INSTITUTE FOR BASIC SCIENCE CENTER FOR GEOMETRY

AND PHYSICS SYMPLECTIC ALGEBRAIC TOPOLOGY OH YONG-GEUN

GABRIEL C. DRUMMOND-COLE

1. September 10, 2013

Okay, I’m going to give a lecture in English. The grade will be based on pass or fail, but as long as you do a few homeworks, that will be fine.

I’m going to talk about some algebraic structures that appear in the topology of symplectic manifolds. That algebraic structure is quite complex, so it requires more complicated homological algebra than classical algebraic topology. I’m not going to talk very much about symplectic manifolds. I’ll talk more about homological algebra of A_{∞} structures.

Let me start by motivating where this kind of structure arises. The story begins with Stasheff or maybe some earlier work, but Stasheff is the one who introduced the precise concept.

So this is Stasheff’s A_{n} structures. In the first few lectures I’m going to explain what these A_{n} structures are about. Everything begins with a peculiar structure in the based loop space. Let’s say X is a topological space and x_{0} some base point. You’re given some distinguished point. Let me denote, we have LX, the free loop space, paths from the unit interval to X such that the initial and final point coincide:

L(X) = {γ : [0, 1] → X|γ(0) = γ(1)}

Then I’ll denote by Ω(X) the subset of the free loop space such that the loops start and end at the basepoint.

Ω(X) = {γ ∈ L(X)|γ(0) = x_{0}}

This is the “based loop space.” It has a product. You can concatenate two loops.

α  •  β

The product ∗ : ΩX × ΩX → ΩX works as follows:

α ∗ β = \{ \begin{matrix}α(2t) & 0 ≤ t ≤ \frac{1}{2} \\ β(2t − 1) & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

This is not commutative. If you do this with the other order, you will not get the same loop.

If we have a third loop γ, we could compare (α∗β)∗γ, that’s one way, that gives

= \{ \begin{matrix}(α ∗ β)(2t) & 0 ≤ t ≤ \frac{1}{2} \\ γ(2t − 1) & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

and this unravels as

= \{ \begin{matrix}α(4t) & 0 ≤ t ≤ \frac{1}{4} \\ β(4t − 1) & \frac{1}{4} ≤ t ≤ \frac{1}{2} \\ γ(2t − 1) & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

I divided the unit interval in half and then the left side in half again.

Now we consider α ∗ (β ∗ γ). We get

= \{ \begin{matrix}α(2t) & 0 ≤ t ≤ \frac{1}{2} \\ β(4(t − \frac{1}{2})) & \frac{1}{2} ≤ t ≤ \frac{3}{4} \\ γ(4(t − \frac{3}{4})) & \frac{3}{4} ≤ t ≤ 1 \end{matrix}

in general, (α ∗ β) ∗ γ ≠ α ∗ (β ∗ γ) as an element in ΩX but thy ar homotopic and the homotopy can be given by an explicit reparameterization. That is, there exists a homotopy

Γ : \underbrace{[0, 1]}_{s} × \underbrace{[0, 1]}_{t} → X

such that Γ(0, ·) ≡ (α ∗ β) ∗ γ and Γ(1, ·) = α ∗ (β ∗ γ).

So this is a path in the loop space, and the path is given by connecting the speeds

like this  ^{s s = 1 s = 0  t}  We have s=0 at the top.

Your first homework is to find the explicit formula for Γ(s, ·). The middle interval of s is given by \frac{1}{4} + \frac{s}{4} ≤ t ≤ \frac{1}{2} + \frac{s}{4}, as a hint.

You can keep doing this with many loops and of course there are many different ways of taking the product.

This homotopy depends on α, β, and γ and the parameter s. We formalize the explicit definition of Γ, this homotopy, and define m_{3} : ΩX×ΩX×ΩX×[0, 1] → ΩX by m_{3}(α, β, γ, s) = Γ(s, ·).

Similarly we can try to construct a homotopy of homotopies. Consider four loops α, β, γ, δ, and consider their quadruple product α ∗ β ∗ γ ∗ δ, and there are many ways of doing this, to find a parameterized based loop, you need to use some parentheses, for example

(1) ((α ∗ β) ∗ γ) ∗ δ
(2) (α ∗ (β ∗ γ)) ∗ δ
(3) α ∗ ((β ∗ γ) ∗ δ)
(4) α ∗ (β ∗ (γ ∗ δ))
(5) (α ∗ β) ∗ (γ ∗ δ)

We can see that these each share one set of parentheses with the next one and we can draw the so-called pentagon. I’ll put each of these quadruple products as a vertex of the pentagon.

(a(bc))d  a((bc)d)  ((ab)c)d  (a(b(cd))  (ab)(cd)

On each edge we can put the explicit homotopy induced by Γ.

So this defines a loop in Ω(X). Each point on an edge defines an element in the loop space, and this loop is actually contractible in the loop space. Again, this can be done with an explicit contraction to the barycenter of the pentagon. I’ll construct that later, but my claim is that we can construct an explicit piecewise linear construction.

This pentagon is an example of the Stasheff polytope K_{4}. So for example, K_{2} is a point, there’s only one way to multiply. K_{3} is the unit interval, which is where we defined the homotopy. K_{4} is the pentagon there, and generally, they keep going, and in general, K_{n} is an (n − 2)-dimensional polytope which will be constructed inductively by some general properties.

Then, it’s a fact that we have a continuous map m_{n} : (ΩX)^{n} × K_{n} → ΩX that satisfies certain (explicit) properties sometimes called A_{n} properties. All of this will be explained later. Such a space is called an A_{n}-space.

Then Stasheff asked the following question. What conditions on a topological space Y makes it homotopy equivalent to ΩX for some topological space X. It has a remarkable answer. Stasheff found:

Theorem 1.1. The answer to this question is exactly this A_{n} property. A topological space Y is homotopy equivalent to ΩX for some X if and only if Y is an A_{∞}-space (an A_{n}-space for all n).

This structure appeared in symplectic geometry. Well.

You can now apply a cohomological functor. Recall the loop space product ∗. Although it is not associative on the space itself, but it induces a coproduct on cohomology, H^{∗}(ΩX, ℝ) → H^{∗}(ΩX × ΩX, ℝ) ≅ H^{∗}(ΩX) ⊗ H^{∗}(ΩX), so that loop product induced this homomorphism, and this is an example of a so-called coproduct. I’ll explain what coproducts mean explicitly next time. That is coassociative, the dual notion of associative.

In other words, m_{2} induced a natural coproduct m_{2} : H^{∗}(ΩX) → H^{∗}(ΩX) ⊗ H^{∗}(ΩX).

Maybe it would be better to work with homology, where we can get a natural product m_{2} : H_{∗}(ΩX) ⊗ H_{∗}(ΩX). Let’s think about the degree. Let’s look at the induced homomorphism (m_{n})_{#} : (H_{∗}(ΩX))^{⊗n} → H_{∗}(ΩX). Consider p_{1}, …, p_{n}, singular chains in ΩX of dimension or degree d_{1}, …, d_{n}. Denote α_{i} = [p_{i}]. Then (m_{n})_{#}(α_{1}, …, α_{n}), to evaluate this you apply to representatives m_{n}(p_{1}, …, p_{n}), and you have an s which comes from K_{n}, and the dimension of K_{n} is n − 2, and so this gives us a map from something of dimension, there is a resulting chain, that dimension is d_{1} + ⋯ + d_{n} + n − 2. Maybe it’s easier to see this on chains. (m_{n})_{#}(p_{1}, …, p_{n}) = m_{n}(p_{1}, …, p_{n} : K_{n}), so recall that m_{n} is a map from (ΩX)^{n} × K_{n} → ΩX. The image of this m_{n} will increase the dimension by n−2. This defines a chain of dimension ∑ d_{j} + (n − 2). So this is sort of a chain map of degree n − 2. This (m_{n})_{#} is a map of degree n − 2. So you regard, well, C_{∗}(ΩX) as a graded vector space. So C_{∗}(ΩX) = ⨁ C(ΩX). This map satisfies, the A_{n} relations induce certain relations on these. The second fact is that the (m_{n})_{#} will satisfy certain quadratic relations called A_{∞} relations.

From the A_{∞} relations, we can derive many facts. It induces, for example, an associative product on H_{∗}(ΩX). This is called an A_{∞} algebra. The conclusion is that (H_{∗}(ΩX)) with m_{n} forms an A_{∞} algebra.

Maybe I can write the quadratic relations. Next time maybe I’ll explain all these unexplained terminologies. Maybe I’ll finish today’s lecture by saying one word, which is that such an A_{∞} algebra also occurs in symplectic geometry for the Floer cohomology of Lagrangian submanifolds.

2. September 12, 2013

Missed

3. September 24, 2013

Maybe I should briefly remind you what we did last time. Recall that we talked about ribbon trees and also looked at rooted ribbon trees and actually the way how we defined ribbon trees is as embedded, you draw this tree but we sort of regard some extra edges as lying on the disk. We call this a ribbon tree because you construct this fictitious disk by thickening the tree. You are given the tree and you can thicken this by gluing several, so, the thickening of trees, what I mean by this, you look at the edges, instead of edges you replace them with bands, and then you are given bands like this [picture] and then what you can do is you identify the thin boundaries by taking the barycenter and then identify the half bands and do the same thing, and you will get some kind of space which is homeomorphic to the disk. By this thickening process, you give the flat metric, and since these are flat, you can identify them, and you will get some singularities, you won’t have a flat metric at these barycenters. Then when you are given this metric, it canonically defines a conformal structure. The conformal structure uniquely extends over the isolated singular points. This is the way we realize this fictitious disk. That’s why we call these things ribbon trees, these are the ribbons.

Anyhow, the rooted ribbon trees, you identify one distinguished vertex v_{0} on the boundary, and all other exterior vertices are outgoing vertices and there is an automatic orientation on each edge toward v_{0}. We denoted the set of isotopy classes of stable (no vertices of valence two) ribbon trees (or rooted ribbon trees) by G_{n+1} (or G_{n,1}).

We gave a partial ordering, well, the ribbon tree is an embedding (T, i) of the tree T into the disk D^{2} and i^{−1}(∂D^{2}) is the exterior vertices of T. In the rooted case we have (T, v_{0}), i. Denote by [T, i] or [(T, v_{0}), i] the isotopy class of (T, i) or ((T, v_{0}), i).

We denote for simplicity t as [T, i] or [(T, v_{0}), i]. We say t < t′ if t′ is obtained by collapsing a sequence of interior edges of t. The maximal element, we call the unique maximal element a corolla. There are no interior edges. An n-corolla or n + 1-corolla or (n, 1)-corolla has no interior edges.

Then we introduced the grafting operation, which I denote as, let t_{1} be [(T, v_{0}), i] (or let me skip this i, assume it) and t_{2} = [(T′, v′)]. This operation ∗_{i} : G_{(n_{1},1)} × G_{(n_{2},1)} → G_{n_{1}+n_{2}−1,1} occurs by the following picture (this is a little confusing)

v′_{0}  v_{i}  v_{1} v_{0}

and you glue these together with some renumbering:

v_{0} v_{1} v_{i−1} v_{i}  v_{i+n_{1}−1} v_{n_{1}+n_{2}−1} v_{i+n_{1}−1}

We now go back to the axioms of the associahedron. We had two, so the third axiom is that K_{n} is a CW-complex. The set of cells in K_{n} is in one to one correspondence with G_{(n,1)}. Axiom four is as follows. Denote by F (t) the face associated to t ∈ G_{(n,1)}. Then F (t) is an open cell of codimension (# of internal edges). The corolla is the generic cell, and if the tree has more interior edges, then the associated cell has higher codimension. Note that this codimension is the same as the number of parentheses.

[Darko Milinkovic: What are the faces of a CW complex? Maybe I’m missing axioms one and two.] K_{n} is an n − 2 dimensional polyhedron. K_{n} has a cell decomposition where each cell is the cone on its boundary.

Axiom five is as follows, maybe this is a proposition, well, given the topology of K_{n}, we have the following: The closure \overline{F (t′)} is ⋃_{t≤t′} F (t) and the boundary is the same except with scrict inequality: ∂\overline{F (t′)} = ⋃_{t<t′} F (t).

There is one more axiom which I’ll state later. But maybe I need to state axiom six to make this proposition true so I’ll hold off on it.

Proposition 3.1. K_{n} is homeomorphic to an (n − 2)-cell.

I’ll give the proof of this proposition later, but I’ll give some examples. K_{2} is a point. There’s only one isotopy class.

[Some discussion of K_{1} and K_{0}] Let me give axiom six. There exist maps ∘_{i} : K_{r} × K_{s} → ∂K_{n} ⊂ K_{n} where n + 1 = r + s. so that the image ∘_{i}(K_{r} × K_{s}) ofrms a facet of K_{n}.

The boundaries of K_{n} is the union over 2 ≤ s ≤ n − 1 and i from 1 to s of the closures of ∘_{i}(K_{(}n + 1 − s) × K_{s}). These are all possible ways of putting one parenthesis in a word of length n so that it’s not the whole word.

That’s the axioms I’ll use to describe these things.

Let’s go back and consider ∂K_{3}. How many boundaries? This is a zero dimensional set. This will be, r + s = n + 1 = 4, and s must be 2. So there are two ways of putting one parentheses: ((ab)c) or (a(bc)). There should be two points. So let’s see what ((ab)c) corresponds to in the graph. They should be ∘_{1}(K_{2} × K_{2}) and ∘_{2}(K_{2} × K_{2}).

v_{1} v_{2}  v_{0}  v_{1}  v_{0} v_{2}

v_{1} v_{2}  v_{0} v_{3}

That’s one. The other is:

v_{1} v_{2}  v_{0}  v_{2}  v_{1} v_{0}

v_{2} v_{3}  v_{1} v_{0}

You can think of this as one of these turning ot the other by passing through a corolla:

Now I’ve done something axiomatic and I want to turn instead to some geometric realizations of K_{n}. One could be metric ribbon trees. Let me set some notation. For a given t = (T, i) ∈ G_{n+1}, we denote by C^{0}(t) as the set of vertices, C^{1}(t) as the set of edges, and C_{int}^{0}(t) and C_{int}^{1}(t) as the interior vertices and edges. Now for t in G_{n+1}, we associate the open cell

Gr(t) = {ℓ : C′_{int}(t) → (0, ∞)}

This is homeomorphic to (0, ∞)^{#C_{int}^{1}(t)} = (0, 1)^{#C_{int}^{1}(t)}. Then we let \overline{Gr(t)} to be the same thing including lengths zero and infinity:

{ℓ : C′_{int}(t) → [0, ∞]}

This is homeomorphic to [0, 1]^{#C_{int}^{1}(t)}

We can make a grafting map for this. We provide a CW-structure on the union of the isotopy classes of Gr(t). By the definition, Gr_{(n,1)} is the resulting complex, which is homeomorphic to K_{n}.

Okay, for the second one, the moduli space \overline{M}_{n+1}^{main}(D^{2}). Consider n + 1 distinct points on S^{1} = ∂D^{2}. Denote this by

Conf_{n+1}(S^{1}) = {(z_{0}, …, z_{n}) ∈ (S^{1})^{n+1}|z_{0}, …, z_{n} are distinct and arranged counterclockwise}.

Then P SL(2, ℝ) acts on S^{1}. Recall that SL(2, ℝ) is two by two matrices of determinant one and SL(2, ℝ) is P SL(2, ℝ) modulo the relation that a matrix is equal to its negation.

Identify D^{2}\{1} with the upper half-plane, z ∈ ℂ such that Im z ≥ 0. These are equivalent via a conformal diffeomorphism.

Then P SL_{2}(ℝ) acts on the upper half-plane by z ↦ \frac{az+b}{cz+d} if the element of

P SL(2, ℝ) is the class of ( \begin{matrix}a & b \\ c & d \end{matrix} ). This action acts the same by an element or

its negation and preserves the upper half plane and the real line union {∞}. For a given (z_{0}, …, z_{n}), we may assume z_{0} = ∞. Then we arrange z_{1} < z_{2} < · · ·.

So you can identify z_{1}, …, z_{n} around the circle. We can realize, or identify, these points, in Conf_{n+1}(S^{1}) as a subset, if you put z_{0} at 1 you still have two dimensions. You can identify the point as a subset of n points in (0, 1). I’ll make this more precise next time.

Definition 3.1. We define M_{n+1}(D^{2}) = Conf_{n+1}(S^{1})/P SL(2, ℝ) where P SL(2, ℝ) acts on (z_{0}, …, z_{n}) by (Az_{0}, …, Az_{n}).

We identify two points if all coordinates are moved by the same fractional linear transformation. This quotient space has a natural differential structure and the dimension is (n + 1) − 3 = n − 2. I’ll explain the cell structure of this which tells you this is another geometric realization of the Stasheff polytope after you compactify, next time I’ll talk about compactification.

4. September 26, 2013

Maybe just for those who are interested in looking at more details of what I’m talking about, here’s a reference, the original paper of Stasheff. This is very explicit and does not use any machinery for graphs or moduli space. It’s hard to motivate where these relations come from. This is the most basic reference: Homotopy Associativity of H-spaces, I and II, in the transactions of the AMS, 1963, volume

108. My presentation is more modern.

Axiom six is a little more involved, which I will explain using the realization of the Stasheff polytope. This is very much like the relationship between degeneracy and face maps for the simplex, so this is an analog for Stasheff polytopes.

There exist a family of maps called face operators. I already introduced these with another notation, but I want to make this more formal. This is ∂_{k}(r, s), but I used ∘_{k} before. I want to make every data clearly now. This is a map from K_{r} × K_{s} → K_{n}, and r + s = n + 1. So s is the only free parameter. n is given so r is determined. You should think of K_{n} as being like a simplex. For this case, there is a map, the face map, in all dimensions. This is the analog of the inclusion of low dimensional faces into the n-simplex. Here s runs from 2 to n − 1 and k ranges from 1 to s. This should properly be K_{s} × K_{r}. Maybe the parentheses way is better:

a_{1} ⋯ a_{k−1}(a_{k} ⋯ a_{k+r−1})a_{k+r} ⋯ a_{n}

I insert K_{r} into the kth spot of K_{s}. This is the spot where you attach the first tree to the second tree.

There are also degeneracy maps s_{j} : K_{n} → K_{n−1}. These range from j from 1 to n.

You have standard relations between the face maps and the degeneracies in the simplex. I’ll write down the relations here but they’re much more complicated.

i s_{j}s_{k} = s_{k}s_{j+1} for k ≤ j (I think this is the same as for the simplex). iia s_{j}∂_{k}(r, s) = ∂_{k−1}(r − 1, s) ∘ (s_{j} × id) for j < k and r > 2. Similarly: iia’ s_{j}∂_{k}(r, s) = ∂_{k}(r, s − 1) ∘ (1 × s_{j−k+1} for s > 2 and k ≤ j ≤ k + s. iib s_{j}∂_{j}(n − 1, 2) = s_{j+1}∂_{j}(n − 1, 2) = π_{j} for 1 < j < n; iib’ s_{1}∂_{2}(2, n − 1) = π_{2} = s_{n}∂_{1}(2, n − 1) iii s_{j}∂_{k}(r, s) = ∂_{k}(r − 1, s)(s_{j−s+1} × id) for k + s ≤ j

The geometric realization makes this very clear. The ribbon graph model makes this clear.

v_{1}  v_{n} v_{0}

The degeneracy map is mapping from Gr(n, 1) to Gr(n−1, 1), just by collapsing the kth exterior edge followed by stabilization (deleting bivalent edges). The face map ∂_{k}(r, s) : \overline{Gr}_{(r,1)} × \overline{Gr}_{(s,1)} → \overline{Gr}_{(n,1)}, which is nothing but the grafting map, and the image lies in the boundary. Once you know those topologies, the relations I wrote down are an immediate consequence of the compactification.

I forgot to mention one remark. By definition, the face operator gives a boundary map and you can see the boundary is

∂K_{n} = ⋃_{s=2}^{n−1} ⋃_{k=1}^{s} K_{r} ⋆_{k} K_{s}

where K_{r} ∗_{k} K_{s} := ∂_{k}(r, s)K_{r} × K_{s}.

The number of faces is \frac{n}{n−1}2 − 1. You can count this. The similar realization of the face operator, well, that’s even easier, remove the jth marked point. That’s a little bit, maybe I’ll say more about that model later.

Let’s look at some special case. Well, maybe in a little bit. I’m ready to define an A_{n}-space. We say that (Y, m, e) satisfies the A_{n} relations [m is a product Y ×Y → Y and e a basepoint of Y] if there is a family of maps M_{i} : K_{i} × Y^{i} → Y for 2 ≤ i ≤ n such that (I’ll interchange Y^{i} and K_{i} factors when it’s convenient for me)

(1) M_{2}(e, y; ∗) = M_{2}(y, e; ∗) = y for y ∈ Y, ∗ ∈ K_{2}. That is, m = M_{2} makes

(Y, m, e) an H-space.
(2) For ρ ∈ K_{r}, σ ∈ K_{s}, r + s = i + 1, we have

M_{i}(y_{1}, …, y_{i}; ∂_{k}(r, s)(ρ, σ)) = M_{r}(y_{1}, …, y_{k−1}, M_{s}(y_{k}, …, y_{k}+s−1; σ), y_{k+s}, …, y_{i}; ρ)

(3) M_{i}(y_{1}, …, y_{j−1}, e, y_{j+1}, …, y_{i}; τ) = M_{i−1}(y_{1}, …, y_{j−1}, y_{j+1}, …, y_{i}; s_{j}(τ)) We call this collection (Y, {M_{i}}) an A_{n}-space. For homework (from Stasheff),

check that this definition is consistent with the relations between ∂_{k}(r, s) and s_{j} in Axiom six.

Let me explain what this relation actually means. The first A_{2}-space is nothing but an H-space. Why is this so? An H-space has M_{2} which is a map K_{2} ×Y^{2} → Y but K_{2} is a point, so if we set m = M_{2}(∗; ), this is a map Y × Y → Y, it defines a multiplication. The relation one basically says that m(e, y) = m(y, e) = y so e plays the role of the identity for this multiplication map.

For general n, consider i = 3. For this, we know K_{3} is homotopic to the unit interval, and M_{3} is a map K_{3} × Y^{3} → Y. So this is a map [0, 1] × Y^{3} → Y. The relations say that M_{3}(y_{1}, y_{2}, y_{3}; ∂_{k}(2, 2)(ρ, σ)) is just a point, this lies in the boundary of K_{3} this is either + or −, and so the second relation says that for k = 1, this is M_{2}(M_{2}((y_{1}, y_{2}; σ), y_{3}; ρ) and for k = 2 it’s M_{2}(y_{1}, M_{2}(y_{2}, y_{3},; σ); ρ). So to simplify the notation here, we denote, well, we can omit ρ and σ since K_{2} is a point. So M_{2}(x, y; ∗) as xy. Then this becomes, the above formula becomes

= \{ \begin{matrix}(y_{1}y_{2})y_{3} \\ y_{1}(y_{2}y_{3}) \end{matrix}

This is the boundary so what this formula says is that this M_{3} which is a map from [0, 1]×Y^{3} → Y, obviously this is a homotopy. It’s a homotopy between M_{3}({0}×•) and M_{3}({1} × •). These two maps are, respectively, (y_{1}y_{2})y_{3} and y_{1}(y_{2}y_{3}). So this M_{3} provides a homotopy between these two different products. In other words, M_{3} provides an associating homotopy between these two different triple multiplications.

Let’s do one more, for i = 4. We know K_{4} is a pentagon and there are five ways to put a product in four letters in parentheses. So we could have an edge that is a homotopy from a(b(cd)) to a((bc)d) and the next goes to (a(bc))d. This is treating bc as one letter. The following one goes from that to ((ab)c)d), then (ab)(cd), and then back to a(b(cd)). So there is an explicit map, restricted to the boundary we get this loop. So we have, concatenating this homotopy, we have a map S^{1} × Y^{4} → Y. That’s provided by the M_{3}. Then the map M_{4} extends this map which is already defined by M_{3} along the boundaries and M_{4} provides an explicit contraction of that map, extending it to K_{4} × Y^{4}. Of course, K_{4} is homeomorphic to D^{2}. This is a homotopy of a homotopy, and the higher things can be regarded as homotopies of homotopies of homotopies.

There’s a trivial way of constructing these A_{n} spaces. An H-space does not have to be associative. Any associative H-space admits an a trivial A_{n} structure for all n by setting M_{i}(y_{1}, …, y_{i}; σ) = y_{1} ⋯ y_{i}. You don’t need any parentheses because the product is associative. The product comes from m = M_{2}. This satisfies, you can check, the A_{n} relations. This map does not depend on the K_{n} element.

Here is the motivating example, of based loop spaces. Let X be a topological space with x_{0} a basepoint. Then Y = ΩX = {γ : [0, 1] → X|γ(0) = γ(1) = x_{0}}. Then ΩX is a (homotopy) H-space. Set the constant loop e and then consider the standard concatenation as the product given by

γ_{1} ∗ γ_{2})(t) = \{ \begin{matrix}γ_{1}(2t) & 0 ≤ t ≤ \frac{1}{2} \\ γ_{2}(2t − 1) & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

Consider i_{L}, i_{R} : Y → Y × Y defined by i_{L}(Y) = (e, γ) and i_{R}(γ) = (γ, e). We want to check that m ∘ i_{L} ∼ id ∼ m ∘ i_{R} as a map Y → Y. So m ∘ i_{L}(γ)(t) is m(e, γ) = e ∗ γ which is

\{ \begin{matrix}∗ & 0 ≤ t ≤ \frac{1}{2} \\ γ(2t − 1) & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

You remain constant for time \frac{1}{2} and then follow γ twice as fast. Similarly, m ∘ i_{R}(γ)(t) follows γ twice as fast and then remains constant for time from \frac{1}{2} to 1:

\{ \begin{matrix}γ(2t) & 0 ≤ t ≤ \frac{1}{2} \\ ∗ & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

We have an obvious homotopy that I don’t want to write from m∘i_{L} to the identity given by changing the \frac{1}{2} to an s:

H(s, γ)(t) = \{ \begin{matrix}∗ & 0 ≤ t ≤ \frac{1−s}{2} \\ γ(\frac{2}{1+s}t − \frac{1−s}{1+s}) & \frac{1−s}{2} ≤ t ≤ 1 \end{matrix}

For the next couple of lectures I’ll prove that (Y, m, e) is an A_{∞} space. I’ll relate this based loop space with another type of path space, Z = Θ(X) which is pairs (r, α)|r ≥ 0, α : [0, r] → X with α(0) = x_{0} = α(r). This is an associative H-space. The multiplication map μ : Z × Z → Z is given by the following. The domain side adds and then we use concatenation. So

μ((r, α), (s, β)) = (r + s),??

and I’ll continue next time.

5. October 1

Let me remind you, let (Y, m, e) be an H-space. We say this satisfies the A_{n} relation if there exists a family of maps M_{i} : K_{i} × Y^{i} → Y for 2 ≤ i ≤ n satisfying the axioms (1)–(3) which I do not want to repeat. We looked at some examples. We considered X a based topological space and denote Y = ΩX and Z = Θ(X). The lemma was

Lemma 5.1. Z is an associative H-space.

Then we have a natural map. Remember that Z = {(r, α)|r ≥ 0, α : [0, r] → X with α(0) = x_{0} = α_{r}}. The product is concatenation without reparameterization. We have a natural map f : Y → Z which is inclusion. We have a map g : Z → Y which is rescaling: g(r, α)(t) = α(rt). These maps have the following properties. If you compose g ∘ f you get the identity. On the other hand, f ∘ g(r, α) = f(α(r·)) = (1, α(r·))

You can easily see that this is homotopic to the identity. What is the homotopy? Indeed, the homotopy is h : [0, 1] × Z → Z and is given by h(s, (r, α))(t) = ((1 − s)r + s, α((1 − s) + sr)t).

Now h(0, ·) = id_{Z} and h(1, ·) = f ∘ g. Noting this, I’m going to derive the following theorems in general.

Theorem 5.1. (ΩX, m, e) (e is the constant path and m the concatenation product) is an A_{∞} space.

This is an immediate corollary of the following theorem:

Theorem 5.2. Suppose Y is a topological space and let Z be an associative H-space with multiplication μ : Z × Z → Z. Suppose you are given maps f : Y → Z and g : Z → Y such that f ∘ g is homotopic to id_{Z}. Define a multiplication m : Y × Y by m(y_{1}, y_{2}) = g(μ(f(y_{1}), f(y_{2}))). Then (Y, m, e) becomes an A_{∞}-space, where e = g(e_{Z}) and e_{Z} is the unit in Z.

We are given a multiplication on Z and want to define one on Y, so we do it like this:

\xymatrix{Z × Z \ar[r]^{μ} & Z \ar[d]^{g} \\ Y × Y \ar[u]^{f×f} \ar[r]_{m} & Y}.

Proof. Denote by μ_{k} the k multiplication map since μ is associative: μ_{k} : Z × ⋯ × Z → Z. Furthermore, for simplicity, we denote z_{1} ⋯ z_{k} = μ(z_{1}, …, z_{k}). This will make it a little more intuitive. We’ll define using μ_{k}, f, g, and the homotopy h, we define an M_{i} : K_{i} × Y^{i} → Y inductively. I’ll describe this precisely up to i = 4.

The M_{2} case, we know K_{2} is a point, we define M_{2} = m = gμ_{2}(f × f). In other words, M_{2}(y_{1}, y_{2}) = g(f(y_{1})f(y_{2})).

Second, let’s see M_{3}. It’s supposed to be K_{3} × Y^{3} → Y. We know K_{3} is the unit interval [0, 1]. What we’ll do is define M_{3}|_{∂K_{3}×Y^{3}} and then extend to the interior. The inductive process you define first on the boundary and then extend it to the interior. You’ll recall ∂K_{3} = K_{2} ∗_{1} K_{2} ∪ K_{2} ∗_{2} K_{2} where K_{2} ∗_{i} K_{2} is the image of ∂_{i}(2, 2)(K_{2} × K_{2}). This lies in the boundary of K_{3}. This is how the face map was defined.

We already defined M_{2}. There are two boundary points. So we define M_{3}(x, y, z; 0) = M_{2}(M_{2}(x, y), z). Similarly M_{3}(x, y, z; 1) = M_{2}(x, M_{2}(y, z)). Now we choose and fix a homotopy h which I assume to exist from f ∘g to Id_{Z}. This will let us interpolate these two maps. So h is a map from f ∘ g to id_{Z}, those are h(0) and h(1). So then we have a map h(t) for time t. So now we interpolate the two maps. Recall how M_{2} was defined. The first map is

g(\underbrace{f(g}_{f∘g}(f(x)f(y)))f(z))

and the second is

g(f(x) \underbrace{f(g}_{f∘g}(f(y)f(z)))).

Then we replace these with h. So consider first

s ↦ g(h_{s}(f · f) · f).

At s = 0 this becomes g(f ∘ g(f · f) · f) which is the first boundary map. At s = 1 this becomes g(f · f · f), which you should regard as the central point. Similarly, the homotopy

g(f · h_{s}(f · f))

connects g(f · f ∘ g(f · f)) to g(f · f · f). So by concatenating these homotopies you get a homotopy from one to the other.

Finally, well, denote these two homotopies as h_{1} and h_{2}. So obviously, we define a concatenation homotopy h from g(f ∘ g(f · f) · f) to g(f · f ∘ g(f · f)) by setting

h(t) = \{ \begin{matrix}h_{1}(2t) & 0 ≤ t ≤ \frac{1}{2} \\ h_{2}(2 − 2t) & \frac{1}{2} ≤ t ≤ 1 \end{matrix}

Pictorially, K_{3} is G_{r_{3,1}}. You should think of this as looking like

where you think of the two endpoints as being K_{2} ∗_{1} K_{2} on the left and K_{2} ∗_{2} K_{2} on the right, with the center point.

We can then look at one more, M_{4} : K_{4} × Y^{4} → Y. Recall that K_{4} is the pentagon

(a(bc))d  ((ab)c)d  (ab)(cd) a(b(cd)) a((bc)d)) •^{v}

Here we first define M_{4}|_{∂K_{4}×Y^{4}} using the given homotopy h. We denote by m the midpoint of an edge and v the barycenter of the K_{4}.

(a(bc))d  ((ab)c)d  (ab)(cd) a(b(cd)) a((bc)d)) • m  v •  • • •  •

We connect m to v and define M_{4}(·, ·, ·, ·, m), which is, say, at the midpoint between ((ab)c)d and (a(bc))d, equal to g(f ∘ g(f · f · f) · f). We have a similar product explicitly given for each midpoint. We get rid of the f ∘g and put g(f ·f ·f ·f) at the center point v, which corresponds to the corolla graph. Now using the homotopy h from f ∘ g to id_{Z}, we define M_{4} on the segment connecting each midpoint m to v. So for instance, M_{4}|_{[m,v]×Y^{4}} sends s to g(h_{s}(f · f · f) · f). In this way we have defined M_{4} restricted to the line segments of this subdivision [ed: the pair subdivision]. Now for any other λv for λ ∈ ∂K_{4}, we interpolate linearly M_{4}(⋯; λ) and M_{4}(⋯, v).

I’m using an axiom of a pentagon. K_{i} is the cone on its boundary. On each line segment I use the homotopy h to interpolate the two maps. The explicit formula is a little bit complicated but that’s the idea. For higher M_{i} we just repeat this process inductively.

The way this was constructed, the A_{n} relations are automatic. These relations say what happened to the restriction of M_{i} to the boundary of K_{i}. But by construction this M_{i} satisfies the relations for any n. The A_{n} relation is nothing but the relation between, well, M_{i}(y_{1}, …, y_{i}; ∂_{k}(r, s)(ρ, σ)) = M_{r}(y_{1}, …, y_{k−1}, M_{s}(y_{k}, …, y_{k} + s − 1; σ), y_{k+s}, …, y_{i}; ρ) for r + s = i + 1. But r and s are already defined. That’s how our inductive construction was defined. □

Corollary 5.1. Then applying the theorem to Y = ΩX, Z = ΘX, we have proved that ΩX is an A_{∞} space.

Not every space carries an A_{∞} structure, only those homotopic to a based loop space do. Now applying the cohomology functor to this space we’ll get an A_{∞} algebra. Let’s apply the cohomology functor to the A_{n} space (Y, {M_{i}}_{i=1}^{n}) So M_{i} : K × Y^{i} → Y. Say the coefficients for cohomology is ℝ. Then m_{i} = (M_{i})^{∗} : H^{∗}(Y) → H^{∗}(K_{i} × Y^{i}). Now K_{i} is contractible so this is canonically isomorphic to H^{∗−(i−2)}(Y) [some discussion, there’s a comment that an explanation will be forthcoming] by integration over the fiber. This integration over the fiber can be defined. If Y is integrated over the fiber, well, we define a shifted complex, the cohomology ring is a graded Abelian group so denote A^{∗} = H^{∗}(Y)[1] by defining A^{k} = H^{k+1}(Y). This is the standard notation. Then define m′_{k}, well, I have a map A^{∗} → H^{∗+1}Y, and well, I want to rewrite m_{i} as m′_{i} : H^{∗}(Y)[1] → (H^{∗}(Y)[1])^{⊗i}. The original m_{i} has degree i − 2 but the degree shifting gives all m_{i} degree 1 for all i. In this business, the keeping track of signs is a complicated matter, very confusing, but if you shift the degree in this way, keeping track of the degree will be much easier because the sign rule will satisfy the Koszul sign. This degree shifting is very important.