SurjunctiveGroup.lean`

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8 declarations, 1 contain sorry, 0 declared as axioms

Gottschalk's surjunctivity conjecture

A group $G$ is surjunctive if every injective, continuous, $G$-equivariant map $A^G \to A^G$ (for any finite alphabet $A$) is surjective.

Here equivariance is with respect to the left shift action of $G$ on $A^G$, defined by $(g \cdot x)(h) = x(g^{-1} h)$, and continuity is with respect to the product topology on $A^G$ (where $A$ carries the discrete topology).

Gottschalk's conjecture (1973) states that every group is surjunctive.

References:

Wikipedia
Gottschalk, W. H. (1973), "Some general dynamical notions"

Imports (1)

FormalConjecturesUtil

def GottschalkSurjunctivity.shift The left shift of x : G → A by g : G, defined by (shift g x)(h) = x(g⁻¹ * h) line 46 source
Source documentation
The left shift of x : G → A by g : G, defined by (shift g x)(h) = x(g⁻¹ * h). This is the standard left shift action of G on A^G. We define it as a plain function rather than a MulAction instance to avoid conflict with the pointwise Pi.instMulAction.

standard axioms: propext what this means
Statement uses: Group
Proof uses: HMul.hMul, Inv.inv
10 instances, coercions and auxiliary constants left out.
```
def shift {A : Type*} (g : G) (x : G → A) : G → A :=
…
```
theorem GottschalkSurjunctivity.shift_apply line 50 source
standard axioms: propext what this means
Statement uses: Group, Eq, GottschalkSurjunctivity.shift, HMul.hMul, Inv.inv
Proof uses: rfl
10 instances, coercions and auxiliary constants left out.
theorem GottschalkSurjunctivity.shift_one line 54 source
standard axioms: Quot.sound, propext what this means
Statement uses: Group, Eq, GottschalkSurjunctivity.shift, OfNat.ofNat
Proof uses: funext, of_eq_true, Eq.trans, True, congrFun', congrArg, HMul.hMul, Inv.inv, inv_one, one_mul, eq_self
17 instances, coercions and auxiliary constants left out.

Tactic trace: 2 steps
Open to load the checked before-and-after goal states.
theorem GottschalkSurjunctivity.shift_mul line 59 source
standard axioms: Quot.sound, propext what this means
Statement uses: Group, Eq, GottschalkSurjunctivity.shift, HMul.hMul
Proof uses: funext, of_eq_true, Inv.inv, Eq.trans, True, congrFun', congrArg, mul_inv_rev, mul_assoc, eq_self
20 instances, coercions and auxiliary constants left out.

Tactic trace: 2 steps
Open to load the checked before-and-after goal states.
def GottschalkSurjunctivity.IsShiftEquivariant A map τ : (G → A) → (G → A) is equivariant (with respect to the left shift action) if τ(shift g x) = shift g (τ x) for all g : G and x : G → A line 65 source
Source documentation
A map τ : (G → A) → (G → A) is equivariant (with respect to the left shift action) if τ(shift g x) = shift g (τ x) for all g : G and x : G → A.

standard axioms: propext what this means
Statement uses: Group
Proof uses: Eq, GottschalkSurjunctivity.shift
```
def IsShiftEquivariant {A : Type*} (τ : (G → A) → (G → A)) : Prop :=
…
```
def GottschalkSurjunctivity.IsSurjunctive A group G is surjunctive if for every finite nonempty type A, every injective, continuous, shift-equivariant map (G → A) → (G → A) is also surjective line 73 source
Source documentation
A group G is surjunctive if for every finite nonempty type A, every injective, continuous, shift-equivariant map (G → A) → (G → A) is also surjective.
Continuity is with respect to the product topology on G → A where A carries the discrete topology.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Group
Proof uses: Fintype, Nonempty, TopologicalSpace, DiscreteTopology, Continuous, GottschalkSurjunctivity.IsShiftEquivariant, Function.Injective, Function.Surjective
1 instances, coercions and auxiliary constants left out.
```
def IsSurjunctive : Prop :=
…
```
theorem GottschalkSurjunctivity.gottschalk_surjunctivity_conjecture Gottschalk's surjunctivity conjecture (1973): every group is surjunctive sorry line 85 source
Source documentation
Gottschalk's surjunctivity conjecture (1973): every group is surjunctive. That is, for every group G and every finite alphabet A, every injective cellular automaton on A^G is surjective.

depends on sorry what this means
Statement uses: Group, GottschalkSurjunctivity.IsSurjunctive
Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous, OfNat.ofNat, Nat
1 instances, coercions and auxiliary constants left out.

Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
```
theorem gottschalk_surjunctivity_conjecture (G : Type) [Group G] :
    IsSurjunctive G := by
…
```
Mentions: GottschalkSurjunctivity.IsSurjunctive
theorem GottschalkSurjunctivity.isSurjunctive_of_finite Every finite group is surjunctive line 92 source
Source documentation
Every finite group is surjunctive. This is a classical result: an injective endomorphism of a finite set is surjective.

standard axioms: Classical.choice, Quot.sound, propext what this means
Statement uses: Group, Finite, GottschalkSurjunctivity.IsSurjunctive
Proof uses: Fintype, Nonempty, TopologicalSpace, DiscreteTopology, Continuous, GottschalkSurjunctivity.IsShiftEquivariant, Function.Injective, Finite.surjective_of_injective
3 instances, coercions and auxiliary constants left out.

Tactic trace: 2 steps
Open to load the checked before-and-after goal states.
```
theorem isSurjunctive_of_finite (G : Type) [Group G] [Finite G] :
    IsSurjunctive G := by
…
```
Mentions: GottschalkSurjunctivity.IsSurjunctive