$SU(2)$

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Section 4.4 $SU(2)$

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Objectives

You should be able to:

Determine the Lie algebra $\mathfrak{su}(2)\text{.}$
Explain the statement that $SU(2)$ is a double cover of $SO(3)\text{.}$

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In this course we will mostly focus on the two most important Lie groups in physics, namely

$SO(3)$ and

$SU(2)\text{.}$ We have already calculated the Lie algebra

$\mathfrak{so}(3)$ associated to the Lie group

$SO(3)$ in Section 4.2. We obtained that

$\mathfrak{so}(3)$ is the three-dimensional vector spanned by elements

$L_1, L_2, L_3$ with the bracket

$\begin{equation*} [L_i, L_j] = i \sum_{k=1}^3 \epsilon_{ijk} L_k . \end{equation*}$

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Subsection 4.4.1 The Lie algebra $\mathfrak{su}(2)$

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Let us now study the Lie algebra

$\mathfrak{su}(2)$ associated to the special unitary group

$SU(2)\text{.}$ The fundamental representation of

$SU(2)$ consists in

$2 \times 2$ unitary matrices

$U$ with determinant one:

$\begin{equation*} U^\dagger U = I, \qquad \det U = 1. \end{equation*}$

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To obtain the Lie algebra, we do a first order expansion

$U = I + i L.$ To first order, the unitarity condition becomes

$\begin{equation*} U^\dagger U \cong (I + i L)^\dagger (I + i L) \cong I - i L^\dagger + i L = I. \end{equation*}$

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Therefore

$L^\dagger = L,$ which says that

$L$ is a Hermitian matrix. We also need to impose the condition that

$\det U = 1\text{.}$ We can write a general

$2 \times 2$ complex-valued matrix as

$\begin{equation*} L = \begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix}, \end{equation*}$

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for

$a,b,c,d \in \mathbb{C}\text{.}$ Then

$\begin{equation*} \det U \cong \det(I+ i L) = \det \begin{pmatrix} 1+a \amp b \\ c \amp 1+d \end{pmatrix} = (1+a)(1+d) - b c. \end{equation*}$

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Since we keep only terms of first order in

$L\text{,}$ we can ignore any terms that are order two in the complex numbers

$a,b,c,d\text{.}$ Thus, to first order, we have

$\begin{equation*} \det U \cong 1 + a + d. \end{equation*}$

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For this to be equal to one we must have

$a=-d\text{,}$ that is, the matrix

$L$ is traceless (has vanishing trace).

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The conclusion is that

$L$ must be a

$2 \times 2$ traceless Hermitian matrix. Any such matrix can be written as

$\begin{equation*} L = \begin{pmatrix} t_1 \amp t_2 + i t_3 \\ t_2 - i t_3 \amp - t_1 \end{pmatrix} \end{equation*}$

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for some real numbers

$t_1, t_2, t_3 \in \mathbb{R}\text{.}$ Indeed, since the matrix is Hermitian, the diagonal terms must be real, and since it is traceless they must add up to zero. The off-diagonal terms must be complex conjugate since the matrix is Hermitian.

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It thus follows that the Lie algebra

$\mathfrak{su}(2)$ is three-dimensional. A natural basis for the vector space consists of the matrices:

$\begin{equation} T_1 = \frac{1}{2} \begin{pmatrix} 0 \amp 1 \\ 1 \amp 0 \end{pmatrix},\qquad T_2 = \frac{1}{2} \begin{pmatrix} 0 \amp -i \\ i \amp 0 \end{pmatrix},\qquad T_3 = \frac{1}{2} \begin{pmatrix} 1 \amp 0 \\ 0 \amp -1 \end{pmatrix}.\label{equation-pauli}\tag{4.4.1} \end{equation}$

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Those are known as Pauli matrices. By calculating the commutators, one finds:

$\begin{equation*} [T_1,T_2] = i T_3, \qquad [T_2, T_3] = i T_1, \qquad [T_3, T_1] = i T_2. \end{equation*}$

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In other words,

$\begin{equation*} [T_i, T_j] = i \sum_{k=1}^3 \epsilon_{ijk} T_k. \end{equation*}$

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What? This is the same abstract algebra as

$\mathfrak{so}(3)\text{!}$ Indeed, it turns out that the Lie algebras

$\mathfrak{so}(3)$ and

$\mathfrak{su}(2)$ are isomorphic. Really, they are the same. This is absolutely fundamental, and, as we will see, is the reason for the appearance of fermions (particles with half-integer spin) in non-relativistic physics.

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This does not mean however that the Lie groups

$SO(3)$ and

$SU(2)$ are isomorphic: in fact they are not. But they are certainly closely related. More precisely, there exists a two-to-one group homomorphism from

$SU(2)$ to

$SO(3)\text{.}$ We say that

$SU(2)$ is a “double covering” of

$SO(3)\text{.}$

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Subsection 4.4.2 $SU(2)$ vs $SO(3)$

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We have seen in the previous section that the Lie algebras

$\mathfrak{su}(2)$ and

$\mathfrak{so}(3)$ are isomorphic. What does this mean for the Lie groups

$SU(2)$ and

$SO(3)\text{?}$

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The fact that their Lie algebras are the same means that the two groups are “locally isomorphic”, which is a rather striking statement. What this means is that, using exponentiation of the Lie algebra, we would recover the same group structure locally near the identity. This does not mean however that the two groups are globally isomorphic: in fact they are not.

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But they are certainly closely related. The precise statement is that there exists a group homomorphism from

$SU(2)$ to

$SO(3)$ that is two-to-one. This is what we show next. We say that

$SU(2)$ is a “double covering” of

$SO(3)\text{.}$

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Let us now construct the group homomorphism

$f: SU(2) \to SO(3)\text{.}$

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Theorem 4.4.1. $SU(2)$ covers $SO(3)$ .

There exists a two-to-one group homomorphism $f: SU(2) \to SO(3)\text{,}$ which means that the Lie groups $SU(2)$ and $SO(3)\text{,}$ which share the same Lie algebras $\mathfrak{su}(2) \cong \mathfrak{so}(3)\text{,}$ are only locally isomorphic.

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We provide a formal proof of this statement using the adjoint representation of the Lie group

$SU(2)$ below. But before we do that, let us write down an informal argument, which may be more enlightening.

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Both

$SU(2)$ and

$SO(3)$ share the same Lie algebra. For

$SU(2)\text{,}$ we obtained the Lie algebra by constructing the generators of

$SU(2)\text{,}$ which is given by the vector space spanned by the Pauli matrices (4.4.1), which we reproduce here for convenience:

$\begin{equation*} T_1 = \frac{1}{2} \begin{pmatrix} 0 \amp 1 \\ 1 \amp 0 \end{pmatrix},\qquad T_2 = \frac{1}{2} \begin{pmatrix} 0 \amp -i \\ i \amp 0 \end{pmatrix},\qquad T_3 = \frac{1}{2} \begin{pmatrix} 1 \amp 0 \\ 0 \amp -1 \end{pmatrix}. \end{equation*}$

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For

$SO(3)\text{,}$ we obtained the algebra by looking at the generators of rotations in three dimensions (4.2.4), which we also reproduce here:

$\begin{equation*} L_1 = -i \begin{pmatrix} 0 \amp 0 \amp 0\\ 0 \amp 0 \amp 1 \\ 0 \amp -1 \amp 0 \end{pmatrix},\qquad L_2 = -i \begin{pmatrix} 0 \amp 0 \amp -1\\ 0 \amp 0 \amp 0 \\ 1 \amp 0 \amp 0 \end{pmatrix},\qquad L_3 = -i \begin{pmatrix} 0 \amp 1 \amp 0\\ -1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 0 \end{pmatrix}. \end{equation*}$

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One can show that all special unitary matrices in

$SU(2)$ can be obtained by exponentiation, and similarly for special orthogonal matrices in

$SO(3)$ (this is because

$SU(2)$ and

$SO(3)$ are compact and connected). Thus we can write an arbitrary

$U \in SU(2)$ as

$\begin{equation*} U(\theta_1, \theta_2, \theta_3) = \exp\left(i \sum_{k=1}^3 \theta_k T_k \right), \end{equation*}$

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and an arbitrary

$R \in SO(3)$ as

$\begin{equation*} R(\theta_1, \theta_2, \theta_3) = \exp\left(i \sum_{k=1}^3 \theta_k L_k \right). \end{equation*}$

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The mapping

$f: SU(2) \to SO(3)$ is then given by

$U(\theta_1, \theta_2, \theta_3) \mapsto R(\theta_1, \theta_2, \theta_3)\text{.}$

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How do we see that the mapping is two-to-one? Let us argue that it is two-to-one by looking at the special case with

$\theta_1=\theta_2=0\text{,}$ but the argument is general (although a bit more tedious) and can be written explicitly for any choice of parameters.

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We know that

$R(0,0,\theta) = R(0,0,\theta+ 2 \pi)\text{,}$ since

$R(0,0,\theta)$ corrrespond to a rotation in three dimensions by angle

$\theta$ about an axis, hence rotating further by

$2 \pi$ does not do anything. On the other hand, since

$T_3$ is diagonal, we can write

$\begin{equation*} U(0,0,\theta) = \exp \left(i \theta T_3 \right) = \begin{pmatrix} e^{i \frac{\theta}{2}} \amp 0 \\ 0 \amp e^{- i \frac{\theta}{2}} \end{pmatrix}. \end{equation*}$

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Thus,

$\begin{equation*} U(0,0,\theta+2\pi) = \begin{pmatrix} e^{i \frac{\theta}{2} + i \pi} \amp 0 \\ 0 \amp e^{- i \frac{\theta}{2} - i \pi} \end{pmatrix} = - U(0,0,\theta). \end{equation*}$

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So shifting

$\theta$ by

$2 \pi$ does not send the unitary matrix

$U(0,0,\theta)$ to itself; rather it sends it to minus itself. Shifting by

$4 \pi$ would bring back

$U(0,0,\theta)$ to itself.

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As a result, we see that both

$U(0,0,\theta)$ and

$U(0,0,\theta+2\pi) = - U(0,0,\theta)$ are mapped to the same rotation matrix

$R(0,0,\theta)\text{,}$ and hence the mapping is two-to-one. The point here is that rotations have period

$2 \pi\text{,}$ while unitary matrices as written in terms of Pauli matrices have period

$4 \pi\text{.}$ This is why

$SU(2)$ covers

$SO(3)$ twice.

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Let us now write down a formal proof of the theorem.

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Proof.

We will construct the group homomorphism explicitly. In fact, it will be given by the adjoint representation of the group $SU(2)\text{,}$ so let us construct this explicitly.

The adjoint representation of a Lie group $G$ is given by a representation of the group elements of $G$ as linear operators acting on the Lie algebra $\mathfrak{g}$ of $G\text{.}$ Since the Lie algebra $\mathfrak{g}$ is a $n$-dimensional vector space, where $n$ is the dimension of the Lie group, the adjoint representation maps the group elements to $n \times n$matrices acting on $\mathfrak{g}\text{.}$ So it gives a representation $T: G \to GL(\mathfrak{g})$ of the same dimension as the dimension of the Lie group $G\text{.}$ In the case of $G=SU(2)\text{,}$ which has dimension $3\text{,}$ the adjoint representation is three-dimensional, and represents the group of elements of $SU(2)$ as $3 \times 3$ matrices in $GL(\mathfrak{su}(2) )\text{.}$ As we will show, it turns out that the image of the adjoint representation is $SO(3) \subset GL(\mathfrak{su}(2) )\text{,}$ and hence it provides the group homomorphism $SU(2) \to SO(3)$ that we are looking for.

Let us now construct the adjoint representation explicitly. We start with an element $X \in \mathfrak{su}(2)\text{,}$ which we think of as a $2 \times 2$ traceless Hermitian matrix. We take an element $U \in SU(2)\text{,}$ which is a $2 \times 2$ special unitary matrix. Then we define:

\begin{equation*} Ad_U(X) = U X U^\dagger, \end{equation*}

which is basically a similarity transformation of $X$ by $U\text{.}$

Let us now show that $Ad_U(X) \in \mathfrak{su}(2)\text{,}$ i.e. $Ad_U(X)$ is traceless Hermitian. The tracelessness property follows since

\begin{equation*} \Tr Ad_U(X) = \Tr (U X U^\dagger) = \Tr (X U^\dagger U) = \Tr X = 0, \end{equation*}

by the cyclic property of traces. As for being Hermitian, we find:

\begin{equation*} (Ad_U(X))^\dagger = (U X U^\dagger)^\dagger = U X^\dagger U^\dagger = U X U^\dagger = Ad_U(X), \end{equation*}

since $X = X^\dagger\text{.}$ Therefore, $Ad_U(X)$ is Hermitian and traceless. In other words, we have a mapping $Ad_U : \mathfrak{su}(2) \to \mathfrak{su}(2)\text{,}$ which takes $X \mapsto Ad_U(X) = U X U^\dagger\text{.}$ That is, $Ad_U \in GL(\mathfrak{su}(2) )\text{.}$

We can now define the adjoint representation of $SU(2)\text{.}$ It is given by the group homomorphism $Ad: SU(2) \to GL(\mathfrak{su}(2))\text{,}$ with $U \mapsto Ad_U\text{.}$ In other words, it takes a special unitary matrix $U \in SU(2)\text{,}$ and maps it to the linear operator $Ad_U$ on $\mathfrak{su}(2)\text{.}$ Since $\mathfrak{su}(2)$ is a three-dimensional vector space, this gives a three-dimensional representation of $SU(2)\text{.}$ Note that it is straightforward to show that $Ad$ is a group homomorphism, since $Ad: U_1 U_2 \mapsto Ad_{U_1 U_2} = Ad_{U_1} \circ Ad_{U_2}$ by definition of $Ad_U\text{.}$

The image of $Ad$ is in $GL(\mathfrak{su(2)})\text{.}$ We can write any $2 \times 2$ Hermitian traceless matrix $X$ as a real linear combination of the Pauli matrices (4.4.1):

\begin{equation*} X = x_1 T_1 + x_2 T_2 + x_3 T_3. \end{equation*}

Thus we can identify $\mathfrak{su}(2) \cong \mathbb{R}^3\text{,}$ where we map an Hermitian traceless matrix $X$ to the vector of coefficients $\vec{x}$ in the basis for $\mathfrak{su}(2)$ given by Pauli matrices. With this identification we can see $Ad: SU(2) \to GL(3,\mathbb{R})\text{.}$ But in fact, the image is in $SO(3) \subset GL(3,\mathbb{R})\text{,}$ as we now show.

We notice that for any Hermitian matrix $X = x_1 T_1 + x_2 T_2 + x_3 T_3\text{,}$ the determinant is calculated as:

\begin{equation*} 4 \det X = (x_3)(-x_3) - (x_1-i x_2)(x_1 + i x_2) = - |\vec{x}|^2. \end{equation*}

So we can identify the determinant of $X \in \mathfrak{su}(2)$ with the length of the vector $\vec{x} \in \mathbb{R}^3.$ Since the determinant is preserved by similarity transformations, it follows that

\begin{equation*} \det X = \det U X U^\dagger = \det Ad_U(X). \end{equation*}

This means that the vectors in $\mathbb{R}^3$ corresponding to $X$ and $Ad_U(X)$ have the same length. In other words, $Ad_U$ is a linear operator on $\mathbb{R}^3$ that preserves the length of vectors: it is an orthogonal transformation in $O(3) \subset GL(3,\mathbb{R})\text{.}$ Thus $Ad : SU(2) \to O(3) \subset GL(3,\mathbb{R})\text{.}$ But in fact, we know even more. From Example 4.1.10, we know that the underlying manifold structure of $SU(2)$ is $S^3\text{.}$ In particular, it is connected (in fact, it is also simply connected). Therefore, the image of the continuous mapping $Ad: SU(2) \to O(3)$ must be connected to the identity in $O(3)\text{,}$ which means that it must be a subgroup of $SO(3) \subset O(3)\text{.}$ Thus $Ad: SU(2) \to SO(3).$

What remains to be shown is that the group homomorphism $Ad: SU(2) \to SO(3)$ is surjective, and that it is two-to-one.

We will leave the proof that $Ad: SU(2) \to SO(3)$ is surjective as an exercise. (Note that there are many ways that this can be done.)

To show that $Ad: SU(2) \to SO(3)$ is two-to-one, pick any unitary matrix $U_1\text{,}$ and set $U_2 := - U_1\text{.}$ Then

\begin{equation*} Ad(U_1) = Ad_{U_1} = Ad_{U_2} = Ad(U_2), \end{equation*}

since for any Hermitian matrix $X\text{,}$

\begin{equation*} Ad_{U_1}(X) = U_1 X U_1^\dagger = (-U_1) X (-U_1)^\dagger = U_2 X U_2^\dagger = Ad_{U_2}(X). \end{equation*}

Thus the mapping $Ad: SU(2) \to SO(3)$ is two-to-one.

Equivalently, we could show that the kernel is non-trivial. The kernel of $Ad$ corresponds to $2 \times 2$ unitary matrices $U$ that are mapped to $Ad_U$ being the identity element of $SO(3)\text{.}$ There are two such matrices,

\begin{equation*} \begin{pmatrix} 1 \amp 0 \\ 0 \amp 1 \end{pmatrix}, \qquad \begin{pmatrix} -1 \amp 0 \\ 0 \amp -1 \end{pmatrix}, \end{equation*}

since in both cases $Ad_U(X) = X\text{,}$ i.e. $Ad_U$ is the identity in $SO(3)\text{.}$ Thus $\ker(f) \cong \mathbb{Z}_2\text{,}$ where $\mathbb{Z}_2$ is realized as $\pm 1$ times the identity matrix. Since the kernel has two elements, this means that $Ad: SU(2) \to SO(3)$ is two-to-one.

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Remark 4.4.2.

We have shown that there is a two-to-one group homomorphism $Ad : SU(2) \to SO(3)\text{.}$ The kernel is $\mathbb{Z}_2\text{,}$ realized as $\pm 1$ times the identity matrix in $SU(2)\text{.}$ Thus, by the first isomorphism theorem, we can write

$\begin{equation*} SO(3) \cong SU(2) / \mathbb{Z}_2. \end{equation*}$

The groups however are said to be “locally isomorphic”, since they have isomorphic Lie algebras. Equivalently, this can be seen since if $U \in SU(2)$ is near the identity, then $-U \in SU(2)$ is not. And hence in a neighborhoud of the identity the map is a local isomorphism.

Section 4.4 SU(2)SU(2)

Objectives

Subsection 4.4.1 The Lie algebra su(2)\mathfrak{su}(2)

Subsection 4.4.2 SU(2)SU(2) vs SO(3)SO(3)

Theorem 4.4.1. SU(2)SU(2) covers SO(3)SO(3).

Proof.

Remark 4.4.2.

Section 4.4 $SU(2)$

Subsection 4.4.1 The Lie algebra $\mathfrak{su}(2)$

Subsection 4.4.2 $SU(2)$ vs $SO(3)$

Theorem 4.4.1. $SU(2)$ covers $SO(3)$ .