⚠️ This module is not part of this project and is only included for reference.
It is either part of the 1lab, the cubical library, or a built-in Agda module.
<!-- ```agda open import 1Lab.Path.Reasoning open import 1Lab.Prelude open import Algebra.Group.Cat.FinitelyComplete open import Algebra.Group.Cat.Base open import Algebra.Group.Homotopy open import Algebra.Group.Ab open import Algebra.Group open import Data.Set.Truncation open import Data.Int.Universal open import Data.Bool open import Data.Int ``` --> ```agda module Homotopy.Space.Circle where ``` # Spaces: The circle {defines="circle"} The first example of nontrivial space one typically encounters when studying synthetic homotopy theory is the circle: it is, in a sense, the perfect space to start with, having exactly one nontrivial path space, which is a [[free group|free group construction]], and the simplest nontrivial [[free group|free group construction]] at that: the integers. ```agda data S¹ : Type where base : S¹ loop : base ≡ base S¹∙ : Type∙ lzero S¹∙ = S¹ , base ``` Diagrammatically, we can picture the circle as being the $\infty$-groupoid generated by the following diagram: ~~~{.quiver} \begin{tikzpicture} \node[draw,circle,label=below:{$\rm{base}$},fill,outer sep=0.1cm, inner sep=0pt, minimum size=0.1cm] (a0) at (0, -1) {}; \draw[->] (0, 0) circle (1cm); \node[] (loop) at (0, 0) {$\rm{loop}\ i$}; \end{tikzpicture} ~~~ In type theory with K, the circle is exactly the same type as `⊤`{.Agda}. However, with `univalence`{.Agda ident=ua}, it can be shown that the circle has at least two different paths: <!-- ``` _ = ⊤ ``` --> ```agda möbius : S¹ → Type möbius base = Bool möbius (loop i) = ua (not , not-is-equiv) i ``` When pattern matching on the circle, we are asked to provide a basepoint `b` and a path `l : b ≡ b`, as can be seen in the definition above. To make it clearer, we can also define a recursion principle: ```agda S¹-rec : ∀ {ℓ} {A : Type ℓ} (b : A) (l : b ≡ b) → S¹ → A S¹-rec b l base = b S¹-rec b l (loop i) = l i ``` <!-- ```agda S¹-elim : ∀ {ℓ} {A : S¹ → Type ℓ} (b : A base) (l : PathP (λ i → A (loop i)) b b) → ∀ s → A s S¹-elim b l base = b S¹-elim b l (loop i) = l i ``` --> We call the map `möbius`{.Agda} a _double cover_ of the circle, since the fibre at each point is a discrete space with two elements. It has an action by the fundamental group of the circle, which has the effect of negating the "parity" of the path. In fact, we can use `möbius`{.Agda} to show that `loop`{.Agda} is not `refl`{.Agda}: ```agda parity : base ≡ base → Bool parity l = subst möbius l true _ : parity refl ≡ true _ = refl _ : parity loop ≡ false _ = refl refl≠loop : ¬ refl ≡ loop refl≠loop path = true≠false (ap parity path) ``` The circle is also useful as a source of counterexamples: we can prove that there is an inhabitant of `(x : S¹) → x ≡ x`{.Agda} which is not constantly `refl`{.Agda}. ```agda always-loop : (x : S¹) → x ≡ x always-loop = S¹-elim loop (double-connection loop loop) ``` ## Fundamental group {defines="loop-space-of-the-circle"} We now calculate the loop space of the circle, relative to an _arbitrary_ implementation of the integers: any type that satisfies their type-theoretic universal property. We call this a _HoTT take_: the integers are the homotopy-initial space with a point and an automorphism. What we'd like to do is prove that the loop space of the circle is _also_ an implementation of the integers, but it's non-trivial to show that it is a set _directly_. ```agda module S¹Path {ℤ} (univ : Integers ℤ) where open Integers univ ``` We start by defining a type family that we'll later find out is the _universal cover_ of the circle. For now, it serves a humbler purpose: A value in $\rm{Cover}(x)$ gives a concrete representation to a path $\rm{base} \equiv x$ in the circle. We want to show that $(\rm{base} \equiv \rm{base}) \simeq \bb{Z}$, so we set the fibre over the basepoint to be the integers: ```agda Cover : S¹ → Type Cover base = ℤ Cover (loop i) = ua rotate i ``` We can define a function from paths $\rm{base} \equiv \rm{base}$ to integers --- or, more precisely, a function from $\rm{base} \equiv x$ to $\rm{Cover}(x)$. Note that the path $\rm{base} \equiv x$ induces a path $\bb{Z} \equiv \rm{Cover}(x)$, and since $\bb{Z}$ is equipped with a choice of point, then is $\rm{Cover}(x)$. ```agda encode : ∀ x → base ≡ x → Cover x encode x p = subst Cover p point ``` Let us now define the inverse function: one from integer to paths. By the mapping-out property of the integers, we must give an equivalence from $(\rm{base} \equiv \rm{base})$ to itself. Since $(\rm{base} \equiv \rm{base})$ is a group, any element $e$ induces such an equivalence, by postcomposition $x \mapsto x \cdot e$. We take $e$ to be the generating non-trivial path, $e = \rm{loop}$. ```agda loopⁿ : ℤ → base ≡ base loopⁿ n = map-out refl (∙-post-equiv loop) n loopⁿ⁺¹ : (n : ℤ) → loopⁿ (rotate .fst n) ≡ loopⁿ n ∙ loop loopⁿ⁺¹ n = map-out-rotate _ _ _ ``` To prove that the map $n \mapsto \rm{loop}^n$ is an equivalence, we shall need to extend it to a map defined fibrewise on the cover. We shall do so cubically, i.e., by directly pattern-matching on the argument $x$. When we're at the base point, we already know what we want to do: we _have_ a function $\rm{loop}^{-} : \bb{Z} \to (\rm{base} \equiv \rm{base})$ already, so we can use that. ```agda decode : ∀ x → Cover x → base ≡ x decode base = loopⁿ ``` For the other case, we must provide a path $\rm{loop}^{-} \equiv \rm{loop}^{-}$ laying over the loop, which we can picture as the boundary of a square. Namely, ~~~{.quiver} \[\begin{tikzcd} {\rm{base}} && {\rm{base}} \\ \\ {\rm{base}} && {\rm{base}} \arrow["{\rm{loop}^n}", from=1-3, to=3-3] \arrow["{\rm{loop}}"', from=3-1, to=3-3] \arrow["{\refl}", from=1-1, to=1-3] \arrow["{\rm{loop}^n}"', from=1-1, to=3-1] \end{tikzcd}\] ~~~ While it doesn't *look* like this square commutes, note that $n$ is an inhabitant of `ua rotate i`{.Agda}, so that its values on the endpoints of `i`{.Agda} are off by one. If we `unglue`{.Agda} $n$, we get an integer whose incarnation at `i = i0`{.Agda} is $1 + n$ (adjusting for the rotation offset) while at `i = i1`{.Agda} it is just $n$. We can provide such a square as sketching out an open _cube_ whose missing face has the boundary above. Here's such a cube: the missing face has a dotted boundary. ~~~{.quiver .tall-2} \[\begin{tikzcd} {\rm{base}} &&&& {\rm{base}} \\ & {\rm{base}} && {\rm{base}} \\ \\ & {\rm{base}} && {\rm{base}} \\ {\rm{base}} &&&& {\rm{base}} \arrow["{\rm{refl}}", from=2-2, to=2-4] \arrow[""{name=0, anchor=center, inner sep=0}, "{\rm{loop}^{1+n}}"', from=2-2, to=4-2] \arrow["{\rm{refl}}"', from=4-2, to=4-4] \arrow[""{name=1, anchor=center, inner sep=0}, "{\rm{loop}^n}", from=2-4, to=4-4] \arrow["{\rm{refl}}"{description}, dashed, from=1-1, to=1-5] \arrow["{\rm{loop}^n}"', dashed, from=1-1, to=5-1] \arrow["{\rm{loop}}"{description}, dashed, from=5-1, to=5-5] \arrow["{\rm{loop}^n}", dashed, from=1-5, to=5-5] \arrow["{\rm{refl}}"{pos=0.2}, from=2-2, to=1-1] \arrow["{\rm{refl}}"'{pos=0.3}, from=2-4, to=1-5] \arrow["{\rm{loop}^{-1}}"'{pos=0.1}, from=4-2, to=5-1] \arrow["{\rm{refl}}"{pos=0.2}, from=4-4, to=5-5] \arrow["{\rm{loop}^{\rm{unglue}(n)}}"{marking, allow upside down}, draw=none, from=0, to=1] \end{tikzcd}\] ~~~ ```agda decode (loop i) n j = hcomp (∂ i ∨ ∂ j) λ where k (k = i0) → loopⁿ (unglue (∂ i) n) j k (i = i0) → ∙→square (loopⁿ⁺¹ n) (~ k) j k (i = i1) → loopⁿ n j k (j = i0) → base k (j = i1) → loop (i ∨ ~ k) ``` We understand this as a particularly annoying commutative diagram. For example, the left face expresses the equation $\mathrm{loop}^n \bullet \mathrm{loop} = \mathrm{loop}^{1+n}$. The proof is now straightforward to wrap up: ```agda encode-decode : ∀ x (p : base ≡ x) → decode x (encode x p) ≡ p encode-decode _ = J (λ x p → decode x (encode x p) ≡ p) $ ap loopⁿ (transport-refl point) ∙ map-out-point _ _ encode-loopⁿ : (n : ℤ) → encode base (loopⁿ n) ≡ n encode-loopⁿ n = p ∙ ℤ-η n where p : encode base (loopⁿ n) ≡ map-out point rotate n p = map-out-unique (encode base ∘ loopⁿ) (ap (encode base) (map-out-point _ _) ∙ transport-refl point) (λ x → ap (encode base) {y = loopⁿ x ∙ loop} (map-out-rotate _ _ _) ·· subst-∙ Cover (loopⁿ x) loop point ·· uaβ rotate (subst Cover (loopⁿ x) point)) n ΩS¹≃integers : (base ≡ base) ≃ ℤ ΩS¹≃integers = Iso→Equiv $ encode base , iso loopⁿ encode-loopⁿ (encode-decode base) open S¹Path Int-integers public ``` It immediately follows from this that the circle is a [[groupoid]], since its loop space is a set. ```agda opaque S¹-is-groupoid : is-groupoid S¹ S¹-is-groupoid = S¹-elim (S¹-elim (is-hlevel≃ 2 ΩS¹≃integers (hlevel 2)) prop!) prop! ``` <!-- ```agda instance H-Level-S¹ : ∀ {k} → H-Level S¹ (3 + k) H-Level-S¹ = basic-instance 3 S¹-is-groupoid ``` --> By induction, we can show that this equivalence respects group composition (that is, $\rm{loop}^{a + b} \equiv \rm{loop}^a \bullet \rm{loop}^b$), so that we have a proper isomorphism of groups. ```agda loopⁿ-+ : (a b : Int) → loopⁿ (a +ℤ b) ≡ loopⁿ a ∙ loopⁿ b loopⁿ-+ a = Integers.induction Int-integers (ap loopⁿ (+ℤ-zeror a) ∙ sym (∙-idr _)) λ b → loopⁿ (a +ℤ b) ≡ loopⁿ a ∙ loopⁿ b ≃⟨ ap (_∙ loop) , equiv→cancellable (∙-post-equiv loop .snd) ⟩ loopⁿ (a +ℤ b) ∙ loop ≡ (loopⁿ a ∙ loopⁿ b) ∙ loop ≃⟨ ∙-post-equiv (sym (∙-assoc _ _ _)) ⟩ loopⁿ (a +ℤ b) ∙ loop ≡ loopⁿ a ∙ loopⁿ b ∙ loop ≃⟨ ∙-post-equiv (ap (loopⁿ a ∙_) (sym (loopⁿ⁺¹ b))) ⟩ loopⁿ (a +ℤ b) ∙ loop ≡ loopⁿ a ∙ loopⁿ (sucℤ b) ≃⟨ ∙-pre-equiv (loopⁿ⁺¹ (a +ℤ b)) ⟩ loopⁿ (sucℤ (a +ℤ b)) ≡ loopⁿ a ∙ loopⁿ (sucℤ b) ≃⟨ ∙-pre-equiv (ap loopⁿ (+ℤ-sucr a b)) ⟩ loopⁿ (a +ℤ sucℤ b) ≡ loopⁿ a ∙ loopⁿ (sucℤ b) ≃∎ π₁S¹≡ℤ : π₁Groupoid.π₁ S¹∙ S¹-is-groupoid ≡ ℤ π₁S¹≡ℤ = sym $ ∫-Path Groups-equational (total-hom (Equiv.from ΩS¹≃integers ∘ Lift.lower) (record { pres-⋆ = λ (lift a) (lift b) → loopⁿ-+ a b })) (∙-is-equiv (Lift-≃ .snd) ((ΩS¹≃integers e⁻¹) .snd)) ``` Furthermore, since the loop space of the circle is a set, we automatically get that all of its higher homotopy groups are trivial. ```agda Ωⁿ⁺²S¹-is-contr : ∀ n → is-contr ⌞ Ωⁿ (2 + n) S¹∙ ⌟ Ωⁿ⁺²S¹-is-contr zero = is-prop∙→is-contr (hlevel 1) refl Ωⁿ⁺²S¹-is-contr (suc n) = Path-is-hlevel 0 (Ωⁿ⁺²S¹-is-contr n) πₙ₊₂S¹≡0 : ∀ n → πₙ₊₁ (suc n) S¹∙ ≡ Zero-group {lzero} πₙ₊₂S¹≡0 n = ∫-Path Groups-equational (Zero-group-is-terminal _ .centre) (is-contr→≃ (is-contr→∥-∥₀-is-contr (Ωⁿ⁺²S¹-is-contr n)) (hlevel 0) .snd) ```