Elie Cartan
1. SU(2), SO(3), and the Double Cover
Last post left us with a contradiction. The Lie algebras are identical:
\[ [T_i, T_j] = \epsilon_{ijk} T_k \tag{1} \]
holds for both \(\mathfrak{so}(3)\) and \(\mathfrak{su}(2)\), with the same structure constants. But the groups obviously aren't the same. One is \(3\times3\) real matrices acting on ordinary space, the other is \(2\times2\) complex matrices acting on \(\mathbb{C}^2\), and post 7 told us \(SU(2)\) is a 3-sphere while \(SO(3)\) is not simply connected. The solution to this problem turns out to be why electrons need \(720\) degrees to come back to themselves.
1.1 Building a map
We need something connecting \(2\times2\) complex matrices to rotations of \(\mathbb{R}^3\). The trick is to find a copy of \(\mathbb{R}^3\) hiding inside the \(2\times2\) matrices. Take a vector \(\mathbf{x} = (x_1, x_2, x_3)\) and build
\[ X = x_1\sigma_1 + x_2\sigma_2 + x_3\sigma_3 = \begin{pmatrix} x_3 & x_1 - ix_2 \\\\ x_1 + ix_2 & -x_3 \end{pmatrix} \tag{2} \]
using the Pauli matrices from last post. This \(X\) is Hermitian and traceless, and conversely every traceless Hermitian \(2\times2\) matrix looks like this for exactly one \(\mathbf{x}\). So the traceless Hermitian matrices are \(\mathbb{R}^3\), just written differently. Where's the length? Take the determinant:
\[ \det X = -x_3^2 - (x_1 - ix_2)(x_1 + ix_2) = -\left( x_1^2 + x_2^2 + x_3^2 \right) = -|\mathbf{x}|^2 \tag{3} \]
The determinant is minus the squared length. Now we have everything we need, because \(SU(2)\) acts on matrices in a way that preserves determinants. Given \(U \in SU(2)\), define
\[ X' = U X U^\dagger \tag{4} \]
and check that \(X'\) is still one of our vectors. It's Hermitian, since \((UXU^\dagger)^\dagger = UX^\dagger U^\dagger = UXU^\dagger\). It's traceless, because trace is cyclic: \(\operatorname{tr}(UXU^\dagger) = \operatorname{tr}(XU^\dagger U) = \operatorname{tr}X = 0\). So \(X'\) corresponds to some vector \(\mathbf{x}'\). And its length is unchanged:
\[ \det X' = \det U \det X \det U^\dagger = \det X \tag{5} \]
since \(\det U = 1\). So \(|\mathbf{x}'| = |\mathbf{x}|\). The map \(\mathbf{x} \mapsto \mathbf{x}'\) is linear and preserves length, so by post 6 it's in \(O(3)\). And since \(SU(2)\) is connected and \(U = I\) gives the identity map, the whole image sits in the component containing the identity, which post 7 identified as \(SO(3)\). Every \(U \in SU(2)\) gives a rotation. Call it \(R(U)\). It's a homomorphism, because
\[ (U_1U_2)X(U_1U_2)^\dagger = U_1 \left( U_2 X U_2^\dagger \right) U_1^\dagger \tag{6} \]
so \(R(U_1U_2) = R(U_1)R(U_2)\).
1.2 Which rotation?
Let's compute one. Take the \(SU(2)\) element generated by \(T_3\):
\[ U(\theta) = \exp\left( -\frac{i\theta}{2}\sigma_3 \right) = \begin{pmatrix} e^{-i\theta/2} & 0 \\\\ 0 & e^{i\theta/2} \end{pmatrix} \tag{7} \]
which is diagonal because \(\sigma_3\) is. Conjugating \(X\) by it, entry by entry:
\[ UXU^\dagger = \begin{pmatrix} x_3 & e^{-i\theta}(x_1 - ix_2) \\\\ e^{i\theta}(x_1 + ix_2) & -x_3 \end{pmatrix} \tag{8} \]
So \(x_3' = x_3\), and reading off the upper right entry,
\[ x_1' - ix_2' = e^{-i\theta}(x_1 - ix_2) = (\cos\theta - i\sin\theta)(x_1 - ix_2) \tag{9} \]
Expanding and matching real and imaginary parts:
\[ x_1' = x_1\cos\theta - x_2\sin\theta, \qquad x_2' = x_1\sin\theta + x_2\cos\theta \tag{10} \]
That's a rotation by \(\theta\) about the \(z\)-axis. The same argument for the other axes gives all rotations, so \(R\) is onto \(SO(3)\). But look at that \(U(\theta)\) and the angle it produces. The matrix has \(\theta/2\) in it. The rotation has \(\theta\).
1.3 The kernel
Post 3 says a homomorphism is determined by its kernel: which \(U\) give the identity rotation? We need \(UXU^\dagger = X\) for every traceless Hermitian \(X\), which means \(U\) commutes with all three Pauli matrices. Write $U =
\begin{pmatrix} a & b \\\\ c & d \end{pmatrix}
$ and impose commutation with \(\sigma_3\):
\[ U\sigma_3 = \begin{pmatrix} a & -b \\\\ c & -d \end{pmatrix}, \qquad \sigma_3 U = \begin{pmatrix} a & b \\\\ -c & -d \end{pmatrix} \tag{11} \]
Equality forces \(b = c = 0\), so \(U\) is diagonal. Now commute with \(\sigma_1\):
\[ U\sigma_1 = \begin{pmatrix} 0 & a \\\\ d & 0 \end{pmatrix}, \qquad \sigma_1 U = \begin{pmatrix} 0 & d \\\\ a & 0 \end{pmatrix} \tag{12} \]
which forces \(a = d\). So \(U = aI\), and \(\det U = a^2 = 1\) gives \(a = \pm 1\). The kernel is
\[ \ker R = \{ I, -I \} \cong \mathbb{Z}_2 \tag{13} \]
Two elements! Both \(U\) and \(-U\) produce the same rotation, which you can see directly: \((-U)X(-U)^\dagger = UXU^\dagger\), since the minus signs cancel. Now apply the first isomorphism theorem from post 3:
\[ SU(2) / \{ \pm I \} \cong SO(3) \tag{14} \]
There it is. \(SU(2)\) is not isomorphic to \(SO(3)\). In fact, it's twice \(SO(3)\). Every rotation corresponds to exactly two elements of \(SU(2)\), differing by a sign. This is the famous double cover!
1.4 Going round twice
The half-angle now does something quite weird. Set \(\theta = 2\pi\) in \(U(\theta)\), i.e. a full rotation, which in \(SO(3)\) is the identity:
\[ U(2\pi) = \begin{pmatrix} e^{-i\pi} & 0 \\\\ 0 & e^{i\pi} \end{pmatrix} = \begin{pmatrix} -1 & 0 \\\\ 0 & -1 \end{pmatrix} = -I \tag{15} \]
Not the identity. A full rotation lifts to minus one in \(SU(2)\). Only at \(\theta = 4\pi\) do we get
\[ U(4\pi) = I \tag{16} \]
You have to go round twice to come back. This also explains the topology from post 7 exactly. Watch what happens to paths. As \(\theta\) runs from \(0\) to \(2\pi\), the corresponding path in \(SO(3)\) is a closed loop, so you end up back at the identity rotation. But upstairs in \(SU(2)\) the path runs from \(I\) to \(-I\): not a loop, an open path with distinct endpoints. A loop that lifts to a non-loop can't be shrunk. Shrinking it continuously would drag the endpoints of the lift together, and they can't meet because they're \(I\) and \(-I\). That's the unshrinkable loop in \(SO(3)\) we claimed existed three posts ago. Run \(\theta\) from \(0\) to \(4\pi\) and the lift is a loop, from \(I\) back to \(I\). Since \(SU(2)\) is a 3-sphere and therefore simply connected, that loop shrinks, so its projection shrinks too. Geometrically, \(SO(3)\) is a solid ball of radius \(\pi\): each point is an axis and an angle, and antipodal points on the boundary surface are the same rotation, so they're glued. A path leaving through the boundary re-enters at the opposite side.
1.5 Why this is physics
\(SU(2)\) has a two-dimensional representation - it's made of \(2\times2\) matrices, so it acts on \(\mathbb{C}^2\) by definition. Does that action descend to \(SO(3)\)? It can't. To be a representation of \(SO(3)\) it would have to assign one matrix to each rotation, but the identity rotation has two preimages, \(I\) and \(-I\), and they act differently on \(\mathbb{C}^2\): one does nothing, the other flips the sign. There's no consistent choice. So \(\mathbb{C}^2\) carries a representation of \(SU(2)\) that is not a representation of \(SO(3)\). Its elements are spinors, and under a full \(2\pi\) rotation
\[ \psi \to -\psi \tag{17} \]
An electron rotated once through a complete turn is not in the state it started in. Its wavefunction has changed sign. This sounds like it should be unobservable, since \(|\psi|^2\) is unchanged, and for a single isolated electron it is. But put a particle in superposition, rotate one branch and not the other, and recombine, and the relative sign is an interference pattern. That experiment was done with neutrons in 1975, and the fringes shifted exactly as the \(4\pi\) periodicity requires. The shape of a group is measurable! That is a very strange thing to think about. Platonists would be very happy!
1.6 End of the series
We come to the end of the beginning. Here’s a quick summary. The isomorphism theorem from post 3 gave \(SU(2)/\{\pm I\} \cong SO(3)\). The kernel-is-normal result made the quotient allowed. Connectedness from post 7 forced the image in \(SO(3)\) rather than \(O(3)\). The Lie algebra from post 9 explained why the two groups look identical up close. Every tool got used. The main idea is that the Lie algebra tells you the group locally, and no more. \(\mathfrak{su}(2)\) and \(\mathfrak{so}(3)\) are the same algebra, and it doesn't determine which group you have. Global structure is genuinely extra information, and here it's the difference between a particle that returns to itself after one turn and one that needs two. Among all groups sharing a given algebra there's a distinguished one, the simply connected one, called the universal cover, which here is \(SU(2)\). Everything else is a quotient of it. That's why quantum mechanics uses \(SU(2)\) rather than \(SO(3)\).