Felix Hausdorff
1. Continuous Symmetry and Topological Groups
So last time we established that \(SO(2)\) is a circle, and \(SU(2)\) is a 3-sphere. Not just "resembles”, but in fact the set of group elements, with its natural parametrisation, is that shape.
Weird! \(D_3\) has six elements and there's nothing more to say about how they sit relative to each other; asking whether two of them are "close" is meaningless. But in \(SO(2)\), a rotation by \(0.001\) radians is obviously near the identity. So these groups carry two structures at once.
1.1 Nearness without distance
To say what "close" means, the minimal structure is a topology: a rule specifying which subsets count as open. In \(\mathbb{R}^n\) an open set is one where every point has a little ball around it still inside the set. If you have not studied topology before, this is a surprisingly accurate definition. We won’t get too much into compactness, openness, and simplicity here, but maybe in another post. The point is that the idea of "openness” is enough to define continuity without ever measuring a distance. A map \(f\) is continuous if the preimage of every open set is open. Nearby inputs give nearby outputs without any metric. In other words, this is an intrinsic geometric quality that does not require any notion of distance or measurement.
Our matrix groups get their topology nicely. \(GL(n,\mathbb{R})\) sits inside the space of all \(n \times n\) matrices, which is just \(\mathbb{R}^{n^2}\) with the entries as coordinates, and every subgroup inherits the topology of that surrounding space. So two matrices are close when their entries are close.
1.2 Topological groups
Now the compatibility condition. A topological group is a group that's also a topological space, in which the two operations
\[ (g, h) \mapsto gh \qquad \text{and} \qquad g \mapsto g^{-1} \tag{1} \]
are both continuous.
Not a huge demand, but it is very important. Basically move \(g\) slightly and \(gh\) moves slightly, move it and \(g^{-1}\) moves slightly.
We can check it on \(SO(2)\). Multiplication is \(\theta_1 + \theta_2\) and inversion is \(\theta \mapsto -\theta\), both continuous. For a general matrix group, multiplication is polynomial in the entries and inversion is rational with nonvanishing denominator (the determinant), so continuity is automatic. Every matrix group is a topological group.
Here's the first thing that gets us. Fix \(g\) and consider left multiplication:
\[ L_g : G \to G, \qquad L_g(x) = gx \tag{2} \]
It's continuous, and \(L_{g^{-1}}\) is a continuous inverse for it, so it's a homeomorphism (it moves the group around without tearing it). Which means a topological group looks the same near every point. Whatever the neighbourhood of the identity is like, every other neighbourhood is like that too, because you can slide one onto the other. Hopefully now you can see slightly why topology talks about holes, why a donut and a mug are the same shape to a topologist, and so on.
1.3 Connectedness, and a group that isn't
A space is connected if you can't split it into two disjoint nonempty open pieces. Path-connected is the more intuitive version: any two points are joined by a continuous path inside the space. For the groups here the two notions agree, so use whichever one you prefer. \(SO(2)\) is connected, since it's a circle, and you can walk around it. So is \(SO(3)\), and so is \(SU(2)\), which is a sphere. But look at \(O(n)\). Every element has \(\det = \pm 1\), and the determinant is continuous. Suppose there were a path from some \(A\) with \(\det A = 1\) to some \(B\) with \(\det B = -1\). Then \(\det\) along that path would be a continuous function starting at \(+1\) and ending at \(-1\), so by the intermediate value theorem it would take the value \(0\) somewhere. But a matrix with zero determinant isn't invertible, so it isn't in the group.
No such path exists! \(O(n)\) falls into two disconnected pieces, and \(SO(n)\) is the piece containing the identity.
This gives a much better answer to a question you might have had earlier: why do physicists always work with \(SO(3)\) rather than \(O(3)\)? Not because reflections are unimportant, but because you cannot reach a reflection continuously from doing nothing. A symmetry you can turn on gradually has to live in the connected piece containing the identity, and that piece is \(SO(3)\). The component containing the identity is always a normal subgroup, and the components are its cosets. For \(O(n)\) there are two cosets and the quotient is \(\mathbb{Z}_2\), exactly as before.
1.4 Compactness
Uh oh. Compactness. The enemy of any math student! Luckily, we are physicists today, so we won’t get too bogged down by definitions. I’ll try to keep it simple, mostly because, I don’t know if I really understand compactness yet! Anyway, a subset of \(\mathbb{R}^N\) is compact if it's closed and bounded. So continuous functions on compact spaces attain their bounds, and you can integrate over a compact group with a finite answer. \(SO(n)\) is compact. Boundedness follows from \(A^TA = I\): every column has unit length, so no entry can exceed 1 in absolute value, and the whole group sits inside a box. Closedness follows because \(A^TA = I\) is a set of polynomial equations, and solution sets of continuous equations are closed. Same argument gives compactness of \(SU(n)\), which we already knew was a sphere. \(GL(n,\mathbb{R})\) is not compact; the matrices \(\operatorname{diag}(t, 1, \dots, 1)\) run off to infinity as \(t\) grows.
Neither is the Lorentz group. Since a boost is parametrised by rapidity, which is unbounded, the group is non-compact even though the corresponding velocities are all under \(c\). Rotations are compact and boosts aren’t. This is eventually why finite-dimensional representations of the Lorentz group can't be unitary, which is behind a lot of what makes relativistic quantum theory awkward. WE will get there soon enough. I don’t know it all too well yet:)
1.5 The thing a circle can do that a sphere cannot
Take a loop in the group, i.e. a path starting and ending at the same point. Can it be continuously shrunk to a point without leaving the group? On a sphere, always! Any loop drawn on a sphere's surface can be reeled in. A space where every loop shrinks is called simply connected.
On a circle, no. A loop that wraps once around cannot be shrunk, because there's nowhere to pull it through, since the hole is in the way. And a loop wrapping twice is different from one wrapping once. So loops in \(SO(2)\) are classified by an integer, the winding number. Needham has some great content on this in his Visual Complex Analysis book iirc. So \(SU(2)\) is a 3-sphere and simply connected. \(SO(3)\) is three-dimensional, same as \(SU(2)\), but it is not simply connected, since there's a loop in it that can't be shrunk. Spoiler! We are going into the territory of spinors here. Eigenchris has some great material on these to understand them (somewhat) intuitively.