# Can all n-manifolds be obtained by gluing finitely many blocks?

Fix a dimension $n\geqslant 2$. Let $S= \{M_1,\ldots, M_k\}$ be a finite set of smooth compact $n$-manifold with boundary. Let us say that a smooth closed $n$-manifold is generated by $S$ if it may be obtained by gluing some copies of elements in $S$ via some arbitrary diffeomorphisms of their boundaries.

For instance:

• Every closed orientable surface is generated by a set of two objects: a disc and a pair-of-pants $P$,
• Waldhausen's graph manifolds are the 3-manifolds generated by $D^2\times S^1$ and $P\times S^1$,
• The 3-manifolds having Heegaard genus $g$ are those generated by the handlebody of genus $g$ alone,
• The exotic $n$-spheres with $n\geqslant 5$ are the manifolds generated by $D^n$ alone.

A natural question is the following:

Fix $n\geqslant 3$. Is there a finite set of compact smooth $n$-manifolds which generate all closed smooth $n$-manifolds?

I expect the answer to be ''no'', although I don't see an immediate proof. In particular, I expect some negative answers to both of these questions:

Is there a finite set of compact 3-manifolds which generate all hyperbolic 3-manifolds?

and

Is there a finite set of compact 4-manifolds which generate all simply connected 4-manifolds?

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Bruno: this is closely related to this question: mathoverflow.net/questions/30567/level-sets-of-morse-functions/… In particular, I think the same argument shows that there is not such set of finitely many compact 3-manifolds generating all hyperbolic 3-manifolds. One could likely make another argument using TQFT's. The value of a unitary TQFT invariant made from finitely many manifolds is bounded by the norm of the manifolds in the vector space associated to the boundary. So if one had manifolds with arbitrarily large TQFT invts., then this would be impossible. –  Ian Agol Oct 18 '10 at 16:38
Thanks! On using TQFT: the number of blocks is however not bounded because of repetitions. For instance, graph manifolds have arbitrarily large TQFT invariants (for instance, they contain the connected sums of $S^2\times S^1$). Moreover, I think it is still unknown whether the invariants of graph manifolds cover all possible invariants of 3-manifolds. –  Bruno Martelli Oct 18 '10 at 17:07
Bruno - you're right, scratch the TQFT comment (I hadn't thought it through carefully). However, I'm sure that the argument in the other question works. The examples are a sequence of small 3-manifolds with Heegaard genus going to infinity. Then any decomposition into manifolds has to have a dual graph a tree, since $b_1=0$. So you can make a Morse function (or generalized Heegaard splitting) out of each piece, and apply the argument in that answer. –  Ian Agol Oct 18 '10 at 19:46
Yes, I am convinced it works. One only needs to fix once for all the same Morse function (or splitting) on each block and then glue them together (see also Greg Kuperberg below). –  Bruno Martelli Oct 18 '10 at 20:32
Slightly tangential, but: if the blocks allowed to have corners, then there's an interesting "yes" answer (the boring "yes" answer is "chop into simplexes) where the pieces behave like generators of a Hopf algebra. See ldtopology.wordpress.com/2011/06/26/… –  Daniel Moskovich Jul 12 '11 at 16:25

## 4 Answers

Thanks to Ian Agol for pointing out this question and a related one on levels of Morse functions - /305067/. In both case, for smooth manifold of dim $> 3$, as expected, there is no finite list of blocks ( or regular level components.) The idea is that one may define the "width" of a group, by representing the group G as the fundamental group of some complex K and then slicing K into "levels". The game to to arrange the slices so that the image of $\pi_1$ of each component of each level set maps a subgroup of small rank under the inclusion into $\pi_1(K)$. width( G) is defined as a Minmax over all slicings of all complexes K with $\pi_1 K = G$ of the rank of these image subgroups. I wrote a few pages to show that width( $\mathbb{Z}^k$) $= k-1$. The only slightly technical ingredient is Lusternick-Schnirelmann category. This answers negatively these finiteness questions since there are $d$ manifolds with $\pi_1 =\mathbb{Z}^k$ all $k$, as long as $d>3$. As soon as the notes are teXed, I can post them on the arxiv or math overflow.

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If you put the teXed notes on any website and indicate enough of how we could find that, somebody (me for example) could edit your post by making a genuine link to the pdf or postscipt file or what have you. The arXiv would be fine. –  Will Jagy Nov 2 '10 at 1:48
That being said, MO does accept Latex in answer windows, there are just a few tricks such as triple backslashes in matrices. So if it is really short, you can paste in the code, or email the .tex file to Ian or me and we will figure it out. –  Will Jagy Nov 2 '10 at 1:58
Thank you, I would be very much interested in seeing these notes. As far as I understand, the question remains open if restricted to simply connected closed 4-manifolds. –  Bruno Martelli Nov 6 '10 at 15:38
c.f. Michael's second answer below, with the link to the arXiv paper. –  Scott Morrison Nov 11 '10 at 21:29

I posted a paper on the arXiv, Group Width which answers this question for manifolds of dim $>3$ with sufficiently complicated fundamental group (there will be no finite set of blocks). As Greg Kuperberg said, there are many interesting variations which remain open and are a nice challenge to technique, e.g. the case of simply connected manifolds.

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Here is the link: arxiv.org/abs/1011.2460 . –  Joseph O'Rourke Nov 11 '10 at 2:00

It is not entirely clear in the question whether a "block" is a manifold with smooth boundary, or perhaps a manifold that is allowed to have ridges or more complicated corners. Let's assume that the boundary has to be smooth and that the blocks have to be glued along connected boundary components, because at the other extreme you can make any PL manifold from copies of a simplex.

If so, then Ian Agol's comment explains everything in dimension 3. As he explained, it's Ian's theorem that there is a non-Haken manifold $M$ of Heegaard genus $g' \ge g$ for every $g$. And, it follows from work of Scharlemann-Thompson and Casson-Gordon that a Morse function on such an $M$ must have a level set with a connected component of genus $\ge g'$. (And equality is trivial because a Heegaard surface is always a fattest Morse level set.)

If you have your blocks, you can always arrange them as a collection of "cups", i.e., you can pick a relative Morse function which is $0$ on every boundary component and negative in the interior. Then you can glue the boundary components of the blocks in pairs with "caps", which are copies of $\Sigma \times I$ with increasing Morse functions that begin at $0$ on their boundaries. (Or, equivalently, you could have a bipartite collection of blocks.) Since you can reuse the same Morse function on each cap or cup of a given type, having finitely many types implies a global bound on the genus of a connected component of a level set.

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On the other hand, I don't see a complete argument using TQFTs, because even if you have finitely many block types you could still get large values for a TQFT invariant. –  Greg Kuperberg Oct 18 '10 at 20:04
Thanks, that's pretty clear. –  Bruno Martelli Oct 18 '10 at 20:34
I think you can do this in dimension 3. In jtopol.oxfordjournals.org/content/1/3/703.refs Costantino and Thurston define a universal set of cusped hypebolic manifolds that are constructed from copies of a single block $B$ with ridges. The block $B$ (also used in other papers by Agol, Minsky and others) is a 3-handlebody having 6 closed embedded curves as ridges. It is a hyperbolic manifold with geodesic boundary consisting of 4 pair of pants and having 6 annular cusps (the ridges). It is obtained by mirroring a checkerboard-coloured regular ideal octahedron along black triangles. –  Bruno Martelli Oct 19 '10 at 7:09
Constantino and D. Thurston, to be precise. It seems that there is a lot left in this question. Even if one found an argument in higher dimensions, you could still ask for which $k$ could the blocks avoid $\le k$-dimensional faces. –  Greg Kuperberg Oct 19 '10 at 7:41
Hmm, interesting. I wonder if the block we described is the "simplest" in some sense. In 3-D, the elementary block needs to have positive hyperbolic volume in some suitable sense. The paper was motivated by 4-manifolds, and that block can be filled in to give a block for 4-manifolds that can be used to avoid 0-faces. (Also, note the spelling of Costantino's name.) –  Dylan Thurston Oct 20 '10 at 12:15

As Bruno rightly points out, my first answer (below) is nonsense. So let me try to say something else vaguely useful.

In the Spring, I heard Ian Biringer talk about his recent work with Juan Souto. I'm pretty sure he stated the following theorem:

Theorem: Let $\epsilon>0$ and $r\in\mathbb{N}$. There are finitely many hyperbolic 3-manifolds with boundary $M_1,\ldots,M_k$ with the property that any hyperbolic 3-manifold $M$ with injectivity radius greater than $\epsilon$ and $\mathrm{rank}\pi_1M\leq r$ can be obtained by gluing the $M_i$ together along their boundaries.

This would imply that there is a finite generating set of blocks for hyperbolic manifolds with appropriate restrictions on the injectivity radius and rank. I believe the proof is non-constructive, so one doesn't actually know what the $M_i$ are.

Poking around on the archive and Ian's web page, I don't see the result in question, so I'm at a loss to provide a reference! But if this sounds useful, then no doubt one could contact Ian or Juan and get the details.

I'm somewhat out of my comfort zone here, but I think this is right.

The figure-8 knot complement $M_8$ is universal, meaning that every closed 3-manifold arises as a Dehn filling on a finite-sheeted covering space of $M_8$. So the family $\lbrace M_8, D^2\times S^1\rbrace$ generates all 3-manifolds.

I don't have time to look at the references right now, but I'll try to get back to it later.

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You are right that $M_8$ is universal, but its coverings are not obtained by gluing copies of $M_8$. If you construct a manifold by gluing copies of $M_8$ then you never get a hyperbolic manifold, because it contains many incompressible tori. (On the other hand, all coverings of $M_8$ are hyperbolic.) –  Bruno Martelli Oct 18 '10 at 15:26
It might be worth saying that if $S= \{M_1,\ldots , M_k\}$ consists of manifolds with toric boundaries, then any hyperbolic manifold generated by $S$ has volume not bigger than the maximum Gromov norm of the $M_i$'s. Therefore you cannot get all hyperbolic 3-manifolds with finitely many cusped ones. Things however are more complicate if we admit (as we do) higher-genus boundary. –  Bruno Martelli Oct 18 '10 at 15:32
He's not allowing gluing with ridges, only gluing along boundary. –  Greg Kuperberg Oct 18 '10 at 15:56