The Groove of Math

or in other words, the Devil is in the Details

Filling the Details #2: Representing 2-homology classes of a 4-manifold.

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For the second article under the FtD tag, I’ve chosen a nice proposition which bothered me some times ago. It’s intuitively easy to believe but I found a scarce amount of details spelled out in the proofs one can find on the classical references. So here we have the following result:

Proposition: Every 2-homology class of a smooth 4-manifold is generated by a surface

The classical proof goes as follows

Proof: (from Kirby’s Topology of 4-Manifolds Theorem 1.1 page 20) There is an isomorphism:

H^2(M, \mathbb{Z})\cong [M,\mathbb{C}P^{\infty}]

so letting \hat{\alpha} being the Poincaré dual of a chosen \alpha \in H_2(M, \mathbb{Z}), there is an homotopy class of maps [f]\colon M \to \mathbb{C}P^{\infty} corresponding to \hat{\alpha}. By cellular approximation, we can homotopy (a representative of) f in order to obtain a map f\colon M \to \mathbb{C}P^2. In fact the 4-skeleton of \mathbb{C}P^{\infty} is \mathbb{C}P^{2} and cellular approximation tells you that the image of M lies there. Make f smoothly transverse to \mathbb{C}P^1\subset \mathbb{C}P^2. Consider f^{-1}(\mathbb{C}P^1), this will be an oriented surface representing \alpha.

yes but… why That’s the real question, why f^{-1}(\mathbb{C}P^1) represents our homology class

we need the following facts:

Claim 1: Let N\to M be a submanifold and let [N] the homology class it defines in M. Then in M the Thom class of the normal bundle is dual to [N].

Proof: Have a look here 🙂.

Back to our original exercise. First of all we need to fix some notations: Let us denote with j^*Th (\nu_1) \in H^2(M) the Thom Class associated to the normal bundle of the inclusion f^{-1}(\mathbb{C}P^1)\to M. Similarly, let j^*Th(\nu_2)\in H^2(\mathbb{C}P^2) be the Thom class associated to the normal bundle of \mathbb{C}P^1\to \mathbb{C}P^2. With PD(y) we will mention the Poincaré Dual of y (both in the case of homology and cohomology).

So let \alpha be our homology class, as we said before PD (\alpha)= f^*\imath, where f is a map as above and \imath \in H^2(\mathbb{C}P^2)=H^2(\mathbb{C}P^1) is  the fundamental class associated to \mathbb{C}P^{\infty} ( or \mathbb{C}P^{1} in our case, doesn’t change anything by cellular approximation). Using the fact that \imath = PD[\mathbb{C}P^{1}] (see the addendum below), we have:

PD(\alpha) = f^{*}(PD[\mathbb{C}P^{1}])

By Claim 1 and naturality of the Thom classes (the normal bundle over f^{-1}(\mathbb{C}P^1) is the pullback via f of the one over \mathbb{C}P^1)

PD(\alpha) = f^{*}(Th(\nu_{2})) = Th ( \nu_{1})  \in H^{2}(M)

Using PD again,

\alpha = PDPD(\alpha)  = PD  Th (\nu_1) \in H_2(M)

Using Claim 1 again,

\alpha = [f^{-1}(\mathbb{C}P^1)]

Which was exactly what we needed to show.



ADDENDUM: Let us recall briefly the definition of the fundamental class. Let (X,\phi) be a polarised Eilenberg-MacLane space of type (A,n) (i.e. we fixed an isomorphism \phi from the homotopy group to A. Using UCT (with map \Phi) there is a unique class \imath \in H^n(X;A), called the fundamental class s.t. the composite

\pi_n(X) \xrightarrow{UCT} H_n(X, Z) \xrightarrow{\Phi(\imath)} A

is the chosen \phi. (it depends on the chosen iso!). In fact X is n-1 connected, and A abelian, the Hurewicz map is an isomorphism and therefore

\beta \colon H_n(X;Z)\to \pi_n(X) \to A

Is a group homomorphism (an iso actually). Since UCT in degree n is an iso, we define \imath := \Phi^{-1}(\beta).

In our case, X = \mathbb{C}P^{\infty}, and since \hom_Z(H_2(\mathbb{C}P^{\infty}),Z)=\hom_Z(Z,Z)\cong Z, the map \beta is a generator of it, and UCT maps generator to generator. So up to a sign, we can assume \imath= PD[\mathbb{C}P^1].


Author: RicPed

Young mathematician :)

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