Configuration Spaces in Algebraic Topology

Home , Homotopy, Lens space

Harvard Math 252x

Conﬁguration Spaces in Algebraic Topology Taught by Ben Knudsen

Notes by Reuben Stern

September 4, 2017

Contents

0.1 Preliminaries ...... 1

1 Wednesday, August 301 1.1 A bunch of examples...... 3

2 Friday, September 13

0.1 Preliminaries This is Math 252x, configuration spaces in algebraic topology, taught by Ben Knudsen. The goal is to understand the homotopy and cohomology of configuration spaces. Office hours will be on Thursdays from 1-3 pm.

1 Wednesday, August 30

Deﬁnition 1.1. The configuration space of k ordered points in the topological space X is

k Confk(X) = {(x1, . . . , xn) ∈ X : xi 6= xj if i 6= j}.

The unordered conﬁguration space is

Bk(X) := Confk(X)/Σk.

m ∼ n Aside 1.2. As we’ll see, B2(R ) = B2(R ) if and only if n = m, so conﬁguration spaces are

1 sensitive to the dimension of manifolds. A homology computation shows

2 2 0 B2(T \ {∗}) 6' B2(R \ S ), even though the two have the same dimension and share S1 ∨ S1 as a common deformation retract.

Example 1.3. The (p, q)-lens space is the quotient

3 L(p, q) := S /Cp,

3 2 2πi/p 2πiq/p where Cp acts on S ⊆ C by multiplication by (e , e ).

2 Theorem 1.4 (Reidemeister). L(p, q1) ' L(p, q2) if and only if q1q2 ≡ ±n mod p, and ∼ ±1 L(p, q1) = L(p, q2) if and only if q1 ≡ ±q2 mod p. For instance, L(7, 1) and L(7, 2) are homotopy equivalent but not homeomorphic.

Theorem 1.5 (Longoni-Salvatore). B2(L(7, 1)) 6' B2(L(7, 2)). So there are two compact manifolds that have the same homotopy type but diﬀerent conﬁgura- tion spaces.

Question 1.6. How much do conﬁguration spaces know of the background manifold?

Perspective 1.7. Let’s consider braids: they can wind around each other, but cannot intersect. 2 It follows that π1(Bk(R )) is the braid group on k strands. Artin showed that

2 ∼ π1(Bk(R )) = hσ1, . . . , σk−1|σiσj = σjσi if |i − j| > 1; σiσj + ...i

Braid-like groups are fundamentally important in knot theory, mapping class groups, quantum groups, category theory (braided monoidal categories), and motion planning and robotics.

Perspective 1.8. (Embeddings). There is a space of framed embeddings of a manifold with a fr ` n n n ﬁxed trivialization of the tangent bundle, i.e., Emb ( k R , R ), which maps to Confk(R ), and in fact is a homotopy equivalence. Since embeddings can be composed, the collection

n {Confk(R )}k≥0 of homotopy types has a hidden algebraic structure of an En-operad.

There are a lot of things to say now:

1. Iterated loop spaces a la May. It turns out that ! a Embfr Rn, Rn × (ΩnX)k −→ ΩnX k

n tells us that Ω X is an En-algebra. This has two consequences:

n n • H∗(Confk(R )) provides operations in H∗(Ω X). • From the recognition theorem, there is an algebraic classiﬁcation of n-fold loop spaces.

2 2. Factorization homology (Lurie, Ayala-Francis), also known as topological chiral homology. R R Suppose I have an En-algebra A; this gives a manifold invariant M A. Roughly, M A is a space of conﬁgurations of points in M, labeled by A.

R n • M Ω X ' Mapc(M,X) if X is (n − 1)-connected (called non-abelian Poincaré duality, due to Lurie and Salvatore). • Applications to conﬁguration spaces. 3. Embedding calculus (Goodwillie, Weiss). Suppose you want to understand Emb(M,N). We get a tower of approximations

. .

T2Emb(M,N)

Emb(M,N) T1Emb(M,N).

1.1 A bunch of examples

Example 1.9. Take a look at the empty manifold. Bk(∅) = Confk(∅) ' {∗} if k = 0, and it is 0 empty otherwise. This follows from the convention ∅ = {∗}. Example 1.10. Now look at R0.  0 {∗} k = 0, 1 Confk(R ) = ∅ k > 1.

Example 1.11. R1. The conﬁguration space of 2 points is just what you think it would look like: 1 ∼ 2 the disjoint union of the interiors of two 2-simplices. And B2(R ) = ∆ is contractible. Now what if k > 2? We have a map π0Confk((0, 1)) → Σk, which is actually a bijection. Each component is homeomorphic to Bk((0, 1)). There is a map

k Bk((0, 1)) −→ ∆ taking (x1, . . . , xk) to (x1, x2 − x1,..., 1 − xk). This is a homeomorphism. 1 1 2 1 Example 1.12. Now consider S . Conf2(S ) = T \ S , which is a cylinder, so is homotopy 1 1 equivalent to S . Similarly, the unordered conﬁguration space B2(S ) is the mobius band. n n n−1 n Example 1.13. Now look at R . Conf2(R ) maps to S ×R>0×R , which is a homeomorphism. So the (x , x ) 7→ x2−x1 is a homotopy equivalence. Gauss map 1 2 kx2−x1k

2 Friday, September 1

Recall the deﬁnition of a conﬁguration space from last time:

3 Deﬁnition 2.1. Let X be a topological space. The (ordered) configuration space of k k points in X is the set Confk(X) = {(x1, . . . , xk) ∈ X : xi = xj iﬀ i = j}. The unordered configuration space Bk(X) is the quotient of Confk(X) by the natural action of the symmetric group Σk.

One thing we can ask is about functoriality of this construction: is Confk(X) functorial? It is to a certain extent: if f : X → Y is an injective map of topological spaces, then there is a map Confk(f) ﬁlling in the dashed arrow in the commutative diagram

f k Xk Y k ⊆ ⊆

Confk(X) Confk(Y ), Confk(f) given by applying f coordinate-wise. It is clear that this induced map is Σk-equivariant, so it descends to a map Bk(f): Bk(X) → Bk(Y ). Thus we have two functors:

inj Σk Confk(−):Top −→ Top inj Bk(−):Top −→ Top.

Here, Topinj is the category of topological spaces and injective maps, and TopΣk is the category of topological spaces with a Σk-action, and Σk-equivariant maps.

Proposition 2.2. If f : X → Y is an open embedding, then so are Confk(f) and Bk(f).

k k Proof. For Confk(f), note that f is an open embedding, and Confk(X) is open in X . For Bk(f), the natural map Confk(X) → Bk(X) is a quotient map, and thus an open embedding. Open embeddings compose.

Deﬁnition 2.3. An injective, continuous map f : X → Y is called an isotopy equivalence if there is an injective, continuous map g : Y → X and homotopies

0 H : f ◦ g ⇒ 1Y and H : g ◦ f ⇒ 1X

0 such that Ht and Ht are injective for all t ∈ [0, 1]. Basically, it is a “homotopy equivalence through embeddings”.

Proposition 2.4. If f : X → Y is an isotopy equivalence, the induced maps Confk(f) and Bk(f) on conﬁguration spaces are homotopy equivalences.

Proof. Consider the diagram

(H0)k (X × I)k Xk

H∆ Xk × Ik i

i×∆

Confk(X) × I Confk(X)

4 ∆ k ∆ 0 0 The composite map H : Confk(X) × I → X acts by H (x1, . . . , xk, t) = (Ht(x1),...,Ht(xk)). 0 k Because Ht is injective, this map pulls back along the inclusion i : Confk(X) → X to give a well-deﬁned homotopy Confk(X) × I → Confk(X).

Example 2.5. Let M be a manifold with nonempty boundary. Then ∂M has a collar neighborhood U =∼ ∂M × (0, 1]. By retracting along the (0, 1] coordinate, the inclusion M,˚ → M is an isotopy equivalence, so ' Confk(M˚) Confk(M).

Notation 2.6. Fix a basis B of the topology of X. Deﬁne two partially-ordered sets Bk and Σ Bk as follows:

( open k ) ∼ G 1. Bk = Uk ⊆ X : U = Ui,Ui ∈ B . We order this by saying U ≤ V if U ⊆ V and the i=1 inclusion i : U,→ V induces a bijection on π0.

Σ n ' o 2. Bk = (U, σ): U ∈ Bk, σ : {1, . . . , k} −−→ π0(U) . This is ordered by saying (U, σ) ≤ (V, τ) if U ≤ V in Bk and i ◦ σ = τ, where i : U → V is the inclusion.

Now let O(X) denote the poset of open subsets of X. We will deﬁne two functions,

◦ Bk :Bk ,−→O(Bk(X)) and ◦ Σ Confk :Bk ,−→O(Confk(X)) as follows:

◦ Bk(U) = {x ∈ Bk(U): x intersects each component of U nontrivially} ◦ n o Confk(U, σ) = (x1, . . . , xk) ∈ Confk(U): xi ∈ Uσ(i) .

k ◦ ∼ Y ∼ ◦ Lemma 2.7. There are homeomorphisms Confk(U, σ) = Uσ(i) = Bk(U). i=1 Proof. The proof starts by considering the diagram

k X Confk(X) Bk(X)

k Y ◦ ◦ Uσ(i) Confk(U, σ) Bk(U). i=1

The dashed arrows can easily be seen to be ﬁlled in. ﬁnish this proof Σ Proposition 2.8. The set Bk is a basis for the topology of Confk(X). The set Bk is a basis for the topology of Bk.

Proof. Finish later.

5 Corollary 2.9. The projection π : Confk(X) → Bk(X) is a covering space map.

Proof.

−1 ◦ ∼ G ◦ π (Bk(U)) = Confk(U, σ) σ ∼ ◦ = Bk(U) × Σk.

Corollary 2.10.

k G =∼ Confi(X) × Confk−i(Y ) ×Σi×Σk−i ΣK Confk(X t Y ). i=0

In particular, we have a homeomorphism

k G =∼ Bi(X) × Bk−i(Y ) Bk(X t Y ). i=0

Now, let ` ≥ k be integers. The projection map X` → Xk onto the ﬁrst k coordinates induces a dashed map on conﬁguration spaces:

` Conf`(X) X (x1, . . . , x`)

k Confk(X) X (x1, . . . , xk).

The ﬁber over (x1, . . . , xk) ∈ Confk(X) is the conﬁguration space Conf`−k(X − {x1, . . . , xk}).

Reminder 2.11. Let f : A → B be a continuous map. Recall that the mapping path space of I f is the homotopy pullback of f along the evaluation map ev0 : B → B:

I Ef B p y ev0

f A B p(0)

There is a homotopy equivalence A → Ef , and the “evaluation at 1” map Ef → B is a ﬁbration. The homotopy fiber of f is the ﬁber of ev1 : Ef → B. A square A A0 f f 0 B B0

6 is said to be homotopy cartesian if the induced map hoﬁb(f) → hoﬁb(f 0) is a weak equivalence.

Now let M be a manifold.

Theorem 2.12 (Fadell–Neuwirth). The square

Conf`−k(M − {x1, . . . , xk}) Conf`(M)

(x1, . . . , xk) Confk(M) is homotopy cartesian.

Corollary 2.13. If M is a simply-connected n manifold with n ≥ 3, then Confk(M) is simply- ∼ connected for all k. In particular, π1(Bk(M)) = Σk.

Proof. Associated to the ﬁbration from Fadell–Neuwirth (and taking ` = k, k = k − 1), we have a long exact sequence in homotopy