Lecture 1.6
Group Theory II: Homogeneous Spaces
To understand the "All You Need" theorem in the next lecture, we need to formalize how groups relate to the spaces they act on. This introduces the concepts of Transitivity, Homogeneous Spaces, and Quotient Spaces.
1. Transitive Actions & Orbits
A group $G$ acts transitively on a space $X$ if for any pair of points $x_1, x_2 \in X$, there exists a group element $g \in G$ such that $g \cdot x_1 = x_2$.
Basically, a transitive action means I can reach anywhere in the space from anywhere else using the group.
- Transitive: Translation group on $\mathbb{R}^2$. I can translate the origin to any point $(x,y)$.
- Not Transitive: Rotation group $SO(2)$ on $\mathbb{R}^2$. Rotating the origin keeps it at the origin. Rotating a point $(1,0)$ keeps it on the unit circle. I cannot reach $(2,0)$ from $(1,0)$ by rotation.
2. Homogeneous Spaces
A space $X$ is called a Homogeneous Space of $G$ if $G$ acts transitively on $X$. Intuitively, the space looks "the same" everywhere from the perspective of the group.
3. The Stabilizer Subgroup
If we fix a reference point (origin) $x_0 \in X$, we can ask: which group elements leave this point untouched?
The stabilizer of $x_0$ in $G$ is the subgroup $H = \text{Stab}_G(x_0) = \{ h \in G \mid h \cdot x_0 = x_0 \}$.
Example: For the roto-translation group $SE(2)$ acting on the plane $\mathbb{R}^2$:
Origin: $\mathbf{0}$.
The translations move $\mathbf{0}$ away.
The rotations around $\mathbf{0}$ leave it fixed.
So, the stabilizer is the rotation group $SO(2)$.
4. Quotient Spaces
There is a beautiful one-to-one correspondence between homogeneous spaces and quotient spaces.
A homogeneous space $X$ is isomorphic to the quotient space $G/H$, where $H$ is the stabilizer of an origin $x_0$.
$$ X \cong G/H $$This means we can think of points in the plane $\mathbb{R}^2$ as "cosets" of rotations. A point $\mathbf{x}$ corresponds to the set of all roto-translations that move the origin to $\mathbf{x}$.
Why this matters: This structure dictates the constraints on our convolution kernels. If the input feature map lives on $X \cong G/H$, the kernel must be invariant to the action of the stabilizer $H$.