Group action for signal

Question

I am working through the book Geometric Deep Learning (https://arxiv.org/abs/2104.13478) and have hit the following problem (Chapter 3.1, page 14). We have a group $\mathfrak{G}$ and a set $\Omega$ for which a signal $\mathcal{X}(\Omega)$ is defined. Let us consider a group action $\mathfrak{g}.u$. Based on it, we automatically obtain an action of $\mathfrak{G}$ on the space $\mathcal{X}(\Omega)$:

$$(\mathfrak{g}.x)(u) = x(\mathfrak{g}^{-1}u)$$

Two questions:

1. What does automatically obtain mean here? Define?
2. How do I verify that this is a valid group action? The best I have been able to get is: $$(\mathfrak{g}.(\mathfrak{h}.x))(u) = \mathfrak{g}.(x(\mathfrak{h}^{-1}u)) = x(\mathfrak{g}^{-1}x(\mathfrak{h}^{-1}u))$$

and

$$((\mathfrak{gh}).x)(u) = x((\mathfrak{gh})^{-1}u))$$

These are not obviously equal. If it is not apparent, I am struggling a bit with the notation for group actions.

This question is related to, but not the same as How group action of $G$ on set $\Omega$ acts on the vector space induced by $\Omega$, $\mathcal{X}(\Omega)$?

Answer 1

For clarity, it should be made clear that $\mathcal{X}(\Omega)$ is not a signal, but the vector space of all signals $x\colon\Omega\to\mathcal{C}$ to some fixed space $\mathcal{C}$ which is suppressed in the notation. What is claimed is that if you have a group action in a space, then without any extra work you get a group action on the space of functions on that space. So in this context "automatically obtain" is a soft phrase which you can interpret as "define using only what you already have and a simple idea".

Before proving that it is a group action, lets look at a very simple example to see what's going on. Start with $\Omega=\mathbb{R}$ thought of as a 1-dimensional affine space (ie a vector space on which we can also perform translations. Let $\mathfrak{G}=\mathbb{R}$ be the group of real numbers under addition. There is a simple action of $\mathfrak{G}$ on $\Omega$ given by $\mathfrak{g}\cdot \omega=\mathfrak{g}+\omega$ which is just $\mathbb{R}$ acting on itself by addition and geometrically corresponds to shifting the real line by a distance $\mathfrak{g}$.

Not it is claimed there is an induced action of $\mathfrak{G}=\mathbb{R}$ on $\mathcal{X}(\Omega)$. Here let us assume that the image space $\mathcal{C}$ is also $\mathbb{R}$ so that $\mathcal{X}(\Omega)$ is just the space of real functions. Let $x\in\mathcal{X}(\Omega)$ be a real function, say $\sin(\omega)$, then for $g\in \mathfrak{G}$, $\mathfrak{g}\cdot x$ is a new real function defined by $g\cdot x(\omega)=x(\mathfrak{g}^{-1}\cdot \omega)$. For $x=\sin(\omega)$ this gives $$\mathfrak{g}\cdot x(\omega)=\sin(\mathfrak{g}^{-1}\cdot \omega)=\sin(\omega-\mathfrak{g}).$$So this is essentially the same function, we've juts shifted the input by $-\mathfrak{g}$.

Let's prove that this is indeed a group action. For $\mathfrak{g},\mathfrak{h}\in\mathfrak{G}$, \begin{align*} (\mathfrak{g}\cdot(\mathfrak{h}\cdot x))(\omega)&=(\mathfrak{h}\cdot x)(\mathfrak{g}^{-1}\cdot\omega)\\ &=x(\mathfrak{h}^{-1}\cdot(\mathfrak{g}^{-1}\cdot\omega))\\ &=x((\mathfrak{h}^{-1}\mathfrak{g}^{-1})\cdot\omega)\\ &=x((\mathfrak{g}\mathfrak{h})^{-1}\cdot\omega)\\ &=((\mathfrak{g}\mathfrak{h})\cdot x)(\omega) \end{align*}