# group representations and group actions (geometric machine learning)

## 1. group action

- Let \(G\) be a group (algebra)
- Let \(\Omega\) be a domain. I usually think of \(\Omega\) as a (finite) grid of pixels.
- Let \(\cal{X}(\Omega)\) be the space of signals. That is \(\cal{X}(\Omega) = \{x: \Omega \rightarrow \mathcal{C}\}\) where \(\mathcal{C}\) is a vector space. I usually think of \(\cal{X}\) as the set of images on a finite grid. That is, a signal is just an assignment of grid positions to color values.
- a group action is a mapping from group elements of \(G\) to transformations on \(\Omega\). So it is the map \((g,u)\mapsto g.u\). Where \(u\in \Omega\) and \(g\in G\). Note that this map is defined pointwise over \(\Omega\). This isn't super important, but tripped up my initial reading.
- To be a group action, this mapping must be compatible with the group operation. So \(g.(h.u) = (gh).u\) for all \(g,h \in G\) and \(u\in \Omega\)
- If we have a group action on \(\Omega\), then we automatically have a group action on \(\mathcal{X}(\Omega)\) given by
- \((g.x)(u) = x(g^{-1} u)\)

- Some things that help make this make sense
- How do you transform a function \(x\)? You should change how it maps inputs to outputs. The transformed \(x\) will still take the same inputs, but it will take it to different places
- Imagine you want to know what an image looks like after being rotated 90 degrees clockwise. You have the original mapping of grid positions to color values. So you should find the "original" grid positions, before rotations, this is given by \(g^{-1}u\). Then you should find the color value at the original position. This is \(x(g^{-1} u)\). Then, this gives you the value of the color of the pixel position of the "new" transformed image.

## 2. group representation

- A special type of group action is a group representation, which is a linear group action
- linear group actions can be described as a map \((g,x) \mapsto g.x\) that is linear in \(x\) (note that we are now talking about signals \(x\) and not points in the domain \(u\))
- a linear group action can also be described by a map \(\rho: G \rightarrow R^{n\times n}\) from group elements to invertible matrices
- We should have that \(\rho(gh) = \rho(g)\rho(h)\). That is, the group operation can be represented by matrix multiplication. This comes from the fact that the representation is a group action.

## 3. side question

- When dihedral groups are presented in textbooks, a polygon with its vertices labeled is often shown, in all its orientations. Edges between two polygons are meant to represent transformations. My question is: what are the symmetries? What are the group elements? The symmetries are transformations. Between two different labelings of vertices, there may be many compositions of transformations that take labeling A to labeling B, but they are all identified with each other. Each symmetry is a group element.