Every 4-manifold has a $\operatorname{Spin}^c$ Structure

Question

$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\SO{SO}$I'm having trouble understanding the proof given in Morgan's The Seiberg–Witten Equations and Applications to the Topology of Smooth Four-Manifolds that every 4-manifold $X$ admits a $\Spin^c$ structure (Lemma 3.1.2). One can easily see from the exact sequence: \begin{equation*} H^1(X;\Spin^c) \to H^1(X; \SO(n)) \oplus H^1(X;\mathbb{Z}) \xrightarrow{c_1+w_2} H^2(X;\mathbb{Z}_2) \end{equation*} that a $\Spin^c$ structure will exist iff $w_2(TX)$ lifts to an integral class, which we can check using Bockstein homomorphisms. After that, I'm lost; I'm not sure if these are theorems, or whether they are supposed to be obvious:

1. In what sense is every $\mathbb{Z}/2^k \mathbb{Z}$ 3-class represented by a mapping from a smooth $\mathbb{Z}/2^k \mathbb{Z}$-manifold into $X$?
2. Why are integral 2-classes that represent torsion elements necessarily represented by embedded oriented surfaces?

EDIT: Since the proof from Morgan's book is quite short, I may as well write out the whole thing here:

"We need only see that $w_2(X)$ lifts to an integral class $c \in H^2(X;\mathbb{Z})$ in order to prove the existence of a $Spin^c$ lifting. But for any class $x \in H_2(X;\mathbb{Z}/2 \mathbb{Z})$ the value of $w_2(X)$ on $x$ is given as follows: represent $x$ as an embedded (possibly non-orientable) closed surface in $X$ and take the self-intersection of this surface modulo two. To see that $w_2(X)$ lifts to an integral class, we must see that its integral Bockstein $\delta w_2(X)$ is zero. But this torsion integral class is zero if and only if it evaluates trivially on every $\mathbb{Z}/2^k \mathbb{Z}$ class of dimension three. Any such class is represented by a mapping of a smooth $\mathbb{Z}/2^k \mathbb{Z}$-manifold into $X$. The value of $\delta w_2(X)$ on such a class is equal to the value of $w_2(X)$ on the Bockstein of this $\mathbb{Z}/2^k\mathbb{Z}$-manifold. Thus, we need only see that $w_2(X)$ vanishes on integral classes which represent torsion elements in $H_2(X;\mathbb{Z})$. But this is clear, any such class is represented by a smoothly embedded oriented surface with zero self-intersection".

I suppose what I'm really asking is which statements in this proof are non-trivial theorems about the topology of 4-manifolds, and which ones are supposed to be obvious?

Sorry, I don't seem to be able to comment, so I'll just say here: Ryan Budney, I hope this makes the question less vague, and Anton Fetisov, yes, there are other proofs of this fact that I do understand, but I'm specifically trying to understand this proof, because it seems very slick.

Answer 1

First let me stress the importance of Anton Fetisov's comment that $X$ must be orientable. Indeed, this seems to be used at nearly every step of the proof you cite.

For example, consider the claim "for any class $x\in H_2(X;\mathbb{Z}/2\mathbb{Z})$ the value of $w_2(X)$ on $x$ is given as follows: represent $x$ as an embedded (possibly non-orientable) closed surface in $X$ and take the self-intersection of this surface modulo two." This can fail if $X$ is non-orientable. For example, take $X=\mathbb{R}P^2\times\mathbb{R}P^2$, which has $w_2(X)=a^2\times 1 + a\times a + 1\times a^2$ where $a\in H^1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$ is the generator. Consider the homology class $x\in H_2(X;\mathbb{Z}/2\mathbb{Z})$ represented by the embedding $\iota:\mathbb{R}P^2\hookrightarrow \mathbb{R}P^2\times\mathbb{R}P^2$ of the first factor. Then $$ \langle w_2(X),x\rangle =\langle w_2(X),\iota_*[\mathbb{R}P^2]\rangle = \langle \iota^*w_2(X),[\mathbb{R}P^2]\rangle = \langle a^2,[\mathbb{R}P^2]\rangle \equiv 1, $$ whereas it is clear that the self-intersection is zero in this case.

Now to your actual questions:

1. Edit: the answer below is incorrect (even with my assumed definition of $\mathbb{Z}/2^k\mathbb{Z}$-manifold), as pointed out by Gustavo Granja in the comments below. Here a $\mathbb{Z}/2^k\mathbb{Z}$-manifold means a manifold which is oriented with respect to singular homology with $\mathbb{Z}/2^k\mathbb{Z}$ coefficients. When $k=1$ this is all manifolds, and the claim reduces to Thom's positive answer to the mod $2$ Steenrod realizability problem. When $k>1$, a $\mathbb{Z}/2^k\mathbb{Z}$-manifold is an orientable manifold $M^m$ with a choice of fundamental class $[M]\in H_m(M;\mathbb{Z}/2^k\mathbb{Z})$. By Thom's results, a homology class $a\in H_3(X;\mathbb{Z}/2^k\mathbb{Z})$ is represented by an embedded $\mathbb{Z}/2^k\mathbb{Z}$-manifold if and only if its Poincaré dual in $H^1(X;\mathbb{Z}/2^k\mathbb{Z})$ is induced from the universal Thom class $t_{\mathbb{Z}/2^k\mathbb{Z}}\in H^1(MSO(1);\mathbb{Z}/2^k\mathbb{Z})$ under some map $f: X\to MSO(1)$. Now note that $MSO(1)\simeq S^1$, and that the Thom class induces an isomorphism $$t^*_{\mathbb{Z}/2^k\mathbb{Z}}:H^1(MSO(1);\mathbb{Z}/2^k\mathbb{Z})\cong H^1(K(\mathbb{Z}/2^k\mathbb{Z},1);\mathbb{Z}/2^k\mathbb{Z}).$$ It follows that every $a\in H_3(X;\mathbb{Z}/2^k\mathbb{Z})$ is represented by an embedded $\mathbb{Z}/2^k\mathbb{Z}$-manifold.

(Note that orientability of $X$ was used here, to get Poincaré duality with $\mathbb{Z}/2^k\mathbb{Z}$ coefficients.)

2. Again by orientability and Thom's results, every class in $H_2(X;\mathbb{Z})\cong H^2(X;\mathbb{Z})$ is represented by an embedded oriented surface $\Sigma\hookrightarrow X$. The torsion assumption is just used to get the final clause, that $\Sigma$ has zero self-intersection. This is because the self-intersection is Poincaré dual to cup square, and the square of a torsion class in $H^2(X;\mathbb{Z})$ is a torsion class in $H^4(X;\mathbb{Z})\cong \mathbb{Z}$, therefore is zero.

Answer 2

This is an answer to 1. It is an edit of a previous answer based on the incorrect assumption that $\mathbb Z/k$-manifold means a closed manifold with a $\mathbb Z/k$-orientation. Thanks to Danny Ruberman for pointing out in the comments that a $\mathbb Z/k$-manifold is a type of manifold with singularities introduced by Morgan and Sullivan in The transversality characteristic class and linking cycles in surgery theory.

Roughly, a $\mathbb Z/k$-manifold of dimension $n$ is obtained from an oriented $n$-manifold with boundary whose boundary is partitioned into $k$ disjoint isomorphic closed $(n-1)$-manifolds by identifying the boundary pieces by orientation preserving diffeomorphisms. See section 2 of $\mathbb Z_k$-stratifolds by Andrés Angel, Carlos Segovia and Arley Fernando Torres for a more precise definition.

A $\mathbb Z/k$-manifold has a fundamental class in mod $k$ homology whose Bockstein is an integral fundamental class for the codimension 1 submanifold along which it is singular.

Given a CW complex $X$, any element of $H_n(X;\mathbb Z/2^k)$ is the image of the fundamental class of a $\mathbb Z/2^k$-manifold $M$ under some map $f \colon M \to X$. Here is a (possibly overcomplicated) justification:

Morgan and Sullivan show (or at least state, bottom of p. 471) that the bordism of $\mathbb Z/k$-manifolds is the homology theory represented by $MSO\wedge M\mathbb Z/k$ classifying cobordism with $\mathbb Z/k$-coefficients.

The spectrum $MSO_{(2)}$ is a wedge of suspensions of $H\mathbb Z/2$ and $H\mathbb Z_{(2)}$ and the canonical map $MSO_{(2)} \to H\mathbb Z_{(2)}$ splits. Smashing with the Moore spectrum $M\mathbb Z/2^k$ we see that the same is true of the map $MSO\wedge M\mathbb Z/2^k \to H\mathbb Z/2^k$. The (split surjective) induced map on homology sends a bordism class $[f \colon M \to X]$ to the image $f_*[M] \in H_*(X;\mathbb Z/2^k)$ of the fundamental class of the $\mathbb Z/2^k$-manifolds $M$. This completes the proof.

Note that this argument would not work for mod $l$ homology with $l$ not a power of $2$. See the paper by Angel, Segovia and Torres linked to above for more information and examples of $p$-torsion classes (with $p$ an odd prime) not represented by maps from $\mathbb Z/p$-manifolds.

Finally, see also Proposition 5.7.4 in Gompf and Stipsicz, 4-manifolds and Kirby calculus, for the original Hirzebruch–Hopf proof as well as a description of the Teichner–Vogt proof in All 4-manifolds have $\operatorname{Spin}^c$-structures for arbitrary oriented $4$-manifolds linked to in the comments on the question (1 2 3). Edit: See also an interesting recent proof of LeBrun's in Twistors, Self-Duality and Spin^c structures which is quite different from the proofs above. As written it only applies to compact $4$-manifolds but it is not hard to slightly adapt it so that it works for all $4$-manifolds.