}
e^{i\theta/2} & 0\\
0 & e^{-i\theta/2}
\end{pmatrix},\ \rho(e^{\sigma X_1})=R_\theta^\eta=\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\theta & -\sin\theta\\
0 & \sin\theta & \cos\theta
\end{pmatrix}
\end{align*}
one can find similar formulas for $\pi(X_2)$ and $\pi(X_3)$:
\begin{align*}
\pi(X_2)&=y\frac{\partial}{\partial \eta}+\eta\frac{\partial}{\partial y}\\
\pi(X_3)&=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}
\end{align*}
It turns out that
\begin{align*}
i\hbar\pi(X_1)&=L_y=i\hbar\left(\eta\frac{\partial}{\partial x}+x\frac{\partial}{\partial \eta}\right)\\
i\hbar\pi(X_2)&=L_x=i\hbar\left(y\frac{\partial}{\partial \eta}+\eta\frac{\partial}{\partial y}\right)\\
-i\hbar\pi(X_3)&=L_\eta=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}
are an analogue of the quantum angular momenta about the $y$-axis, $x$-axis and $\eta$-axis respectively. To see this, in $\mathbb{R}^{2+2}$ the momentum operator $p$ is given by

$$p^\mu =i\hbar \nabla^\mu =i\hbar\left(\frac{\partial}{\partial\eta},-\nabla\right)=i\hbar\left(\frac{\partial}{\partial\eta},-\frac{\partial}{\partial x},-\frac{\partial}{\partial y}\right) \ \ \ \ \ \ (1)$$
What may be called quantum angular momentum in $\mathbb{R}^{2+2}$ may be foramlly found by $L=r_\mu\times p^\mu$ with $p^\mu$ in (1). The cross product $v\times w$ in $\mathbb{R}^{2+1}$is given by
$$v\times w:=\left|\begin{array}{ccc} e_0 & e_1 & -e_2\\ v^1 & v^2 & v^3\\ w^1 & w^2 & w^3 \end{array}\right|$$
where $e_0=(1,0,0)$, $e_1(0,1,0)$, and $e_2=(0,0,1)$. Let $r_\mu=(\eta,x,y)$. Then
\begin{align*}
L&=r_\mu\times(i\hbar\nabla^\mu)\\
&=\left|\begin{array}{ccc}
e_0 & e_1 & -e_2\\
\eta & x & y\\
\frac{\partial}{\partial\eta} & -\frac{\partial}{\partial x} & -\frac{\partial}{\partial y}
\end{array}\right|\\
&=i\hbar\left[\left(-x\frac{\partial}{\partial y}+y\frac{\partial}{\partial x}\right)e_0+\left(y\frac{\partial}{\partial\eta}+\eta\frac{\partial}{\partial y}\right)e_1+\left(\eta\frac{\partial}{\partial x}+x\frac{\partial}{\partial\eta}\right)e_2\right]
\end{align*}
If we write $L=(L_\eta,L_x,L_y)$, then
\begin{align*}
L_\eta&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\\
L_x&=i\hbar\left(y\frac{\partial}{\partial\eta}+\eta\frac{\partial}{\partial y}\right)\\
L_y&=i\hbar\left(\eta\frac{\partial}{\partial x}+x\frac{\partial}{\partial\eta}\right)
\end{align*}
In quantum mechanics in $\mathbb{R}^{2+2}$, the conservation of the above quantum angular momentum $L$ is expected.

References:

 Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer-Verlag 2000

 F. Reese Harvey, Spinors and Calibrations, Academic Press 1990

 Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

# Quantum Angular Momentum and $\mathfrak{su}(2)$ Representation

In classical mechanics, the angular momentum of a body is given by
$$L=r\times p$$ where $r$ and $p$ denote radius arm and linear momentum respectively. In quantum mechanics, the angular momentum of a spinning particle can be obtained by replacing linear momentum $p$ by momentum operator $-i\hbar\nabla$. As a result, the components of quantum mechanical angular momentum $L$ is given by
\begin{align*}
L_x&=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\
L_y&=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\
L_z&=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}

It is interesting to see that the quantum mechanical angular momentum can be obtained purely from algebra, more specifically from representation theory. The relevant representations are the representations of the special unitary group $\mathrm{SU}(2)$ and its Lie algebra $\mathfrak{su}(2)$. This hints us that the symmetry of the background space plays a crucial role in quantum mechanics.

The rotations in $\mathbb{R}^3$ form the special orthogonal group $\mathrm{SO}(3)$. $\mathrm{SO}(3)$ is not simply connected (see  for detalis) and the special unitary group $\mathrm{SU}(2)$ is the universal covering group of $\mathrm{SO}(3)$. ($\mathrm{SU}(2)=S^3$ so it is simply connected.) The covering map $\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a $\mathrm{SU}(2)$ representation. To see this, note that Euclidean 3-space $\mathbb{R}^3$ can be identified with the set of $2\times 2$ hermitian matrices of the form
$$\underline{X}=\begin{pmatrix} z & x-iy\\ x+iy & -z \end{pmatrix}$$
with the inner product $\langle\ ,\ \rangle$ defined by
$$\langle\underline{X},\underline{Y}\rangle=\frac{1}{2}\mathrm{tr}(\underline{X}\cdot\underline{Y})$$
In particular
$$|\underline{X}|^2=\frac{1}{2}\mathrm{tr}\underline{X}^2=-\det\underline{X}$$
Here the hermitian matrix $\underline{X}$ is identified with the vector $(x,y,z)\in\mathbb{R}^3$. Since $|X|^2=-\det\underline{X}$, the identification is an isometry. $\mathrm{SU}(2)$ acts on $\mathbb{R}^3$ isometrically by the action
$$\mathrm{SU}(2)\times\mathbb{R}^3\longrightarrow\mathbb{R}^3;\ (U,X)\longmapsto U^{-1}XU$$
For a fixed $U\in\mathrm{SU}(2)$, the map
$$\mathbb{R}^3\longrightarrow\mathbb{R}^3;\ X\longmapsto U^{-1}XU$$
is an orientation preserving isometry of $\mathbb{R}^3$. Thus the Lie group action induces a Lie group representation $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$. Since both $U$ and $-U$ result the same isometry, the representation $\rho$ is a 2:1 map. The kernal of $\rho$ is $\mathbb{Z}_2=\{\pm I\}$, so we have $\mathrm{SU}(2)/\mathbb{Z}_2=\mathrm{SO}(3)$. The quotient group $\mathrm{SU}(2)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(2)$ is called the projective special unitary group. The double cover of the special orthogonal group $\mathrm{SO}(n)$ is called the spin group and is denoted by $\mathrm{Spin}(n)$. Hence, the double cover $\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is the spin group $\mathrm{Spin}(3)$.

Let $\mathcal{H}$ be the Hilbert space of states $\psi$ as smooth functions on $\mathbb{R}^3$. Define a map $\Pi:\mathrm{SU(2)}\longrightarrow\mathrm{GL(\mathcal{H})}$ as follows: For each $U\in\mathrm{SU}(2)$, $\Pi(U):\mathcal{H}\longrightarrow\mathcal{H}$ is an isomorphism defined by
$$[\Pi(U)\psi](v)=\psi(\rho(U)^{-1}v),\ v\in\mathbb{R}^3$$
where $\rho$ is the universal covering map $\rho: \mathrm{SU}(2)\stackrel{2:1}{\longrightarrow}\mathrm{SO}(3)$. $\Pi$ is indeed a group homomorphism: For $U_1,U_2\in\mathrm{SU}(2)$,
\begin{align*}
\Pi(U_1)[\Pi(U_2)\psi](v)&=\Pi(U_2\psi)(\rho(U_1)^{-1}v)\\
&=\psi(\rho(U_2)^{-1}\rho(U_1)^{-1}v)\\
&=\psi((\rho(U_1)\rho(U_2))^{-1}v)\\
&=\psi(\rho(U_1U_2)^{-1}v)\\
&=[\Pi(U_1U_2)\psi](v)
\end{align*}
Hence, $\Pi$ is an infinite dimensional real representation of $\mathrm{SU}(2)$. Here the fact that $\rho$ is a group homomorphism is used. We can also obtain the corresponding representation $\pi$ of the Lie algebra $\mathfrak{su}(2)$. $\pi$ can be computed as
$$\pi(X)=\frac{d}{dt}\Pi(e^{tX})|_{t=0}$$
So,
\begin{align*}
[\pi(X)\psi](v)&=\frac{d}{dt}[\Pi(e^{tX})\psi](v)|_{t=0}\\
&=\frac{d}{dt}\psi(\rho(e^{tX})^{-1}v)|_{t=0}
\end{align*}
The Lie algebra $\mathfrak{su}(2)$ has the canonical basis
\begin{align*}
X_1&=\frac{i}{2}\sigma_1=\frac{i}{2}\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}\\
X_2&=\frac{i}{2}\sigma_2=\frac{i}{2}\begin{pmatrix}
0 & i\\
-i & 0
\end{pmatrix}\\
X_3&=\frac{i}{2}\sigma_3=\frac{i}{2}\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
\end{align*}
Let us calculate $\pi$ for the basis member $X_3$. $e^{\theta X_3}=\begin{pmatrix} e^{i\theta/2} & 0\\ 0 & e^{-i\theta/2} \end{pmatrix}$ and $\rho(e^{\theta X_3})=R_\theta^z$ where $R_\theta^z=\begin{pmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}$ is rotation in $\mathbb{R}^3$ about the $z$-axis by angle $\theta$. Let $v(\theta)$ be a curve in $\mathbb{R}^3$ defined by
$$v(\theta)=\rho(e^{\theta X_3})^{-1}v=(R_\theta^z)^{-1}v$$ so that $v(0)=v$. Write $v(\theta)=(x(\theta),y(\theta),z(\theta))$ and $v=(x,y,z)$. Then by the chain rule,
\begin{align*}
[\pi(X_3)\psi](v)&=\frac{\partial\psi}{\partial x}\frac{dx}{d\theta}|_{\theta=0}+\frac{\partial\psi}{\partial y}\frac{dy}{d\theta}|_{\theta=0}+\frac{\partial\psi}{\partial z}\frac{dz}{d\theta}|_{\theta=0}\\
&=y\frac{\partial\psi}{\partial x}-x\frac{\partial\psi}{\partial y}
\end{align*}
Hence,
$$\pi(X_3)=y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$$
Using
\begin{align*}
e^{\phi X_2}&=\begin{pmatrix}
\cos\frac{\phi}{2} & -\sin\frac{\phi}{2}\\
\sin\frac{\phi}{2} & \cos\frac{\phi}{2}
\end{pmatrix},\ \rho(e^{\phi X_2})=R_\phi^y=\begin{pmatrix}
\cos\phi & 0 & -\sin\phi\\
0 & 1 & 0\\
\sin\phi & 0 & \cos\phi
\end{pmatrix}\\
e^{\sigma X_1}&=\begin{pmatrix}
\cos\frac{\sigma}{2} & i\sin\frac{\sigma}{2}\\
i\sin\frac{\sigma}{2} & \cos\frac{\sigma}{2}
\end{pmatrix},\ \rho(e^{\sigma X_1})=R_\sigma^x=\begin{pmatrix}
1 & 0 & 0\\
0 & \cos\sigma & -\sin\sigma\\
0 & \sin\sigma & \cos\sigma
\end{pmatrix}
\end{align*}
one can find similar formulas for $\pi(X_2)$ and $\pi(X_1)$:
\begin{align*}
\pi(X_2)&=z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\\
\pi(X_1)&=z\frac{\partial}{\partial y}-y\frac{\partial}{\partial z}
\end{align*}
Note that
\begin{align*}
i\hbar\pi(X_1)&=L_x=-i\hbar\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\\
-i\hbar\pi(X_2)&=L_y=-i\hbar\left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}\right)\\
i\hbar\pi(X_3)&=L_z=-i\hbar\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)
\end{align*}
i.e. the angular momenta about the $x$-axis, $y$-axis and $z$-axis respectively.

References:

 Walter Greiner, Quantum Mechanics, An Introduction, 4th Edition, Springer-Verlag 2000

 Brian C. Hall, Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer-Verlag 2004

 Shlomo Sternberg, Group Theory and Physics, Cambridge University Press 1994

# Lie Group and Lie Algebra Representations

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 $W=V_m$.