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tidy up the action topology file
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‎FLT/HIMExperiments/right_module_topology.lean

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import Mathlib.RingTheory.TensorProduct.Basic -- we need tensor products of rings at some point
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import Mathlib.Topology.Algebra.Module.Basic -- and we need topological rings and modules
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import Mathlib.Tactic
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import Mathlib
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/-
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@@ -36,100 +36,123 @@ functions from `M` (now considered only as an index set, so with no topology) to
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-/
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section noncommutative
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-- Let A be a ring, with a compatible topology.
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variable (A : Type*) [Ring A] [TopologicalSpace A] [TopologicalRing A]
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class IsActionTopology (R M : Type*) [SMul R M]
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[τR : TopologicalSpace R] [τM : TopologicalSpace M] : Prop where
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isActionTopology' : τM = ⨆ m, .coinduced (· • m) τR
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/-- The "right topology" on a module `M` over a topological ring `A`. It's defined as
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the finest topology on `M` which makes every `A`-linear map `A → M` continuous. It's called
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the "right topology" because `M` goes on the right. -/
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abbrev Module.rtopology (M : Type*) [AddCommGroup M] [Module A M]: TopologicalSpace M :=
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-- Topology defined as LUB of pushforward.
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⨆ (f : A →ₗ[A] M), TopologicalSpace.coinduced f inferInstance
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-- just to make R and M explicit
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def isActionTopology (R M : Type*) [SMul R M]
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[TopologicalSpace R] [TopologicalSpace M] [IsActionTopology R M ]:=
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IsActionTopology.isActionTopology' (R := R) (M := M)
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def ActionTopology (R M : Type*) [SMul R M] [TopologicalSpace R] :
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TopologicalSpace M := ⨆ m, .coinduced (· • m) (inferInstance : TopologicalSpace R)
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namespace ActionTopology
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namespace type_stuff
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variable {R : Type} [τR : TopologicalSpace R]
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-- let `M` and `N` have an action of `R`
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variable {M : Type*} [SMul R M] [τM : TopologicalSpace M] [IsActionTopology R M]
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variable {N : Type*} [SMul R N] [τN : TopologicalSpace N] [IsActionTopology R N]
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example (L) [CompleteLattice L] (f : M → N) (g : N → L) : ⨆ m, g (f m) ≤ ⨆ n, g n := by
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exact iSup_comp_le g f
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theorem map_smul_pointfree (f : M →[R] N) (r : R) : (fun m ↦ f (r • m)) = fun m ↦ r • f m :=
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by ext; apply map_smul
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-- let `M` and `N` be `A`-modules
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variable (M : Type*) [AddCommGroup M] [Module A M]
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variable {N : Type*} [AddCommGroup N] [Module A N]
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/-- Every `A`-linear map between two `A`-modules with the canonical topology is continuous. -/
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lemma Module.continuous_linear (e : M →ₗ[A] N) :
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@Continuous M N (Module.rtopology A M) (Module.rtopology A N) e := by
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sorry -- maybe some appropriate analogue of Hannah/Lugwig's proof will work?
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lemma continuous_of_mulActionHom (e : M →[R] N) : Continuous e := by
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-- Let's turn this question into an inequality question about coinduced topologies
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rw [continuous_iff_coinduced_le]
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-- Now let's use the fact that τM and τN are action topologies (hence also coinduced)
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rw [isActionTopology R M, isActionTopology R N]
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-- There's an already-proven lemma in mathlib that says that coinducing an `iSup` is the
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-- same thing as taking the `iSup`s of the coinduced topologies
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-- composite of the coinduced topologies is just topology coinduced by the composite
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rw [coinduced_iSup]
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simp_rw [coinduced_compose]
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-- restate the current state of the question with better variables
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change ⨆ m, TopologicalSpace.coinduced (fun r ↦ e (r • m)) τR ≤
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⨆ n, TopologicalSpace.coinduced (fun x ↦ x • n) τR
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-- use the fact that `e (r • m) = r • (e m)`
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simp_rw [map_smul]
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-- and now the goal follows from the fact that the sup over a small set is ≤ the sup
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-- over a big set
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apply iSup_comp_le (_ : N → TopologicalSpace N)
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-- A formal corollary should be that
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def Module.homeomorphism_equiv (e : M ≃ₗ[A] N) :
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-- lean needs to be told the topologies explicitly in the statement
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let τM := Module.rtopology A M
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let τN := Module.rtopology A N
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M ≃ₜ N :=
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-- And also at the point where lean puts the structure together, unfortunately
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let τM := Module.rtopology A M
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let τN := Module.rtopology A N
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-- all the sorries should be formal.
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{ toFun := e
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invFun := e.symm
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left_inv := sorry
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right_inv := sorry
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def homeomorph_of_mulActionEquiv (φ : M →[R] N) (ψ : N →[R] M) (h1 : ∀ m, ψ (φ m) = m)
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(h2 : ∀ n, φ (ψ n) = n) : M ≃ₜ N :=
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{ toFun := φ
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invFun := ψ
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left_inv := h1
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right_inv := h2
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continuous_toFun := sorry
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continuous_invFun := sorry
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}
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-- this is definitely true
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lemma unit : (inferInstance : TopologicalSpace Unit) = ActionTopology R Unit := by
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sorry
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-- claim: topology on the 1-point set is the canonical one
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example : (inferInstance : TopologicalSpace Unit) = Module.rtopology A Unit := by
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-- is this true?
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lemma prod : (inferInstance : TopologicalSpace (M × N)) = ActionTopology R (M × N) := by
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sorry
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-- Anything from this point on *might* need commutative.
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-- Just move it to the commutative section and prove it there.
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-- is this true? If not, is it true if ι is finite?
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lemma pi {ι : Type*} {M : ι → Type*} [∀ i, SMul R (M i)] [τM : ∀ i, TopologicalSpace (M i)]
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[∀ i, IsActionTopology R (M i)] :
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(inferInstance : TopologicalSpace (∀ i, M i)) = ActionTopology R (∀ i, M i) := by
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sorry
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-- Claim: topology on A doesn't change
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example : (inferInstance : TopologicalSpace A) = Module.rtopology A A := by
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-- is this true? If not, is it true if ι is finite?
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lemma sigma {ι : Type*} {M : ι → Type*} [∀ i, SMul R (M i)] [τM : ∀ i, TopologicalSpace (M i)]
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[∀ i, IsActionTopology R (M i)] :
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(inferInstance : TopologicalSpace (Σ i, M i)) = ActionTopology R (Σ i, M i) := by
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sorry
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example :
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let _τM : TopologicalSpace M := Module.rtopology A M
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let _τN : TopologicalSpace N := Module.rtopology A N
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(inferInstance : TopologicalSpace (M × N)) = Module.rtopology A (M × N) := by sorry
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end type_stuff
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section R_is_M_stuff
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example :
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let _τM : TopologicalSpace M := Module.rtopology A M
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let _τN : TopologicalSpace N := Module.rtopology A N
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(inferInstance : TopologicalSpace (M × N)) = Module.rtopology A (M × N) := by sorry
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variable {R : Type} [τR : TopologicalSpace R] [Monoid R]
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example (ι : Type*) :
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let _τM : TopologicalSpace M := Module.rtopology A M
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(inferInstance : TopologicalSpace (ι → M)) = Module.rtopology A (ι → M) := by sorry
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-- is this true? Do we need that R is commutative? Does it work if R is a only semigroup?
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example : (inferInstance : TopologicalSpace R) = ActionTopology R R := by
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sorry
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end noncommutative
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end R_is_M_stuff
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section commutative
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section what_i_want
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-- Let A be a commutative ring, with a compatible topology.
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variable (A : Type*) [CommRing A] [TopologicalSpace A] [TopologicalRing A]
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-- let `M` and `N` be `A`-modules
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variable (M : Type*) [AddCommGroup M] [Module A M]
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variable {N : Type*} [AddCommGroup N] [Module A N]
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variable (M : Type*) [AddCommGroup M] [Module A M] [TopologicalSpace M] [TopologicalAddGroup M] [IsActionTopology A M]
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variable {N : Type*} [AddCommGroup N] [Module A N] [TopologicalSpace N] [TopologicalAddGroup N] [IsActionTopology A N]
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-- Remark: A needs to be commutative so that I get an A-action on M ⊗ N.
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-- What generality is this true in? I only need it when M and N are finite and free
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open scoped TensorProduct
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lemma Module.continuous_bilinear :
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let _τM : TopologicalSpace M := Module.rtopology A _
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let _τN : TopologicalSpace N := Module.rtopology A _
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let _τMN : TopologicalSpace (M ⊗[A] N) := Module.rtopology A _
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lemma continuous_of_bilinear :
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letI : TopologicalSpace (M ⊗[A] N) := ActionTopology A (M ⊗[A] N)
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Continuous (fun (mn : M × N) ↦ mn.1 ⊗ₜ mn.2 : M × N → M ⊗[A] N) := by sorry
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-- Now say we have a (not necessarily commutative) `A`-algebra `D` which is free of finite type.
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-- are all these assumptions needed?
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variable (D : Type*) [Ring D] [Algebra A D] [Module.Finite A D] [Module.Free A D]
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instance Module.topologicalRing : @TopologicalRing D (Module.rtopology A D) _ :=
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let _ : TopologicalSpace D := Module.rtopology A D
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{
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continuous_add := by
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sorry
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continuous_mul := by
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sorry
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continuous_neg := by
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sorry }
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end commutative
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variable (D : Type*) [Ring D] [Algebra A D] [Module.Finite A D] [Module.Free A D] [TopologicalSpace D] [IsActionTopology A D]
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theorem needed_for_FLT : TopologicalRing D where
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continuous_add := sorry
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continuous_mul := sorry
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continuous_neg := sorry
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end what_i_want
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-- possible hints at https://github.com/ImperialCollegeLondon/FLT/blob/main/FLT/HIMExperiments/module_topology.lean
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-- except there the topology is *different* so the work does not apply

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