Dennis Sullivan Course Notes March 11, 2005 Nathaniel Rounds, guest speaker

Gabriel C. Drummond-Cole

March 18, 2005

I’m going to talk about cohomology operations, and give the sort of basic example of them, which is Steenrod squares, and then I’ll find the Wu class and the Stiefel-Witten classes, and on Monday I’ll talk about Thom; his Fields medal in 1958 was for showing that a mnanifold was a boundary basically only if these things all vanish.

1 cohomology operations

Definition 1 A cohomology operation is a natural transformation of functors Θ : H^{m}(·, G) → H^{m}(·, G′) so that the appropriate diagram commutes:

\xymatrix{H^{m}(X, G) \ar[r]^{Θ} & H^{n}(X, G′) \\ H^{m}(Y, G) \ar[u]_{f^{∗}} \ar[r] & H^{n}(Y, G′)} .

There are a few obvious ones and they’re actually the only obvious ones.

Example 1 • a group homomorphism ϕ : G → G′.

• δ : H^{n} → H^{n+1}

• x ↦ x ∪ x.

Note that as in the last case, these need not be ring homomorphisms.

Now I want to give a more concrete example. Those who took Dusa’s class last spring will remember that we had this as the definition of cohomology.

H^{n}(X, G) is set-isomorphic to [X, K(G, n)] where π_{n}(K(G, n)) = G and the other homotopy groups are zero. G is a finitely generated Abelian group. This space is unique up to homotopy type. The only example I’m interested in is G = ℤ/2. Brackets denote homotopy classes of maps from the first to second argument. I’ll give the correspondence but will not prove it to be a bijection.

We have the Hurewicz homomorphism h : G = π_{n}(K(G, n)) → H_{n}(K(G, n)), which is an isomorphism because there is no other homotopy, i.e., it’s n − 1-connected. Since this an isomorphism its inverse h^{−1} ∈ Hom(H_{n}(H(G, n), G)) ≅ H^{n}(K(G, n), G). Now if I call ω = [h^{−1}] then f corresponds to [f^{∗}ω].

So now I claim I don’t have to define this operation for all spaces and all coefficients, I just have to define it for ω on K(G, n). That is, there exists a bijection between cohomology operations and H^{n}(K(G, n), G′).

How do I want to prove this? If I have a cohomology operation Θ it acts on ω to give Θ(ω). I claim that this is a bijection, and well-defined. Suppose α ∈ H^{m}(X, G), that’s the same as having f_{α} : X → K(G, m), and Θ_{ω} ∈ H^{n}(K(G, m), G′) so Θ_{ω} : K(G, m) → K(G′, n). Composing these gives me the map I want.

The next claim is that every cohomology operation arises in this way, and that is because of naturality.

If I have a cohomology operation H^{m}(·, G) → H^{n}(·, G′), and β ∈ H^{m}(X, G) then that corresponds to f_{β} : X → K(G, m) and Θ(β) = Θ(f_{β}^{∗}ω) = f_{β}^{∗}(Θ(ω)).

That characterization tells us something about cohomology operations that we didn’t know. They can’t lower dimension, and can only fix dimension if they are a group homomorphism because H^{m}(K(G, m), G′) = Hom(H_{m}(K(G, m)), G′) = Hom(G, G′).

Now I’ll define Steenrod squares: Sq^{i} : H^{n}(X, ℤ/2) → H^{n+i}(X, ℤ/2)

1. naturality

2. stability: σ : H^{n}(X) → H^{n+1}(ΣX)

3. ring homomorphism: Sq^{i}(α+β) = Sq^{i}(α)+Sq^{i}(β); Sq^{i}(α∪β) = ∑_{j} Sq^{j}(α)∪Sq^{i−j}(β)

A few of these are easy. Sq^{i}(x) = x if i = 0; Sq^{i}(x) ∼ δx if i = 1. There is a short exact sequence 0 → ℤ_{2} → ℤ_{4} → ℤ_{2} → 0 We can take Hom and get

0 → Hom(·, ℤ_{2}) → Hom(·, ℤ_{4}), ⋯

[argument ensues.]

It is x^{2} = x ∪ x for i = dim x. It’s 0 for i > dim x.

[Keep going, do whatever you want.] The main thing that I wanted to say was to give some idea of where these things come from. I’ll draw the picture that Dennis would draw. He’s going to draw that picture. [A circle with two hemispheres attached?] [That’s a crappy picture.] I’ll try to explain that a little bit better. I can think of x ⊗ x ∈ H^{n}(X) ⊗ H^{n}(X) so x ⊗ x ∈ H^{2n}(X × X). Now T : X × X → X × X swaps factors; So x ⊗ x corresponds to a homotopy class of maps from X × X → K(2n, ℤ_{2}). I claim that this is homotopic to (x ⊗ x) ∘ T : X × X → K(2n, ℤ_{2}). So then that homotopy is a map I ×X ×X → K(2n, ℤ_{2}). This is a homotopy f and I can also look at f ∘(id×T) Pieces these together and get a map from S^{1} × X × X → K(2n, ℤ_{2}). This is contractible, so I can fill in the homotopy and extend to D^{2} × X × X → K(2n, ℤ_{2}). I use this to prescribe a map on “the other side” by (s, x_{1}, x_{2}) ≅ (−s, x_{2}, x_{1}). This gives me S^{2} × X × X → K(2n, ℤ_{2}). I continue and only run into trouble at 2n. I claim I can get around it and get a map S^{∞} × X × X → K(2n, ℤ_{2}). [How do you get around the obstruction?] It’s not obvious that you can. The homotopy is only ℤ/2, so you can’t miss it by a lot. The original map is injective. [discussion.] Now I’ll say how to use such a map to get Steenrod squares. I defined this so that it respects antipodes so this will pass to the quotient ℝℙ^{∞}.

[figure: S^{∞} × X; S^{∞} × X × X; K(2n, ℤ_{2}); ℝℙ^{∞} × X > @[rru]]

So H^{∗}(ℝℙ^{∞}, ℤ_{2}) ≅ ℤ_{2}[τ], τ of degree one. So α ∈ H^{2n}(ℝℙ^{∞} × X, ℤ/2) we can write f_{α⊗α} = ∑_{j} τ^{2n−j} ⊗ a^{j}; then Sq^{j}(α) = a^{n+j}. α ⊗ α is in the 2n cohomology of X × X. Then the above diagram shows how to get the diagonal map of which I take a Kunneth decomposition. I’m going to write down one thing and then we’ll all leave. I have the pairing on cohomology of a closed manifold, H^{n−i} × H^{i} → H^{n}(M) ≅ ℤ_{2}. So Sq^{i} ∈ Hom(H^{n−i}, ℤ_{2}). Then we can say, there exists V_{i} such that V_{i} ∈ H^{i} with V_{i}∪x = Sq^{i}(x) for all x ∈ H^{n−i}(M).

Then W_{k} = ∑_{i+j=k} Sq^{i}(V_{j})