Camille Jordan
1. Matrix Groups
Everything so far has been finite. But, the groups we actually want (all rotations of a sphere, all Lorentz transformations, etc) have infinitely many elements, and dividing \(|G|\) by \(|H|\) becomes meaningless.
So we need examples that are infinite but still concrete enough to compute with. Matrices!
1.1 The general linear group
Start with the biggest one. Fix a field, say \(\mathbb{R}\), and take all invertible \(n \times n\) matrices:
\[ GL(n, \mathbb{R}) = \{ A : A \text{ is } n \times n \text{ and } \det A \neq 0 \} \tag{1} \]
Check the axioms. Matrix multiplication is associative, the identity matrix is an identity, and invertibility is the condition for inverses to exist. And closure holds because
\[ \det(AB) = \det(A)\det(B) \tag{2} \]
so a product of two matrices with nonzero determinant has nonzero determinant. That's a group, called the general linear group.
It's infinite, and it's also non-abelian for \(n \geq 2\), which is easy to see:
\[ \begin{pmatrix} 1 & 1 \\\\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\\\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\\\ 1 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 0 \\\\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\\\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\\\ 1 & 2 \end{pmatrix} \tag{3} \]
\(GL(n, \mathbb{R})\) is the group of all invertible linear maps on \(\mathbb{R}^n\). Every matrix group we care about is a subgroup of it.
1.2 Preserving a length
Suppose we don't want all linear maps, only the ones that don't distort. Distances in \(\mathbb{R}^n\) come from the dot product
\[ \mathbf{x} \cdot \mathbf{y} = \mathbf{x}^T \mathbf{y} \tag{4} \]
so demand that \(A\) leave it alone:
\[ (A\mathbf{x})^T (A\mathbf{y}) = \mathbf{x}^T \mathbf{y} \tag{5} \]
The left side is \(\mathbf{x}^T A^T A \mathbf{y}\). For this to hold for every \(\mathbf{x}\) and \(\mathbf{y}\) we need
\[ A^T A = I \tag{6} \]
Matrices satisfying that are orthogonal, and they form the group \(O(n)\). It really is a group: if \(A^TA = I\) and \(B^TB = I\) then
\[ (AB)^T(AB) = B^T A^T A B = B^T B = I \tag{7} \]
so it's closed, and \(A^{-1} = A^T\) is orthogonal too.
Now take determinants of \(A^TA = I\). Since \(\det(A^T) = \det(A)\),
\[ (\det A)^2 = 1, \qquad \det A = \pm 1 \tag{8} \]
So \(O(n)\) splits into two pieces. The determinant is a homomorphism \(O(n) \to \{+1, -1\}\), and by post 3 its kernel is a normal subgroup; the matrices with \(\det = +1\), called the special orthogonal group \(SO(n)\). These are the rotations. The other piece, \(\det = -1\), contains the reflections, and it isn't a subgroup because it doesn't contain the identity.
This is the same structure as \(D_3\) in post 3, where the sign homomorphism split the triangle's symmetries into rotations and reflections with \(D_3 / \{e,r,r^2\} \cong \mathbb{Z}_2\). Here it's
\[ O(n) / SO(n) \cong \mathbb{Z}_2 \tag{9} \]
Same thing, just infinite.
1.3 \(SO(2)\) explicitly
Let's actually solve the constraint in the smallest interesting case. Write a general \(2 \times 2\) matrix and impose \(A^TA = I\):
\[ A = \begin{pmatrix} a & b \\\\ c & d \end{pmatrix}, \qquad A^T A = \begin{pmatrix} a^2 + c^2 & ab + cd \\\\ ab + cd & b^2 + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\\\ 0 & 1 \end{pmatrix} \tag{10} \]
Three equations: \(a^2 + c^2 = 1\), \(b^2 + d^2 = 1\), and \(ab + cd = 0\).
The first says \((a, c)\) sits on the unit circle, so \(a = \cos\theta\) and \(c = \sin\theta\) for some angle. The second puts \((b,d)\) on the circle too, say \(b = \cos\phi\), \(d = \sin\phi\). The third then reads
\[ \cos\theta\cos\phi + \sin\theta\sin\phi = \cos(\theta - \phi) = 0 \tag{11} \]
so \(\phi = \theta \pm \pi/2\). Taking \(\phi = \theta + \pi/2\) gives \(b = -\sin\theta\), \(d = \cos\theta\), and
\[ A = \begin{pmatrix} \cos\theta & -\sin\theta \\\\ \sin\theta & \cos\theta \end{pmatrix} \tag{12} \]
with \(\det A = \cos^2\theta + \sin^2\theta = 1\). The rotation matrices! Again, the other sign choice gives \(\det = -1\), the reflections.
This group is labelled by a single continuous parameter \(\theta\), and the parameter lives on a circle, since \(\theta\) and \(\theta + 2\pi\) give the same matrix. The group is a circle. So now the group isn't just a set with a multiplication, it's a set with a shape. The abstract idea of groups now have geometric interpretations. Composition is addition of angles:
\[ A(\theta_1) A(\theta_2) = A(\theta_1 + \theta_2) \tag{13} \]
which you can verify by multiplying out and using the angle-addition formulas. So \(SO(2)\) is abelian. It's the only one of these that is.
1.4 Over the complex numbers
Repeat the construction with \(\mathbb{C}^n\). The natural inner product now involves a conjugate,
\[ \langle \mathbf{x}, \mathbf{y} \rangle = \mathbf{x}^\dagger \mathbf{y} \tag{14} \]
where \(\dagger\) means transpose and conjugate. Preserving it gives
\[ U^\dagger U = I \tag{15} \]
the unitary group \(U(n)\). Same argument as before shows it's a group.
Determinants again: \(\det(U^\dagger) = \overline{\det U}\), so \(|\det U|^2 = 1\), meaning \(\det U\) is a complex number of modulus 1. That's a weaker condition than \(\pm 1\), in fact it's a whole circle of possibilities rather than two points. Demanding \(\det U = 1\) picks out the special unitary group \(SU(n)\).
This group matters a lot so let’s go through it. Write
\[ U = \begin{pmatrix} \alpha & \beta \\\\ \gamma & \delta \end{pmatrix} \tag{16} \]
The condition \(\det U = 1\) plus \(U^\dagger U = I\) forces \(\gamma = -\bar\beta\) and \(\delta = \bar\alpha\), giving
\[ U = \begin{pmatrix} \alpha & \beta \\\\ -\bar\beta & \bar\alpha \end{pmatrix}, \qquad |\alpha|^2 + |\beta|^2 = 1 \tag{17} \]
Now count parameters. Two complex numbers is four real numbers, minus one constraint, leaves three. And the constraint \(|\alpha|^2 + |\beta|^2 = 1\) is precisely the equation of the unit sphere in \(\mathbb{R}^4\).
\(SU(2)\) is a 3-sphere. (a.k.a a ball, or a 4D hollow sphere).
Compare \(SO(3)\), the rotations of ordinary space. A rotation needs an axis, which is a point on a sphere, and an angle - that's also three parameters. So we have two different groups, both three-dimensional, but one is a perfectly round sphere while the other isn't. They have an interesting relationship.
1.5 The idea
Every group here was built by taking all invertible matrices, then demand that something specific be preserved.
\[ \begin{array}{ll} \text{dot product on } \mathbb{R}^n & O(n) \\\\ \text{dot product and orientation} & SO(n) \\\\ \text{Hermitian product on } \mathbb{C}^n & U(n) \\\\ \text{Hermitian product and volume} & SU(n) \end{array} \tag{18} \]
Now, try to preserve the Minkowski form \(-t^2 + x^2 + y^2 + z^2\) instead of the dot product and you get the Lorentz group \(O(1,3)\), which is where we’re going. But the new fact in this post isn't any particular group. It's that \(SO(2)\) turned out to be a circle and \(SU(2)\) turned out to be a sphere. These groups have a shape, so you can ask whether a path between two elements exists, whether the group is connected, whether a loop in it can be shrunk to a point. None of those questions make sense for \(D_3\).