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Consider a smooth torus endowed with the non-bounding spin structure. Pick a basis of its first homology and a diffeomorphism inducing the S-transformation

$\left(\begin{array}{cc} 0 & 1 \\-1 & 0\end{array}\right)$

in the chosen basis. Construct the mapping torus $U$ of this diffeomorphism. Take the non-bounding spin structure on the $S^1$ base. This induces a spin structure on $U$. As the 3rd spin cobordism group vanishes, there is a smooth spin 4-manifold $W$ whose boundary is $U$.

The question is: How can one concretely construct such a $W$?

More generally, when given a manifold that is known to bound, what are the tools available to construct a concrete bordism to the empty manifold?

Update: I explain why Marco's answer below does not provide 4-manifolds bounding the mapping torus above with the required spin structure. It is not difficult to see that $H^1(U;\mathbb{Z}_2) = \mathbb{Z}_2^2$, with one generator associated to the base circle and one generator associated to the cycle on the torus fiber left invariant under the S transformation. The required spin structure restricts to the non-bounding spin structure on representative loops in both classes. This means that these loops cannot become homotopically trivial in the 4-manifold. As the 4-manifolds Marco constructs are all simply connected, they cannot carry a spin structure restricting to the one I am interested in on the boundary.

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    $\begingroup$ I don't think your update is convincing. Kaplan's algorithm provides, for any spin 3-manifold, a simply connected spin 4-manifold that it bounds. This can also be seen more abstractly; take any spin 4-manifold with given spin boundary, and do (spin) surgery on generators of the fundamental group. to make it simply connected. What your argument might show is that there is no single 4-manifold over which every spin structure on U extends. But there's no reason to expect that there would be such a 4-manifold. $\endgroup$ – Danny Ruberman Sep 28 '16 at 12:19
  • $\begingroup$ Indeed, it turns out that there is no natural way of restricting the spin structure from the 3-manifold to obtain a spin structure on the loop, unlike what I assumed in the update. $\endgroup$ – Samuel Monnier Sep 28 '16 at 20:49
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I think that there are three spin structure on T^2 that do not bound, and that we have three candidates spin structures on U that we don't know how to fill, a priori. (Thanks Danny for clarifying this.)

In any case, for 3-manifolds there is an algorithm due to Kaplan that settles your question.

Steve J. Kaplan, Constructing framed 4-manifolds with given almost framed boundaries, Trans. Amer. Math. Soc. 254 (1979), 237-263.

This is described pretty well in Gompf and Stipsicz's book 4-manifolds and Kirby calculus, but let me show how it works in practice, at least in this case. I will assume some familiarity with (basic) Kirby calculus.

According to Neumann's plumbing calculus, $U$ is the boundary of the plumbed 4-manifold $P$ associated to the graph:

The plumbing graph.

$P$ is not spin, since its intersection form is not even: the class represented by the central vertex has odd square. So, we'll try to make it even by looking for characteristic sublinks: if we call $u, v, w$ the three 'leaf' vectors, we observe that $u$, $v$, $w$, and $u+v+w$ are all characteristic in the intersection lattice of the graph/plumbed 4-manifolds. This means that for every vertex $y$, whenever $x$ is one of the four vectors above, $x\cdot y \equiv y\cdot y \pmod 2$.

Now, Kaplan's algorithm tells us to blow up each component of the characteristic sublink until we get to a $\pm1$, and then blow it down. (Well, at least if we're in a sufficiently nice case: if the characteristic sublink contains actual knots, or if the link is more complicated, things become a bit more annoying - but let's focus on this simple case here.)

Let's try with one of the three simplest ones, the one corresponding to the vertex of weight $-2$: we blow it up negatively once, and then we blow it down, thus obtaining the following plumbing graph:

The spin plumbing.

This is now a plumbing graph of a spin 4-manifold (with a unique spin structure), whose boundary gives a spin structure on $U$. You can play the game for the other characteristic sublinks, that correspond to distinct spin structures on $U$, and you obtain in each case a spin filling $W$ of $U$.

What I haven't tried to do is determine which of the four spin structures I've exhibited is the one that comes from your construction, but maybe one can play around with the Rokhlin invariant. (Two of them have the same Rokhlin invariant, though.)

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    $\begingroup$ I think there is just one non-bounding spin structure on the torus. Here are 3 bounding ones. Consider T in $S^3$, with its induced spin structure; it bounds solid tori on either side (1). Consider T as the boundary of a solid torus, and twist by the $Z_2$ cohomology class that extends over the solid torus (2). Do that with the other solid torus (3). (You have to change the orientation for the 3d one.) These are all products of spin structures on circles where one of them bounds. The 4th nonbounding one is the product of the nonbounding spin structures on the circle. $\endgroup$ – Danny Ruberman Sep 25 '16 at 14:39
  • $\begingroup$ Thanks! I'll need some time to digest this and learn more about Kirby calculus. But it sounds good! Yes indeed, there is only one non-bounding spin structure on the torus, as Danny explained. $\endgroup$ – Samuel Monnier Sep 25 '16 at 18:17
  • $\begingroup$ Unfortunately, it seems that your construction cannot yield the required spin structure, essentially because the 4-manifolds you construct are simply connected. I updated the question, explaining the problem in more detail. Still, the general idea of looking for a solution using Kirby calculus sounds like a good one. $\endgroup$ – Samuel Monnier Sep 28 '16 at 7:21
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    $\begingroup$ I don't see what goes wrong here, but Theorem 5.7.14 in Gompf--Stipsicz states that every spin 3-manifold bounds a simply connected spin 4-manifold. (In fact, it even says something a little stronger.) $\endgroup$ – Marco Golla Sep 28 '16 at 8:13
  • $\begingroup$ The argument in my "update" was incorrect indeed. $\endgroup$ – Samuel Monnier Sep 28 '16 at 20:50
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Marco's answer above comes close to a solution, but as he mentions, one still has to figure out which of the manifolds bounds the mapping torus with the specified spin structure. I want to record this part of the reasoning in this answer.

The possible four manifolds are described by the following diagrams:

  1. Marco's second diagram above.
  2. A diagram with a trivalent node labeled 0, linked to a -4 node, a -2 node and a string of three 2 nodes, obtained from the characteristic sublink given by one of the -4 nodes in Marco's first diagram.
  3. The same diagram as above, obtained from the other -4 node.
  4. An affine E7 Dynkin diagram, obtained from the characteristic sublink given by the three nodes -2, -4, -4 in Marco's first diagram.

The Rokhlin invariant of a spin 3-manifold is given by the signature mod 16 of a spin 4-manifold admitting it as a boundary. When the 4-manifold is specified by a plumbing diagram as above, this is just the signature of the adjacency matrix of the diagram (where the links indicate an off-diagonal value of -1). For reasons that will become clear, we only need Rokhlin invariants mod 8. The Rokhlin invariants mod 8 of the four 4-manifolds above are -1,1,1,-1.

Now we need a way of computing the Rokhlin invariant directly on the mapping torus to compare with the values above. A method for computing the Rokhlin invariant mod 8 in this way has been described by Taylor in

Taylor, L. R, Relative Rochlin invariants, Topology and its Applications, 1984, 18, 259 - 280

The Rokhlin invariant is the Arf invariant of a quadratic refinement of the linking pairing on the torsion part of the first integral homology. The value of the quadratic refinement on a given homology class is determined by the restriction of the spin structure of the 3-manifold to a representing cycle. Getting all the sign conventions straight to compare with the values of the Rokhlin invariant computed above looks like a nightmare, but fortunately we can reason as follows.

On the one hand, the first torsion homology can be represented by one of the cycle generating the homology of the fiber of $U$. This means that switching the spin structure on the base circle will not change the Rokhlin invariant. On the other hand, Johannes' solution above corresponds to the affine E7 diagram, as it is the only one that can accomodate the degree 2 cohomology classes that will necessarily appear when resolving the orbifold singularities. As I mentioned in a comment of his solution, he has the right spin structure on the fibers, but not on the base. This means that the 4-manifold that I am after is the one that has the same Rokhlin invariant, namely the one originally proposed by Marco.

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Here is an attempt at the specific rather than the general question.

Let $X_0 := (\Delta \times \mathbb{C}/\Lambda)/\mathbb{Z}_4$, where $\Delta \subset \mathbb{C}$ is the unit disc, $\Lambda$ is the lattice $\mathbb{Z} + i\mathbb{Z}$ and the generator of $\mathbb{Z}_4$ acts by $(z, w+ \Lambda) \mapsto (-iz, iw + \Lambda)$. Note that the group action preserves the obvious $SU(2)$-structure.

The boundary of $X_0$ is the mapping torus $U$ that you are interested in. However, $X_0$ has orbifold singularities: the image of $(0,\frac12 + \Lambda)$ is a $\mathbb{C}^2/{\pm1}$ singularity, while the images of $(0,\Lambda)$ and $(0, \frac{1+i}2 + \Lambda)$ are of the type $\mathbb{C}^2/\mathbb{Z}_4$. But if you let $X$ be a crepant resolution of $X_0$, then $X$ is a coboundary of $U$, and because $X$ has an $SU(2)$-structure it is also spin.

To see which spin structure this induces on $U$, consider the framing on $U$ induced by the $SU(2)$-structure on $X$. The framing has the property that it respects the splitting of the tangent bundle of $U$ into a horizontal and vertical part (with respect to the $T^2$ fibration), and the framing of the vertical part is invariant under the obvious translation action on each fibre. I think that is enough to identify the induced spin structure on $U$ as the one you wanted.

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  • $\begingroup$ Thanks! But unless I'm missing something, I think that the horizontal part of the spin structure is the bounding spin structure on the base circle, instead of the non-bounding one. This is because you started with a disk in the horizontal direction. $\endgroup$ – Samuel Monnier Sep 29 '16 at 7:09

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