Wilhelm Killing
1. Lie Algebras and the Exponential Map
Last post we differentiated a group and got a vector space. For \(SO(3)\) the tangent space at the identity turned out to be the antisymmetric matrices, three-dimensional, and the nonlinear condition \(A^TA = I\) became the linear condition \(X^T = -X\). But something was clearly missing. A vector space knows nothing about the group being non-abelian, and rotations in three dimensions don't commute, but that information has to survive the differentiation somewhere.
1.1 Getting back to the group
First, going back to the group. Given a tangent vector \(X\), define
\[ \exp(X) = I + X + \frac{X^2}{2!} + \frac{X^3}{3!} + \cdots \tag{1} \]
which converges for any matrix. Post 8 showed this works for \(SO(2)\): the single generator \(J\) exponentiated to the full rotation matrix. Does it work in general? Take \(X\) antisymmetric, so \(X^T = -X\), and check whether \(\exp(X)\) is orthogonal. Transposing a power series termwise,
\[ \exp(X)^T = \exp(X^T) = \exp(-X) \tag{2} \]
and since \(X\) commutes with itself, \(\exp(-X)\exp(X) = \exp(0) = I\). So
\[ \exp(X)^T \exp(X) = I \tag{3} \]
The exponential of an antisymmetric matrix is orthogonal. The tangent space maps back into the group. There's a second condition to check, \(\det = 1\), and it comes from a useful identity:
\[ \det \exp(X) = e^{\operatorname{tr} X} \tag{4} \]
An antisymmetric matrix has zeros down the diagonal, so \(\operatorname{tr} X = 0\) and the determinant is \(e^0 = 1\). We end up with \(SO(3)\), not just \(O(3)\). That identity also fixes an issue from last post. For \(SU(n)\) we need \(\det U = 1\), which forces \(e^{\operatorname{tr}X} = 1\), so \(\operatorname{tr} X = 0\). The tracelessness condition on \(\mathfrak{su}(n)\) is in fact the differentiated determinant condition.
1.2 Generators of rotations
Let's build \(SO(3)\). Rotations about the three axes give three antisymmetric generators:
\[ J_x = \begin{pmatrix} 0 & 0 & 0 \\\\ 0 & 0 & -1 \\\\ 0 & 1 & 0 \end{pmatrix}, \qquad J_y = \begin{pmatrix} 0 & 0 & 1 \\\\ 0 & 0 & 0 \\\\ -1 & 0 & 0 \end{pmatrix}, \qquad J_z = \begin{pmatrix} 0 & -1 & 0 \\\\ 1 & 0 & 0 \\\\ 0 & 0 & 0 \end{pmatrix} \tag{5} \]
These span the antisymmetric matrices, so they're a basis for the tangent space. Check that \(J_z\) does what it should. Its square is
\[ J_z^2 = \begin{pmatrix} -1 & 0 & 0 \\\\ 0 & -1 & 0 \\\\ 0 & 0 & 0 \end{pmatrix} \tag{6} \]
so \(J_z^3 = -J_z\), and the powers cycle exactly as in post 8. Summing the series gives
\[ \exp(\theta J_z) = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\\\ \sin\theta & \cos\theta & 0 \\\\ 0 & 0 & 1 \end{pmatrix} \tag{7} \]
a rotation by \(\theta\) about the \(z\)-axis, with the third row and column untouched. Three matrices, and every rotation in three dimensions is \(\exp(\theta_x J_x + \theta_y J_y + \theta_z J_z)\).
1.3 Where the non-commutativity went
Now the point of the post. For ordinary numbers \(e^ae^b = e^{a+b}\). For matrices this fails because they are not commutative. We can measure this using the commutator, which is why it comes up in the BCH formula. More on that in a second. First take two tangent vectors \(X\) and \(Y\) and a small parameter \(\varepsilon\). Go a little way along \(X\), then along \(Y\), then back along \(X\), then back along \(Y\):
\[ C(\epsilon) = \exp(\epsilon X)\exp(\epsilon Y)\exp(-\epsilon X)\exp(-\epsilon Y) \tag{8} \]
If everything commuted this would be exactly \(I\), since the two would cancel. It doesn't, so expand and see what survives. To second order,
\[ \exp(\epsilon X)\exp(\epsilon Y) = I + \varepsilon(X + Y) + \varepsilon^2\left( \frac{X^2}{2} + XY + \frac{Y^2}{2} \right) + O(\varepsilon^3) \tag{9} \]
and the reversed pair is the same thing with \(\varepsilon \to -\varepsilon\). Writing \(A = X+Y\) and \(B = \tfrac{X^2}{2} + XY + \tfrac{Y^2}{2}\), the product is
\[ C(\varepsilon) = (I + \varepsilon A + \varepsilon^2 B)(I - \varepsilon A + \varepsilon^2 B) + O(\varepsilon^3) = I + \varepsilon^2(2B - A^2) + O(\varepsilon^3) \tag{10} \]
The first-order terms cancel. Now compute what's left:
\[ 2B - A^2 = (X^2 + 2XY + Y^2) - (X^2 + XY + YX + Y^2) = XY - YX \tag{11} \]
So
\[ C(\varepsilon) = I + \epsilon^2 (XY - YX) + O(\varepsilon^3) \tag{12} \]
There’s the failure of the group to commute. It shows up at second order, and what it produces is the commutator:
\[ [X, Y] = XY - YX \tag{13} \]
So it is actually built from two tangent vectors and it is itself a tangent vector; the left-hand side \(C(\epsilon)\) is a product of group elements, so it's in the group, and its deviation from the identity is a tangent direction. The tangent space is closed under the commutator. That's what makes it a Lie algebra: a vector space with a bracket that's bilinear, antisymmetric (\([X,Y] = -[Y,X]\), immediate from the definition), and satisfies the Jacobi identity
\[ [X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0 \tag{14} \]
which you can verify by expanding all six products and watching them cancel in pairs.
1.4 The structure constants
Compute the bracket for our rotation generators. Multiplying out \(J_xJ_y\) and \(J_yJ_x\) and subtracting:
\[ [J_x, J_y] = J_z \tag{15} \]
and cyclically for the others. Compactly,
\[ [J_i, J_j] = \varepsilon_{ijk} J_k \tag{16} \]
That’s the Levi-Civita symbol, for those who have studied some tensor calculus. But here those numbers \(\epsilon_{ijk}\) are the structure constants, and they are the entire content of the Lie algebra. Everything about how \(SO(3)\) fails to be abelian is compressed into that one equation. Now do \(SU(2)\). Its algebra is the traceless anti-Hermitian \(2\times2\) matrices, and a convenient basis is built from the Pauli matrices (more on these in another post):
\[ \sigma_1 = \begin{pmatrix} 0 & 1 \\\\ 1 & 0 \end{pmatrix}, \qquad \sigma_2 = \begin{pmatrix} 0 & -i \\\\ i & 0 \end{pmatrix}, \qquad \sigma_3 = \begin{pmatrix} 1 & 0 \\\\ 0 & -1 \end{pmatrix} \tag{17} \]
These are Hermitian and traceless, so \(T_j = -\tfrac{i}{2}\sigma_j\) is anti-Hermitian and traceless in the algebra. Using \([\sigma_1, \sigma_2] = 2i\sigma_3\),
\[ [T_1, T_2] = \left(-\frac{i}{2}\right)^2 [\sigma_1, \sigma_2] = -\frac{1}{4}(2i\sigma_3) = -\frac{i}{2}\sigma_3 = T_3 \tag{18} \]
and in general
\[ [T_i, T_j] = \varepsilon_{ijk} T_k \tag{19} \]
The same structure constants.
1.5 The same algebra, different groups
So we have two groups. \(SO(3)\) consists of \(3\times3\) real orthogonal matrices; \(SU(2)\) consists of \(2\times2\) complex unitary matrices. They act on different spaces and their elements aren't even the same size. And yet
\[ \mathfrak{so}(3) \cong \mathfrak{su}(2) \tag{20} \]
as Lie algebras. Which raises an obvious problem. We said the algebra is enough - that exponentiating the tangent space recovers the group. If the algebras are identical, how can the groups be different? We already have a clue, from two posts back. \(SU(2)\) is a 3-sphere and simply connected. \(SO(3)\) is three-dimensional but has a loop that can't be shrunk. Differentiating at the identity is a purely local operation, and it can't see that: the same reasoning showed \(O(3)\) and \(SO(3)\) share a tangent space despite \(O(3)\) having a whole extra component. So the Lie algebra determines the group near the identity and no further. Two groups with the same algebra look identical up close and can differ globally, and the difference is topological.
Sorting out exactly what that relationship is (and why it means a rotation by \(2\pi\) isn't always the identity) is the last post of this series. Hopefully that last idea gives you a hint at why we use these groups to describe spinors!