MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A graph manifold is a closed 3-manifold $M$ that admits a finite collection of disjoint embedded tori $\mathcal{T}$ so that $M \setminus \mathcal{T}$ is a disjoint union of Seifert fibred spaces (i.e. spaces admitting a foliation with a circle as fibre).

My interest in this class of manifolds lies in its characterization through the methods of geometrization: A 3-manifold is a graph manifold if and only if its simplicial norm vanishes.

Question: What is known about the second homotopy group of graph manifolds? I am particulary interested in an answer to the following question: How many graph manifolds have finite second homotopy group?

The only hint I could find was the following result due to D. Hume (cf. The universal cover of any graph manifold quasi-isometrically embeds in the product of three metric trees.

share|cite|improve this question
up vote 10 down vote accepted

The answer to your question is almost trivial: if the tori/Klein bottles are required to be $\pi_1$-injective, then the manifold will be a $K(\pi,1)$, so $\pi_2$ will vanish. Also, notice in the case of Seifert-fibered spaces, the universal cover is either $\mathbb{R}^3$ or $S^3$, and thus again $\pi_2=0$.

However, there is a bit of discrepancy about what is considered a graph manifold in the literature. For 3-manifold topologists, the manifolds are required to be irreducible (so the tori are $\pi_1$-injective), and thus $\pi_2=0$. However, for geometers, sometimes they allow some of the tori to be compressible, so one obtains solid tori for some of the Seifert pieces. This is useful, because such graph manifolds have $F$-structures, and thus can admit sequences of metrics which collapse in the sense of Cheeger-Gromov. So if you're asking about this flavor of graph manifolds, then the answer is that $\pi_2$ is infinitely generated, unless you happen to have something finitely covered by $S^1\times S^2$, in which case it's $\mathbb{Z}$, or $S^3$ or $R^3$, in which case $\pi_2$ is trivial. In particular, $\pi_2$ is never finite. This kind of graph manifold opens up a can of worms since it is highly non-canonical: there are graph manifold structures on $S^3$ associated to any iterated cable link. See these slides for the classification of universal covers of closed 3-manifolds.

share|cite|improve this answer
You can see the notes to my talk linked above for a geometric proof. The point is that you can put a non-positively curved metric on each Seifert piece with geodesic boundary. Then in the universal cover, you can put these metrics together to get a global npc metric. Alternatively, you can map $S^2$ into the space, and make it transverse to the canonical tori. Then an innermost disk in the preimage must be homotopically trivial, so you can homotope it off, leaving fewer components of the preimage. If the sphere misses the tori, then it is contractible. – Ian Agol Jul 13 '12 at 21:24
@ Vitali: See p. 17 of the slides: You can't necessarily match up the NPC metrics downstairs (although the cases that you can are classified by BKN equations), but upstairs you can, using $GL(R^2)$ is homotopic to $O(2)$. – Ian Agol Jul 13 '12 at 22:24
@Vitali: A different simple proof uses graphs of spaces (as in Scott-Wall). Start from the fact that a Seifert fibered manifold is a $K(\pi,1)$. Any graph manifold with $\pi_1$-injective tori is a graph of spaces whose vertex spaces (Seifert fibered manifolds) are $K(\pi,1)$'s and whose edge spaces (tori) are $K(\pi,1)$'s. It is simple to deduce that the total space must then be a $K(\pi,1)$. – Lee Mosher Jul 13 '12 at 23:08
@Lee Mosher thanks. but my question really was about that last easy to deduce step :=) – Vitali Kapovitch Jul 13 '12 at 23:32
I guess all I am really trying to say is that Agol's second argument can be recast without any 3-manifold topology, as a general statement about graphs of $K(\pi,1)$'s (well, no 3-manifold topology except for the statement that Seifert fibered manifolds are $K(\pi,1)$'s). From a homotopy theory point of view, it might be simpler to think of the proof that way. From a 3-manifold topology point of view, there's no improvement on Agol's wording. – Lee Mosher Jul 14 '12 at 7:03

Have you thought of using the 2-dimensional Seifert-van Kampen theorem?

R. Brown and P.J. Higgins, ``On the connection between the second relative homotopy groups of some related spaces'', Proc. London Math. Soc. (3) 36 (1978) 193-212.

This gives information on the second relative homotopy group $\pi_2(X_2,X_1,x)$ as a crossed module over $\pi_1(X_1,x)$, in the case $X$ is a union of subspaces $U^a,a \in A$, and in terms of the relative homotopy group $\pi_2(X_2 \cap U^a, X_1\cap U^a)$ as a crossed module over $\pi_1(X_1\cap U^a,x)$ under certain connectivity assumptions. Further details, including more modes of computation, are in Part I of the book advertised on

I put forward this as an idea, without having done any work on the problem posed. However, since this 2-d SvKT seems little known, it seems worth putting forward the possibility. Note the point that if the theorem applies, then it determines the 2-type itself as a crossed module, but it can still be a problem to compute the second homotopy group. Some methods are available, but few have worked on the area.

If this idea is inapplicable to this case, I won't be worried. But the question asked is about manifolds which are unions in a prescribed way, so it might work.

A short list of papers with applications of such higher homotopy Seifert-van Kampen theorems is on

share|cite|improve this answer

I thought perhaps I should write up my comments to Agol's answer as a separate answer itself.

Proving asphericity of a graph manifold $M$ with $\pi_1$-injective tori can be done from the point of view of rather easy homotopy theory, using the Scott-Wall concept of graphs of spaces. By splitting along the torus decomposition of $M$, one gets a finite graph of spaces whose vertex spaces are Seifert fibered 3-manifolds and whose edge spaces are toruses, both of which are $K(\pi,1)$'s. Assuming that each inclusion of a torus edge space into an incident Seifert Fibered vertex space is $\pi_1$-injective ---- which as Agol says is always true when the Seifert fibered spaces are not solid tori --- the universal covering of the graph of spaces is a tree of contractible spaces. More precisely, it is a tree of spaces over the Bass-Serre tree $T$ of the splitting, the vertex spaces upstairs are universal covers of the Seifert fibered spaces which are contractible, and the edge spaces upstairs are universal covers of the torusses which are contractible.

Any tree of contractible spaces is contractible, and therefore the downstairs space (in this case the graph manifold) is aspherical. Contractibility of a tree of contractible spaces can be checked by mapping a sphere in, and applying induction on the (finite) number of edges of the subtree $\tau \subset T$ whose edges spaces intersect the image of the sphere. The induction step is to homotope the map of the sphere to push it off an edge space incident to a valence 1 vertex space of $\tau$, reducing the number of edges of $\tau$ by $1$. In the base step, where $\tau$ has one edge, the sphere map is homotopic to a constant within an edge space.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.