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 \section{Label-Dependent Notions of Symmetry} \section{Label-Dependent Notions of Symmetry} \label{sec:labdepnotions} There are various ways to define a notion of symmetry, not all of which are distinct. In each case we need all players to have the same number of strategies, consequently all games are implicitly $m$-strategy games. It is often assumed when defining symmetric games that all players have the same strategy labels and any notion of symmetry will treat the same labels as equivalent. We shall refer to these as \textit{label-dependent} notions. \subsection{Permutations Acting On Strategy Profiles} ... ... @@ -15,6 +15,8 @@ The author notes that our somewhat unintuitive notation has been chosen so that \end{proof} \end{lemma} It might be worth explaining which bijections of $A$ we have from $\{s\mapsto\pi(s): \pi \in S_N\}$?? Same for $\{s\mapsto g(s): g \in \bij(\Gamma)\}$ later. Since $\pi^{-1}(s) = (s_{\pi(i)})_{i \in N}$ for all $s \in A$, $s \mapsto \pi(s)$ and $s \mapsto \pi^{-1}(s)$ are dual to each other. Hence the dual results hold for $\pi^{-1}$. \begin{lemma} ... ... @@ -49,7 +51,9 @@ This gives us $\pi(\sigma_i) = \pi(\sigma)_{\pi(i)} \in \Delta(A_{\pi(i)})$ and \end{proof} \end{proposition} \begin{proposition} Note that since $A_i = A_j$ for all $i, j \in N$, the following groupoids in Propositions \ref{prop:firstgroupoidprop} and \ref{prop:secondgroupoidprop} are all groups (or are they? the first yes but what about the second with regards to order/positions of the tuples?). (should they be rephrased, perhaps even with different notation?? what to denote the $A_i$'s as though? Maybe $A_0$? or $A'$ or $\overline{A}$ or $\underline{A}$ or do strategy profiles as $\mathcal{A}$ but would confuse people, as would using $\mathcal{A} = A^n$) \begin{proposition} \label{prop:firstgroupoidprop} $\set{\sigma_i \mapsto \pi(\sigma_i): i \in N, \pi \in S_N}$ is a subgroupoid of $\set{\bij(\Delta(A_i), \Delta(A_j)): i, j \in N}$. \begin{proof} ... ... @@ -64,7 +68,7 @@ This gives us $\pi(\sigma_i) = \pi(\sigma)_{\pi(i)} \in \Delta(A_{\pi(i)})$ and \end{proof} \end{proposition} \begin{proposition} \begin{proposition} \label{prop:secondgroupoidprop} $\set{\sigma_{-i} \mapsto \pi(\sigma_{-i}): i \in N, \pi \in S_N}$ is a subgroupoid of $\set{\bij({\nabla(A)}_{-i}, {\nabla(A)}_{-j}): i, j \in N}$. \begin{proof} ... ... @@ -78,7 +82,7 @@ This gives us $\pi(\sigma_i) = \pi(\sigma)_{\pi(i)} \in \Delta(A_{\pi(i)})$ and &= (\sigma_1, \ldots, \sigma_{i-1}, \sigma_{i+1}, \ldots, \sigma_n) \\ &= \sigma_{-i}. \end{align*} Hence $\id_{{\nabla(A)}_{-i}} = \sigma_{-i} \mapsto \id_N(\sigma_{-i}) \in Y$ (check this??); Hence $\id_{{\nabla(A)}_{-i}} = \sigma_{-i} \mapsto \id_N(\sigma_{-i}) \in Y$; \item Let $i \in N$, $\pi \in S_N$ and $\sigma_{-i}, \sigma'_{-i} \in {\nabla(A)}_{-i}$, then: \begin{align*} \pi(\sigma_{-i}) &= \pi(\sigma'_{-i}) \\ ... ...