Local vs Global Nash Equilibrium intuition

Question

A Nash Equilibrium is a point in the space of strategies of a game where both players can not change their respective strategy without reducing their individual chances/winrates/rewards.

What is the intuition behind a local Nash Equilibrium? Does a local Nash Equilibrium prohibit a "jump" (i.e non-smooth change) in the chosen strategy of a single player?

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Definition of a weak local equilibrium (Source, Slide 5)

Let $(S_1, d_1), \cdots ,(S_n, d_n)$ be metric spaces and $p_1, \cdots , p_n : S_1 × \cdots × S_n → R$ real functions. We say that $\tilde{s} = (\tilde{s}_1, \cdots , \tilde{s}_n) \in S$ is a weak local equilibrium (abbrev., w.l.e.) for $p_1, \cdots , p_n$ if for all $\epsilon > 0$ there exists $\delta > 0$ such that $p_i(\tilde{s}_1, \cdots , \tilde{s}_{i−1}, s_i , \tilde{s}_{i+1}, \cdots , \tilde{s}_n) ≤ p_i(\tilde{s}) + \epsilon d_i(s_i, \tilde{s}_i), \forall s_i \in B(\tilde{s}_i , \delta), i = 1, 2, \cdots , n.$

Answer 1

Full disclosure I've never encountered this solution concept before, but just analytically I'd view it as a localized, tempered version of the usual notion of $\varepsilon$-equilibrium (which gets at your idea of jumps: an $\varepsilon$-equilibrium is one in which no player has a deviation that makes them strictly $\varepsilon > 0$ better off than playing their part in the tuple itself). Using your notation:

A tuple $s = (s_i, s_{-i}) \in \prod_i S_i$ is an $\varepsilon$-equilibrium if: $$ p_i(\hat{s}_i, s_{-i}) \le p_i(s_i, s_{-i}) + \varepsilon $$ for all $i, \hat{s}_i \in S_i$.

A weak local equilibrium $\tilde{s}$ asks that, for all $\varepsilon > 0$, that there exists some neighborhood of $\tilde{s}$ on which $\tilde{s}$ is a local $\varepsilon$-equilibrium, but now with the caveat that the bound on how much any agent can gain from a unilateral deviation before the condition is violated is non-uniform across this neighborhood: strategies further away face a higher threshold to count as a profitable unilateral deviation. That's what the $\varepsilon d(s_i, \tilde{s}_i)$ term captures. In other words, for some fixed neighborhood, strategies further from $\tilde{s}_i$ under the metric on $S_i$ must provide a (linear in $\varepsilon$) higher return than closer alternatives to count as a viable deviation.