Lie Groups

Sophus Lie

1. Lie Groups

Sophus Lie wasn't trying to invent group theory for physics. He was trying to solve differential equations.

Galois had shown that whether a polynomial can be solved by radicals depends on a finite group attached to it. Lie wanted the same for differential equations — attach a group of symmetries to an equation and read off whether it can be integrated. But the symmetries of a differential equation aren't a finite shuffle of roots. They're continuous: translate the solution, scale it, rotate it, by any amount you like.

So he needed a theory of continuous groups, and what he found was that the continuity does most of the work for you.

1.1 Smooth, not just continuous

Last post we asked only that multiplication and inversion be continuous. Now demand more: that the group be a smooth manifold and that the operations be smooth. That's a Lie group.

Manifolds get a proper treatment later in this series. For now the working picture is enough: a space that looks like \(\mathbb{R}^n\) near any point, with enough consistency that you can differentiate. A circle qualifies — zoom in far enough and any arc looks like a line. A sphere qualifies. Something with a sharp corner does not.

Every matrix group we've met is a Lie group. They sit inside \(\mathbb{R}^{n^2}\), they're cut out by polynomial equations like \(A^TA = I\), and multiplication is polynomial in the entries, so smoothness is inherited rather than imposed.

The step from continuous to smooth looks small. It isn't. Continuity lets you ask whether elements are near each other; smoothness lets you differentiate, and differentiating the group is what turns an infinite object into a finite one.

1.2 Differentiating a group

Here's the whole idea, and it's worth doing concretely before saying it in general.

Take \(SO(2)\) and sit at the identity, \(\theta = 0\). Ask which direction the group is heading as \(\theta\) increases:

\[ A(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\\\ \sin\theta & \cos\theta \end{pmatrix} \tag{1} \]

Differentiate entry by entry and evaluate at zero:

\[ \left. \frac{dA}{d\theta} \right|_{\theta = 0} = \left. \begin{pmatrix} -\sin\theta & -\cos\theta \\\\ \cos\theta & -\sin\theta \end{pmatrix} \right|_{\theta=0} = \begin{pmatrix} 0 & -1 \\\\ 1 & 0 \end{pmatrix} \tag{2} \]

Call that matrix \(J\). It is not in the group — it isn't orthogonal, and \(\det J = 1\) but \(J^TJ = I\) fails, since \(J^T = -J\). It's a velocity, not a position. It lives in the tangent space to the group at the identity.

And the group is one-dimensional, so that's the only direction available. The entire infinite group has been reduced to a single matrix.

Nothing has been lost, either. Expand \(A(\theta)\) as a Taylor series about the identity:

\[ A(\theta) = I + \theta J + \frac{\theta^2}{2} J^2 + \cdots \tag{3} \]

Compute the powers: \(J^2 = -I\), so \(J^3 = -J\), and \(J^4 = I\). The series is

\[ A(\theta) = \left( 1 - \frac{\theta^2}{2!} + \cdots \right) I + \left( \theta - \frac{\theta^3}{3!} + \cdots \right) J = \cos\theta \cdot I + \sin\theta \cdot J \tag{4} \]

which is exactly the rotation matrix we started with. Hopefully this looks familiar to you. J behaves just like the imaginary unit i! And we have that one matrix, exponentiated, regenerates the whole group. A Lie group is infinite, but its tangent space at the identity is a finite-dimensional vector space, and the group can be recovered from it. Post 5 ended by noting that counting arguments die for infinite groups. This is the replacement: don't count elements, differentiate.

1.3 Why the identity is enough

Two things justify looking only at the identity, and one of them we already proved.

From the last post: left multiplication \(L_g(x) = gx\) is a homeomorphism, so the group looks the same near every point. With smoothness it's a diffeomorphism, so the tangent space at any \(g\) is just the tangent space at the identity slid over by \(g\). Nothing new lives anywhere else.

The second reason is generation. Take any element \(g\) that can be joined to the identity by a path. Chop the path into \(N\) tiny steps; each step is a group element very close to \(I\), so it looks like \(I + \epsilon X\) for some small \(\epsilon\) and some tangent direction \(X\). Multiply them all together and you recover \(g\). Everything in the connected component containing the identity is a product of things infinitesimally close to the identity.

Which is why the disconnectedness result from last post matters here. Differentiating at the identity tells you about the identity component only. It can say nothing whatsoever about the reflections in \(O(3)\), because no path reaches them. The tangent space at the identity is the same for \(O(3)\) and \(SO(3)\) — those groups have identical local structure and different global structure.

1.4 Doing it again, with a constraint

\(SO(2)\) was almost too easy, since we had an explicit formula. Try \(SO(3)\), where we don't.

Take a smooth path \(A(t)\) in \(SO(3)\) with \(A(0) = I\). Every point on it satisfies the defining condition:

\[ A(t)^T A(t) = I \tag{5} \]

Differentiate both sides. The right side is constant, so its derivative vanishes, and the product rule gives

\[ \dot{A}^T A + A^T \dot{A} = 0 \tag{6} \]

Evaluate at \(t = 0\), where \(A = I\). Writing \(X = \dot{A}(0)\) for the tangent vector:

\[ X^T + X = 0, \qquad \text{that is} \qquad X^T = -X \tag{7} \]

So the tangent space to \(SO(3)\) at the identity is the set of antisymmetric matrices. We didn't need to solve for the group's elements — we differentiated the constraint instead, and a quadratic condition on group elements became a linear condition on tangent vectors.

That linearity is the entire payoff. Count the dimension: a \(3\times3\) antisymmetric matrix has zeros on the diagonal and the lower triangle determined by the upper, so there are 3 free parameters. \(SO(3)\) is three-dimensional, which matches the physical count of axis-plus-angle.

Run the same argument on \(SU(n)\). From \(U^\dagger U = I\) we get \(X^\dagger = -X\), anti-Hermitian; from \(\det U = 1\), a computation we'll do properly next post gives \(\operatorname{tr} X = 0\). For \(SU(2)\): a \(2\times2\) anti-Hermitian traceless matrix has 3 real parameters. Same dimension as \(SO(3)\), again.

1.5 What we've actually got

Step back and look at what just happened, because it's the reason this subject works.

A Lie group is a curved, infinite object. Its tangent space at the identity is a flat, finite-dimensional vector space. And a nonlinear condition defining the group — \(A^TA = I\) — turned into a linear condition defining the tangent space — \(X^T = -X\). Hard problem replaced by an easy one.

Two things are still missing. The tangent space as described is just a vector space, and a vector space has no memory of the group being non-abelian; we know \(SO(3)\) has rotations that don't commute, so that information has to be encoded somewhere. And we've seen the exponential regenerate \(SO(2)\) from \(J\), but not whether that always works, or how far it reaches.

Both are next post's business, and they resolve into the same object.

Next: Lie algebras and the exponential map.