Symmetry of a game

Question

I am now reading Nash's 1951 paper Non-cooperative games and I have a question about the definition of symmetry of a game.

Symmetries of Games(Nash 1951)

If two strategies belong to a single player they must go into two strategies belonging to a single player. Thus if $\phi$ is the permutation of the pure strategies, it induces a permutation $\psi$ of the players.

Each $n$-tuple of pure strategies is therefore permuted into another $n$-tuple of pure strategies. We may call $\lambda$ the induced permutation of these $n$-tuples. Let $\xi$ denote an $n$-tuple of pure strategies and $p_{i}(\xi)$ the payoff to player $i$ when the $n$-tuple $\xi$ is employed. We require that if $j=i^{\psi}$, then $p_{j}(\xi^{\lambda})=p_{i}(\xi)$.

The permutation $\phi$ has a unique linear extension to the mixed strategies. If $$s_{i}=\sum_{\alpha}c_{i\alpha}\pi_{i\alpha}$$We define $(s_{i})^{\phi}=\sum_{\alpha}c_{i\alpha}(\pi_{i\alpha})^{\phi}$

The extension of $\phi$ to the mixed strategies clearly generates an extension of $\lambda$ to the $n$-tuples of mixed strategies. We shall also denote this by $\lambda$.

We define a symmetric $n$-tuple $\mathcal S$ of a game by $\mathcal S^{\lambda}=\mathcal S$ for all $\lambda$.

In Nash's notation, $c_{i\alpha}$ is the probability weight put on action $\pi_{i\alpha}$. i.e $\sum_{\alpha}c_{i\alpha}=1$ and $c_{i\alpha}\geq 0$. $\pi_{i\alpha}$ is the $\alpha$-th strategy in the pure-strategy space of player $i$. i.e $S_{i}=\{\pi_{i\alpha}\}_{\alpha=1}^{N_{i}}$ where $N_{i}$ is finite.

I don't quite understand the definition of a permutation of strategies. To me it seems that there are two interpretations for this.

The first interpretation is that it is a permutation of strategies within player $i$'s strategy space. For example, if a player's strategy space is $S=\{U,D\}$, then a permutation is $S'=\{D,U\}$. But if this is the case, how can it induce a permutation on players?

The second interpretation is that it is a permutation of a strategy profile. To be more specific, suppose in a game with players $1$ and $2$, with $S_{1}=\{L,R\}$ and $S_{2}=\{U,D\}$, and a strategy profile $\xi=(L,U)$. Then $(U,L)$ a permutation of $(L,U)$ as defined above and the induced $\psi$ permutes the players from $\{1,2\}$ to $\{2,1\}$.

However, if the second interpretation is correct, suppose the strategy profile is $(\frac{1}{3}L+\frac{2}{3}R,\frac{1}{4}U+\frac{3}{4}D)$, then a possible permutation of the mixed strategy profile is $(\frac{1}{3}U+\frac{2}{3}D,\frac{1}{4}L+\frac{3}{4}R)$, does it mean that a symmetric tuple of a game only exists when the pure-strategy spaces are the same for all players?

Thanks in advance!

Answer 1

I split my answer in two parts, one about how symmetries are defined and the other what the meaning of a symmetric tuple is.

1. In principle, a symmetry is a permutation of the disjoint union of all pure-strategy spaces of all players. One can then derive symmetries on the tuples of pure strategies. If a symmetry $\phi$ maps a (pure) strategy of one player, let's call her Ann to a strategy of another player, Bob, then Ann and Bob must have the same number of strategies. To see this, note that $\phi$ maps by assumption any two strategies of Ann to strategies of Bob. Since $\phi$ is injective, Bob must have at least as many strategies as Ann. But $\phi^{-1}$ maps some, and hence all, strategies of Bob to strategies of Ann. Since $\phi^{-1}$ is injective, Ann must have at least as many strategies as Bob. Together, they must have the same number of strategies. The strategies need not have the same label. Also, a symmetry can also switch strategies of a single player.

2. A tuple is symmetric if it is mapped to itself by every induced symmetry on tuples. Now, the more "symmetric" a game is, the more symmetries it has. But a game can be so asymmetric that the only symmetry is the trivial one that maps every strategy to itself. In such a game, every tuple is mapped by all symmetries (really, the only one) to itself. So every tuple in such a game is symmetric. If a symmetry in your example permutes $U$ and $L$, then it must also permute $R$ and $D$. Let's assume that these symmetries together with their inverses and the identity are the only symmetries. The profile $(\frac{1}{3}L+\frac{2}{3}R,\frac{1}{4}U+\frac{3}{4}D)$ is not a symmetric tuple, because it is not mapped to itself. However, the tuple $(\frac{1}{3}L+\frac{2}{3}R,\frac{1}{3}U+\frac{2}{3}D)$ would be symmetric.