What does $R^G$ mean if $G$ is a group? The following is from the paper The Modern Mathematics of Deep Learning page 45.
I want to know what the notation $\mathbb{R}^{G}$ means. For example, if $d=3$, I think $G=\{0,1,2\}$, then I confuse about $a, b \in \mathbb{R}^{G}$. What is the dimension of vector a? How do I understand $\mathbb{R}^{G}$?

A convolutional $NN$ corresponds to multiple convolutional blocks, which are special types of layers. For a group $G$, which typically is either $[d] \cong \mathbb{Z} /(d \mathbb{Z})$ or $[d]^{2} \cong(\mathbb{Z} /(d \mathbb{Z}))^{2}$ for $d \in \mathbb{N}$, depending on whether we are performing one-dimensional or two-dimensional convolutions, the convolution of two vectors $a, b \in \mathbb{R}^{G}$ is defined as
$$
(a * b)_{i}=\sum_{j \in G} a_{j} b_{j^{-1} i}, \quad i \in G
$$

 A: I think is just the space of tuples with indexes in $G$, that is, vectors in the form $$(v_g)_{g\in G}$$or, equivalently, the set of functions $G\rightarrow\mathbb R$. For example, if $G$ is a finite group with elements $\{g_1,\cdots,g_n\}$, then there is a bijection $\mathbb R^G\rightarrow\mathbb R^n$ defined as $$(v_g)_{g\in G}\mapsto(v_{g_1},\cdots,v_{g_n})$$ so $\mathbb R^G$ (in the case of a finite group) is $\mathbb R^{|G|}$. The difference is that in $\mathbb R^G$ you can use the group structure on $G$ to define another operation $*:\mathbb R^G\times\mathbb R^G\rightarrow\mathbb R^G$ by \begin{align}
(v*w)_{g}=\sum_{ij=g}v_iw_j\tag{1}\end{align}
Edit I don't know if $G$ is supposed to be finite. Clearly yes if $\mathbb R^G$ is the set of all functions $G\rightarrow\mathbb R$, otherwise $(1)$ may be infinite. There is an alternative definition of $\mathbb R^G$ that works for an infinite $G$ as well and is equivalent to the definition above when $G$ is finite: $$\mathbb R^G=\{f:G\rightarrow\mathbb R|f(g)\neq0\text{ for finitely many }g\}$$ then you can still define $*:\mathbb R^G\times\mathbb R^G\rightarrow\mathbb R^G$ as in $(1)$ and it's always defined. Regardless, I think in the context of the article, you can simply assume $G$ finite.
