Doubles of 2-handlebodies

Question

Let $X$ denote a $4$-manifold with boundary obtained by adding $k_1$ $1$-handles to $B^4$ and $k_2$ many $2$-handles to the resulting manifold i.e. $X$ is an arbitrary $4$-dimensional $2$-handlebody. Let $D(X)$ denotes its double. Evidently $D(X)$ is nullbordant as it is the boundary of $X\times [0,1]$.

My question is which nullbordant $4$-manifolds arise as doubles of $2$-handlebodies? If $M$ is an arbitrary closed $4$-manifold, can $M \# -M$ be written as such a double? Or more generally does there exist an $N$ such that $M\# N$ can be written as such a double?

Answer 1

Let's call a double of a 2-handlebody a 2-double for simplicity.

Your first question is interesting, I only have some partial remarks. As you noticed if $X^4$ is a double then it must be null-cobordant hence $\sigma(X) = 0$, and we can also add that the Euler characteristic $\chi(X) \in 2\mathbb Z$ has to be even.

At this point we can already conclude that: the only simply connected rational surfaces which are 2-doubles are $\mathbb S^2\times \mathbb S^2$ and $n\mathbb {CP}^2\#n \overline{\mathbb{CP}}^2$. Indeed the constraint on $\sigma$ and $\chi$ imply that these are the only possibilities and they clearly are 2-doubles since $\mathbb S^2\times \mathbb S^2$ and $\mathbb {CP}^2\# \overline{\mathbb{CP}}^2$ are the doubles of the two $\mathbb D^2$-bundles over $\mathbb S^2$ and connected sums of two 2-doubles is a 2-double.

If we consider the case $b^+\geq 2$ then we have another constraint from Seiberg-Witten theory: by Kirby calculus it is not difficult to see that if a 2-double has $b_2(X)>0$ then exist an embedded 2-sphere $S\subset X$ such that $S^2 = 0$ and $[S]\neq 0 \in H_2(X)$. To see this, let $X = D(Z)$, $Z$ a 2-handlebody and consider a Kirby diagram for $Z$. Since $H_2(X)\neq 0$, up to handle-slides we can assume that exists a 2-handle of $Z$, $h$ representing a non-trivial homology class. To construct the double we glue 2-handles along 0-framed meridians of each 2-handle attaching circle (and then we add some 3-handles). The 0-framed meridian associated to $h$ will induce a non-trivial homology class in $H_2(X)$ (represented by a sphere) because it cannot be in the image of the three-handles for otherwise $h$ would not be a 2-cycle of $C_2(Z;\mathbb Z)$.

Now it follows from Fintushel & Stern. Immersed spheres in 4-manifolds and the immersed Thom conjecture, Turkish J. Math., 19(1995), no. 2, 145–157 that if $X$ is a 2-double and $b^+(X)\geq 2$ then $X$ has vanishing Seiberg-Witten invariants. Consequently, by the work of Taubes we obtain that simplectic 4-manifolds and complex surfaces with $b^+\geq 2$ cannot be 2-doubles (note this works regardless of $\pi_1(X)$).

Regarding your last question we can show the following: for any simply connected $X^4$ exists a simply connected $N$ such that $X\#N$ is a 2-double.

Proof: by blowing up and down $X$ we can make its intersection form isomorphic to the intersection form of $n\mathbb {CP}^2\#n \overline{\mathbb{CP}}^2$, thus by Freedman's theorem is also homeomorphic to it. Moreover by Wall's theorem we can stabilize with a certain number of $\mathbb S^2\times \mathbb S^2$ and obtain diffeomorphic manifolds. But the result of this stabilization is a double for the first proposition above.

Answer 2

A hyperbolic 4-manifold has zero signature and hence is null-cobordant. However there exist hyperbolic 4-manifolds with trivial isometry group, and hence which cannot be a double of a 2-handlebody (such a double would admit an orientation-reversing isometry by Mostow’s theorem). This gives a partial answer to your first question (showing that there exists manifolds which are nullcobordant and not the double of a 2-handlebody).

Answer 3

Let $X$ denote a $2$-handlebody. I claim that the inclusion $\partial X \to X$ induces a surjection on fundamental groups. Indeed, let $Y\subset X$ denote the underlying $1$-handlebody (i.e. the union of the 0 handle and all 1 handles), then the inclusion of $Y$ into $X$ induces a surjection on fundamental groups. Additionally the inclusion $\partial Y \to Y$ induces a surjection on fundamental groups as well. Now any curve on $\partial Y$ is up to homotopy disjoint from the attaching circles of the $2$-handles of $X$, in particular the inclusion of $\partial X \to X$ also induces a surjection on fundamental groups.

With that knowledge at hand let $f \colon D(X) \to X$ denote the folding map. By Seiffert Van Kampen and the aforementioned surjectivity and because the kernels of bot maps in the corresponding pushout agree (since they are the same map) we get that $f$ induces an isomorphism on fundamental groups. In particular the map $D(X) \to B\pi_1(D(X))$ classifying the universal covering of $D(X)$ factorizes through $f$. Since $f$ has vanishing $4$-th homology this shows that $D(X)$ has to be inessential.

Being essential is preserved under connected sums, hence any aspherical manifold (or any other essential manifold) will never be such a double, even up to connected sum with other $4$-manifolds.

This argument also works in higher dimensions and should show that the double of any manifold, where the inclusion of the boundary is $\pi_1$-surjective has to be inessential.