# The Representation of $\mathrm{SU}(2)$

Let $\mathcal{H}_j$ be the space of polynomial functions on $\mathbb{C}^2$ that are homogemeous of degree $2j$. An element in $\mathcal{H}_j$  is a polynomial in complex variables $x$ and $y$ that is a linear combination of polynomials $x^py^q$ where $p+q=2j$. $\mathcal{H}_j$ has dimension $2j+1$ since it has a basis given by
$$x^{2j},x^{2j-1}y,x^{2j-2}y^2,\cdots,y^{2j}.$$
For any $g\in\mathrm{SU}(2)$, let $U_j(g)$ be the linear transformation of $\mathcal{H}_j$ given by
$$U_j(g)f(v)=f(g^{-1}v)$$
for $f\in\mathcal{H}_j$ and $v\in\mathbb{C}^2$. Then $U_j$ is a representation: $U_j(I)$ is the identity. For any $g,h\in\mathrm{SU}(2)$,
\begin{align*}
U_j(g)U_j(h)f(v)&=U_(h)f(g^{-1}v)\\
&=f(h^{-1}g^{-1}v)\\
&=f((gh)^{-1}v)\\
&=U_j(gh)f(v)
\end{align*}
for $f\in\mathcal{H}_j$ and $v\in\mathbb{C}^2$.

Physicists call $U_j$ spin-$j$ representation. Since $2j+1$ has to be a positive integer, we have spin-0 representation, spin-$\frac{1}{2}$ representation, spin-1 representation, etc. It is interesting to see the correspondence between spin-$j$ representation and particles. The only known spin-0 particle is Higgs-boson, the so-called God particle, which is responsible for giving masses to bosons. The Higgs-boson appears to have been discovered recently by the LHC (Large Hadron Collider) at CERN. Spin-$\frac{1}{2}$ particles are fermions which include all quarks and leptons. Spin-1 particles are gauge bosons (force-carrying particles) such as photons, W bosons, Z bosons, gluons. Curiously there are currently no spin-$\frac{3}{2}$ particles predicted in particle physics. The hypothetical gravitons are believed to be spin-$2$ particles.

Proposition. The spin-0 representation of $\mathrm{SU}(2)$ is equivalent to the trivial representation in which every element of the group acts on $\mathbb{C}$ as the identity.

Proposition. The spin-$\frac{1}{2}$ representation of $\mathrm{SU}(2)$ is equivalent to the fundamental representation in which every element $g\in\mathrm{SU}(2)$ acts on $\mathbb{C}^2$ by matrix multiplication.

Note that the $U_j$ are irreducible and that they are all of the irreducible representations.

$\mathbb{R}^3$ can be identified with the set of $2\times 2$ Hermitian matricies of the form
\begin{align*}
X&=\begin{pmatrix}
z & x-iy\\
x+iy & -z
\end{pmatrix}\\
&=z\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}+x\begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}+y\begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}\\
&=x\sigma_1+y\sigma_2+z\sigma_3,
\end{align*}
where $\sigma_1,\sigma_2,\sigma_3$ are called the Pauli spin matrices in physics. Define an inner product $\langle\ ,\ \rangle$ on the Hermitian matrices by
$$\langle X, Y\rangle=\frac{1}{2}\mathrm{tr}(XY).$$
In particular,
$$|X|^2=\frac{1}{2}\mathrm{tr}(X^2)=-\det X.$$
With this inner product, the identification is an isometry. $\mathrm{SU}(2)$ acts on $\mathbb{R}^3$ via the representation
$$\rho:\mathrm{SU}(2)\longrightarrow\mathrm{GL}(3,\mathbb{R})$$
defined by
$$\rho(U)X=UXU^{-1}$$
for $U\in\mathrm{SU}(2)$ and $X\in\mathbb{R}^3$. It turns out that $X\longmapsto UXU^{-1}$ is an orientation preserving isometry of $\mathbb{R}^3$, so
$$\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3).$$
Since both $U$ and $-U$ result the same isometry, the representation $\rho:\mathrm{SU}(2)\longrightarrow\mathrm{SO}(3)$ is a $2:1$ map. Since $\mathrm{SU}(2)=S^3$ is simply-connected, $\rho$ is a universal convering map and that we have
$$\mathrm{SU}(2)/\mathbb{Z}_2=\mathrm{SO}(3).$$
The quotient group $\mathrm{SU}(2)/\mathbb{Z}_2$ is denoted by $\mathrm{PSU}(2)$ and called the projective special unitary group.

The double cover i.e. $2:1$ cover of $\mathrm{SO}(n)$ is called the spin group and is denoted by $\mathrm{Spin}(n)$. For $n>2$, $\mathrm{Spin}(n)$ is simply-connected so it is the universal cover of $\mathrm{SO}(n)$. Some examples of spin groups are
\begin{align*}
\mathrm{Spin}(1)&=\mathrm{O}(1)=\mathbb{Z}_2=\{\pm I\}\\
\mathrm{Spin}(2)&=\mathrm{U}(1)=\mathrm{SO}(2)\\
\mathrm{Spin}(3)&=\mathrm{SU}(2)\\
\mathrm{Spin}(4)&=\mathrm{SU}(2)\times\mathrm{SU}(2)
\end{align*}
Note that $\mathrm{SO}(3)\subset\mathrm{GL}(3,\mathbb{R})\subset\mathrm{GL}(3,\mathbb{C})$, so for any $g\in\mathrm{SU}(2)$, $\rho(g):\mathbb{C}^3\longrightarrow\mathbb{C}^3$. Hence, $\rho$ is in fact equivalent to the spin-1 representation of $\mathrm{SU}(2)$.

In quantum mechanics, unitary representation is particularly important. Let $\mathcal{H}$ be the Hilbert space of states derived from a quantum mechanical system. Let $\rho$ be a representation of a Lie group $G$ on $\mathcal{H}$ i.e. $\rho:G\longrightarrow\mathrm{GL}(\mathcal{H})$. $\rho$ is called a unitary representation if
$$\langle\rho(g)\psi,\rho(g)\phi\rangle=\langle\psi,\phi\rangle$$
for all $g\in G$ and $\psi,\phi\in\mathcal{H}$. Intuitively each $\rho(g)$ may be understood as a rotation. For example, say $G=\mathrm{SO}(3)$. First rotating the particle by some amount $h\in\mathrm{SO}(3)$ and then rotating it by some amount $g\in\mathrm{SO}(3)$ should have the same effect as rotating it by the amount $gh\in\mathrm{SO}(3)$. That is $\rho(g)\rho(h)=\rho(gh)$. This tells why we need a representation in quantum mechanics. In quantum mechanics, the inner product $\langle\ ,\ \rangle$ measures probability. For instance if a particle is in the state $\psi$, then $\langle\psi,\phi\rangle$ is the probability of finding the particle in the state $\phi$. Rotating a particle amounts to a change of coordinates and the state of a particle should not depend on a change of coordinates. Hence, in quantum mechanics we require representation to be unitary.

Lastly I would like to mention the difference between bosons and fermions in terms of representation. Let $f\in\mathcal{H}_j$, the spin-$j$ representation space. Then
$$U_j(-I)f(v)=f(-v)=(-1)^{2j}f(v)$$
since $f$ is a homogeneous polynomial of degree $2j$. This implies that
$$U_j(-I)=(-1)^{2j}.$$
Hence, $U_j$ maps both $I$ and $-I$ to the identity if $j$ is an integer, while it does not if $j$ is a half-integer.

References:

[1] John Baez, Javier P. Muniain, Gauge Fields, Knots and Gravity, World Scientific 1994

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations, An Elementary Introduction, Springer-Verlag 2003

# Lie Group Actions and Lie Group Representations

$\mathrm{SO}(3)$ acts on $\mathbb{R}^3$ meaning that each element of $\mathrm{SO}(3)$ defines a linear transformation (rotation) of $\mathbb{R}^3$. So we can say that $\mathrm{SO}(3)$ describes the rotational symmetry of $\mathbb{R}^3$.

Definition. A group $G$ is said to act on a vector space $V$ if there exists a map $\rho:G\longrightarrow\mathrm{GL}(V)$ such that
$$\rho(gh)v=\rho(g)\rho(h)v,\ \forall v\in V.$$
We also say that $\rho$ is a representation of $G$ on $V$.

Lie groups are closely related to the fundamental forces of our universe, electromagnetism, weak force, strong force, and gravity. Different Lie groups give different equations called Yang-Mills equations, which describe different forces. Those Lie groups involved with fundamental forces are called symmetry groups (usually by mathematicians) or gauge groups (usually by physicists, in mathematics gauge group means
the group of gauge transformations.). The Lie group $\mathrm{U}(1)=\mathrm{SO}(2)$ is the gauge group for electromagnetism. Yang-Mills equations with $\mathrm{U}(1)$ are simply Maxwell’s equations. (They are linear equations.) $\mathrm{SU}(3)$ is the gauge group for strong force and $\mathrm{SU}(2)\times\mathrm{U}(1)$ is the gauge group for electroweak force. The groups such as $\mathrm{SU}(2)\times\mathrm{U}(1)$ are called direct products. Recall the following propositon from abstract algebra:

Proposition. Given groups $G$ and $H$, the cartesian product $G\times H$ becomes a group with product
$$(g,h)\cdot (g’,h’)=(gg’,hh’),$$
identity $e=(e_G,e_H)$, and inverse $(g,h)^{-1}=(g^{-1},h^{-1})$.

The group $G\times H$ is called the direct product or direct sum of $G$ and $H$. (We use the notation $G\oplus H$ for direct sum.) $G\times H$ is abelian if and only if both $G$ and $H$ are abelian. If $G$ and $H$ are Lie groups, so is $G\times H$.

The gauge group of the entire standard model (physical theory that attempts to unify electromagnetism, strong and weak force) is $\mathrm{SU}(3)\times\mathrm{SU}(2)\times\mathrm{U}(1)$. Physicists have been looking for a nicer group which has $\mathrm{SU}(3)\times\mathrm{SU}(2)\times\mathrm{U}(1)$ as its subgroup. The simplest choice is $\mathrm{SU}(5)$. The $\mathrm{SU}(5)$ model was proposed by Sheldon Glashow and it predicts the decay of protons. However there has been no sign of proton decay after many years of experiments and most physicists do not believe protons would decay. The standard model does not include gravity. Physicists do not know which Lie group is involved with gravity. Possible candidates are $\mathrm{SO}^+(3,1)$ or $\mathrm{SL}(2,\mathbb{C})$. So far all the attempts to come up with a gauge theory of gravity have failed.

There are many examples that hint us that symmetry plays a crucial role in particle theory. Representation is closely related to symmetry and there is a wonderful connection between the charge of a particle and the representation of gauge group, namely the charge of a particle amounts to a choice of a representation for the gauge group. This will be discussed in a separate post. Let us study a bit more about representations.

Equivalent Representations

Two representations
$$\rho: G\longrightarrow\mathrm{GL}(V),\ \rho’: G\longrightarrow\mathrm{GL}(V)$$
are equivalent if there exists an isomorphism $T:V\longrightarrow V$ with
$$\rho(g)T=T\rho’(g),\ \forall g\in G.$$

Getting New Representation from Old Ones

Direct Sum

Let $G$ be a (Lie) group and $\rho: G\longrightarrow\mathrm{GL}(V)$, $\rho’:G\longrightarrow\mathrm{GL}(V’)$ representations. The direct sum $\rho\oplus\rho’:G\longrightarrow\mathrm{GL}(V\oplus V’)$ of $\rho$ and $\rho’$ is a representation defined by $$(\rho\oplus\rho’)(v,v’)=(\rho(g)v,\rho’(g)v’),\ \forall v\in V,v’\in V’.$$

Tensor Product

Let $V$ and $V’$ be finite dimensional vector spaces. Pick a basis $\{e_i\}$ for $V$ and a basis $\{e’_j\}$ for $V’$. Then the tensor product $V\otimes V’$ is the vector space whose basis is given by all expressions of the form $e_i\otimes e’_j$. If $v=v^ie_i\in V$ and $v’=v’^je’_j\in V’$, then
$$v\otimes v’=v^iv’^je_i\otimes e’_j.$$
So clearly we obtain,
$$\dim V\otimes V’=(\dim V)(\dim V’).$$
While this definition of tensor product is easy to understand, it does depend on the choice of bases. There is a general definition of tensor product with universal mapping property.

Theorem. Let $V$, $W$ be finite vector spaces over a field $K$. Then there exists a finite dimensional space $T$ over $K$ and a bilinear map
\begin{align*}
V\times W&\longrightarrow T\\
(v,w)&\longmapsto v\otimes w
\end{align*}
satisfying the following properties

T1. If $U$ is a vector space over $K$ and $g:V\times W\longrightarrow U$ is a bilinear map, then there exists uniquely a linear map $g_\ast:T\longrightarrow U$ such that for every $(v,w)\in V\times W$,
$$g(v,w)=g_\ast (v\otimes w).$$
In other words, the following diagram commutes.
$$\begin{array}{ccc} & & T\\ & \nearrow &\downarrow\\ V\times W & \longrightarrow & U \end{array}$$

T2. If $\{v_1,\cdots,v_n\}$ is a basis of $V$ and $\{w_1,\cdots,w_n\}$ is a basis of $W$, then $v_i\otimes w_j$, $i=1,\cdots,n$, $j=1,\cdots,m$ form a basis of $T$

For the proof of this theorem, see for instance [3]. The vector space $T$ is denoted by $V\otimes W$ and called the tensor product of $V$ and $W$.

Let $\rho:G\longrightarrow\mathrm{GL}(V)$ and $\rho’: G\longrightarrow\mathrm{GL}(V’)$. The tensor product $\rho\otimes\rho’: G\longrightarrow\mathrm{GL}(V\otimes V’)$ is a representation defined by
$$(\rho\otimes\rho’)(g)(v\otimes v’)=\rho(g)v\otimes\rho’(g)v’.$$

Subrepresentations

Let $\rho$ be a representation of a group $G$ on the vector space $V$. Suppose that $V’$ is an invariant space $V$ i.e. $\rho(g)(V’)\subset V’$ for all $g\in G$. Define a representation $\rho’: G\longrightarrow\mathrm{GL}(V’)$ by
$$\rho’(g)v=\rho(g)v,\ \forall v\in V’.$$
$\rho’$ is called a subrepresentation of $\rho$.

References:

[1] John Baez, Javier P. Muniain, Gauge Fields, Knots and Gravity, World Scientific 1994

[2] Brian C. Hall, Lie Groups, Lie Algebras, and Representations, An Elementary Introduction, Springer-Verlag 2003

[3] Serge Lang, Linear Algebra, 2nd Edition, Addison-Wesley Publishing Co. 1972

# 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:

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

[2] F. Reese Harvey, Spinors and Calibrations, Academic Press 1990

[3] 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 [3] 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:

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

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

[3] 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 $W=V_m$.