3-manifold embedded in 1-connected 4-manifold

Question

Assume that 3-manifold $M$ is embedded in 1-connected 4-manifold $N$ and it separates it onto two manifolds $N_1,N_2$ with boundary. In such case from Mayer-Vietoris we have:

$H_2N\to H_1M\to H_1N_1\oplus H_1N_2 \to H_1N $

If $N$ is also 2-connected then $H_1M$ is isomorphic to above sum of $H_1N_1\oplus H_1N_2$. From another algebraic machinery called "universal coefficient theorem" we get that torsion subgroup of $H_1N_1$ is equal to torsion of $H_1N_2$. In this case manifold $N$ is homology sphere.

Can we somehow for chosen manifold $M$ decide on when it is embeddable in homology sphere $N$ or not ? It is probably not easy problem. One condition is that torsion group is of shape $G+G$. I just wonder how I can continue from this point further.

On the other hand is it known procedure for embedding any 3-manifold into 1-connected 4-manifold ?

EDIT 2018-10-05

I would like to add more details about my reasoning. Imagine manifold $M$ is embedded in $R^5$ - every 3-manifold can be embedded there. Assume it is defined by equations $\{f=0,g=0\}$ for smooth real functions $f,g$. Now I would like to embed manifold $M$ into manifold $N^4$ as simple as possible. Is such $N$ exist in $R^5$ ? Yes, we can take e.g. $\{f=0\}$. It seems to me that we can find $N$ 1-connected. Consider following argument. If there is essential loop in $N$ not intersecting $M$ then it can be removed via surgery. The essential loop cannot intersect $M$ in one point, since $M$ separates $N$ (to be fiugred out why). If essential loop intersect $M$ in two points then we can represent it as product of two loops - each in different component of $N$. we can remove these loops via surgery on $N_1$ and $N_2$. I assume that these surgeries can be considered as embedded in $R^5$.

Now the idea is investigate manifold $M$ by possible 1-connected $N$s which can embrace it. The 4-manifolds considered are the ones embeddable in $R^5$ and 1-connected. Do we know anything about such class of 4-manifolds ?

Related ideas are following.

1. Apply physics of minimal soap surface. It works for circle in $R^3$. Can it work for 3-manifold in $R^5$ ? We can assume that $M$ is not knotted.
2. For given $M\subset S^4$ having Betti number $r$ and torsion $G$ splitting sphere onto $N_1,N_2$ what are possibilities for homology of $N_1$ and $N_2$ ?.
3. Is it known when 1-connected 3-complex $K$ is embeddable in $R^5$ ? Boundary of tubular neighborhood of $K$ is 1-connected 4-manifold $N$ with $\pi_2$ the same as $K$ (we select only such $K$ that $N$ is 1-connected).

Thank you for response from Mike Miller.

Answer 1

A 2-connected 4-manifold is homeomorphic to $S^4$. It is a huge open problem to ask whether or not manifolds embed smoothly into $S^4$. It is also a huge open problem whether there are 4-manifolds which are homeomorphic, but not diffeomorphic, to $S^4$ (giving other candidates for $N$ in your question).

I'm going to restrict to the case that $M$ is an integer homology sphere. The question is hard enough in that case.

Anyway, there are two answers.

1) If you care about locally flat embeddings (topological, but not smooth) into $S^4$, Freedman proved that every homology 3-sphere has such an embedding into $S^4$ in his landmark paper The topology of four-dimensional manifolds. This is where you go if you want to know most things about topological (not smooth) 4-manifolds. Presumably people after this have said things about rational homology spheres, but I am out of the loop.

2) If you want the embeddings to be smooth, this is a fantastic and fantastically difficult question. It was a triumph already in the 1960s when Rokhlin showed that "the signature of a spin 4-manifold $Y$ bounds, taken modulo 16" is an invariant of 3-manifolds; one proves that the Poincare sphere $\Sigma$ has Rokhlin invariant $8$, while anything that embeds in $S^4$ has Rokhlin invaraint $0$, so $\Sigma$ does not embed smoothly in $S^4$.

The natural extension of this is the definition of the homology cobordism group, written $\Theta_3$. Its elements are homology 3-spheres, modulo the relation that $Y_1 \sim Y_2$ if there is a compact oriented 4-manifold $W$ with $\partial W = Y_1 \sqcup \overline{Y_2}$ so that the inclusions $Y_i \to W$ induce isomorphisms on homology. The group operation is connected sum, the identity element is $S^3$, and the inversion operation is orientation-reversal.

What one sees from your computations is that if a homology 3-sphere $Y$ embeds in $S^4$, then drilling a 4-ball out from either of the sides you call $N_i$, you get a homology cobordism from $Y$ to $S^3$. In particular, $[Y] = 0 \in \Theta_3$.

What Rokhlin's invariant (divided by 8) does is give a homomorphism $\mu: \Theta_3 \to \Bbb Z/2$. Because $\mu(\Sigma) = 1$, it is not zero in the homology cobordism group, so does not embed smoothly in $S^4$. The race is on! Find as many invariants of $\Theta_3$ as you can, so that you can show that your favorite element is nonzero in that group.

It was proved in the 90s that $\Theta_3$ is infinitely generated (there is an infinite linearly independent subset). Kim Froyshov proved in the early 2000s that there is a homomorphism $h: \Theta_3 \to \Bbb Z$ for which $h(\Sigma) = 1$, so that even $\#^n \Sigma$ does not embed in $S^4$ for any positive integer $n$. In fact, it has been recently announced that there is a surjective homomorphism $\Theta_3 \to \Bbb Z^\infty$, so $\Theta_3$ splits as a direct sum $\Bbb Z^\infty \oplus Q$, where $Q$ is... some mystery abelian group.

Even beyond this, here is a fascinating and wide-open question. Suppose $[Y] = 0 \in \Theta_3$ - that is, $Y$ bounds a 4-manifold with the homology of a point. Is it then true that $Y$ embeds smoothly in $S^4$? This would follow from the smooth Poincare conjecture in 4 dimensions, but nobody's really too sure whether or not they even believe that should be true.

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A great read is Budney-Burton, Embeddings of 3-manifolds in $S^4$ from the point of view of the 11-tetrahedron census.