A simple theorem about vacuum states in spatio-temporal dynamical systems, which include PDE but also things like cellular automata and coupled map lattice

Posted Tuesday November 7 2023.

A “spatio-temporal” dynamical system consists of

• a space X of points such as n-dimensional euclidean space,
• a space Y of states,
• a monoid T representing the time variable,
• an evolutionary operator (a monoid action) Φ : T × Y^X → Y^X acting on maps X → Y

In physics these are called “field theories”. For example, a classical scalar field theory is heat diffusion in 3 dimensions, in which case X = R³, Y = R, T = [0, ∞), and we can actually write down the evolutionary operator Φ^t using the heat kernel in 3 dimensions. On the other hand, we could model Conway’s Game of Life by taking X = Z², Y = 0, 1, and T = N.

These two example systems are pretty different, but they do share one thing in common: they have spatial symmetry, in the sense that the evolutionary operator Φ^t is equivariant with respect to the action of some group G acting on X. For a group element g ∈ G and a point x ∈ X we will denote the action of g on x by g * x. Notably, any group action of G on X extends to a group action on Y^X. So if g ∈ G and u ∈ U we can write g * u for the action of g to u.

Definition 1: Let G be a group, let X be a set, and let the mapping (g, x) ↦ g * x be a group action of G on X. The group action is called transitive if for any x₁, x₂ ∈ X, there exists g ∈ G such that g * x₁ = x₂.

By “vacuum state” we just mean a constant state - that is a function u : X → Y such that there exists some y ∈ Y such that u(x) = y for all y. Some people just write u ≡ y. Anyway, because of the transitive group action, if u is constant, then so is Φ(t, u) for all t ∈ T. Let’s prove it!

Lemma 1: Suppose the group action (g, x) ↦ g * x is transitive. Then u ∈ Y^X is constant iff. u = g * u, ∀g ∈ G.

Proof: For the forward direction, suppose u is constant, u ≡ y, and let g ∈ G. Then u(x) = (g * u)(x) = y for all x ∈ X, so g * u = u. For the other direction, suppose g * u = u for all g ∈ G. We want to show u is constant, so let x₁, x₂ ∈ X. Since G is transitive, we know there exists g ∈ G such that x₂ = g * x₁. Thus u(x₂) = u(g * x₁) = (g * u)(x₁) = u(x₁). Thus, u is constant. QED.

The next lemma shows that if the group action is transitive, and the dynamical system is equivariant with respect to the group action, then constant states will remain constant.

Lemma 2: Suppose the action (g, x) ↦ g * x is transitive, and suppose Φ, the action of T on Y^X, is G-equivariant, i.e. Φ(t, g * u) = g * Φ(t, u) for all u ∈ Y^X, g ∈ G, and t ∈ T. Then if u ∈ Y^X is constant, so is Φ(t, u) for any t ∈ T.

Proof: Suppose u is constant, i.e. u ≡ y for some y ∈ Y. Now let t ∈ T. We wish to show Φ(t, u) is constant. By Lemma 1, it suffices to show g * Φ(t, u) = Φ(t, u) for all g ∈ G. So let g ∈ G. By Lemma 1, since u is constant we know g * u = u. By assumption we have g * Φ(t, u) = Φ(t, g * u). Replacing g * u = u, we have g * Φ(t, u) = Φ(t, u), the desired result. QED.

Corollary: If the action $(g,x) g x $ is transitive and the action of Φ^t on Y^X is G-equivariant, then there is a unique monoid action of ϕ : T × Y → Y satisfying

Φ ∘ (1_T × i) = i ∘ ϕ