Riemann surfaces provide interesting examples of 1-types - interesting as they have roles in diverse areas. However, apart from 2-dimensional lens spaces, I can't readily bring to mind natural examples of spaces with non-trivial first two homotopy groups (non-trivial firt $k$-invariant optional, I suppose). Given a crossed module one can get an interesting 2-type, but this is via geometric realisation, so hardly finite-dimensional and not smooth in the usual sense (maybe in some exotic notion of smoothness).

Do natural examples of spaces with interesting 2-types turn up anywhere?

I tried to find out if I could construct one in a naive way at this question, but it fell over.

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    $\begingroup$ It's been a while but aren't there theorems to this effect: there are no finite-dimensional $K(\pi,n)$ spaces for $n\geq 2$. So if you had a finite-dimensional 2-type, its universal cover would be a finite-dimensional $K(\pi,2)$. If I recall, it's some kind of Serre spectral sequence argument, showing that $H_i(K(\pi,n))$ is non-trivial for infinitely-many $i$. I may be mis-remembering? $\endgroup$ Aug 18, 2010 at 4:14
  • $\begingroup$ I don't need a 2-type per se, but a space with an interesting 2-type. But this is a good point you raise +1. $\endgroup$ Aug 18, 2010 at 4:35
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    $\begingroup$ The wedge $S^1 \vee S^2$ is interesting. $2$-knot complements in $S^4$ have an interesting $2$-type, as well. $\endgroup$ Aug 18, 2010 at 4:38
  • $\begingroup$ "...2-knot complements in S^4..." - cool. This is the sort of thing I was after. If you want to put this as an answer I'll accept it. $\endgroup$ Aug 19, 2010 at 0:29
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    $\begingroup$ The complement of an $n$-knot for $n>1$ is aspherical if and only if the knot complement has the homotopy-type of $S^1$. This is an old result of Dyer and Vasquez, 1973. The reference is in Hillman's book "2-knots and their groups" but Google books isn't bringing up that part of the book, and my actual copy is in my office... Google books does bring up Eckmann's 1976 proof, though. $\endgroup$ Dec 31, 2010 at 22:51

3 Answers 3


The 2-type of a 4-manifold is an extremely interesting invariant. In fact, work of Hambleton and Kreck shows that in many cases it determines the homotopy type (if one adds the intersection form as an obvious additional invariant). As a consequence, such 2-types have a very rich structure.

It's quite tricky to figure out which 2-types appear in this way, just like it is tricky to figure out which groups arise as fundamental groups of 3-manifolds.

This is a very interesting open problem, even if one puts aside the (difficult) 4-manifold questions and just works with 4-dimensional Poincare complexes.


Cappell-Shaneson knots are a special class of knots in homotopy 4-spheres. In a sense they were designed as an example of knots with a well-known but prescribed 2-type.

The definition is like this. Look at bundles over $S^1$ whose fiber is a once-punctured $(S^1)^3$. The monodromy is an element of $A \in GL_3(\mathbb Z)$. A non-trivial (but fun) exercise is to check this manifold $M = \left((S^1)^3\setminus\{*\}\right) \rtimes_A S^1$ is the complement of a smoothly embedded $S^2$ in a homotopy $4$-sphere if and only if $det(A) = \pm 1$.

$\pi_2 M$ has a single generator as a module over $\pi_1 M$, but it's far from a free module over $\pi_1 M$. Perhaps the best way to describe $\pi_2 M$ is that it's a Laurent polynomial ring $\mathbb Z[x^\pm,y^\pm,z^\pm]$ where the $x,y,z$ correspond to the generators of $\pi_1 ((S^1)^3 \setminus\{*\})$ i.e. this is $H_2$ of the universal cover, which has a natural identification with $(\mathbb R^3 \setminus \mathbb Z^3) \times \mathbb R$. The action of $\pi_1 M$ is just the action on this covering space, so you get not just multiplication by units $x^ay^bz^c$ but also the automorphisms of $\mathbb Z^3$ (coming from $A \in GL_3(\mathbb Z))$ acting on the exponents of the polynomials.

The homotopy 4-spheres that contain Cappell-Shaneson knots were once considered possible counter-examples to the smooth 4-dimensional Poincare conjecture. Recent work of Akbulut and Gompf seem to have largely removed this possibility.

  • $\begingroup$ I like this answer, it gives a nice computational flavour to the construction. $\endgroup$ Dec 16, 2012 at 23:18

In low dimensional topology there are host of examples along the lines of the curve complex of a surface. Someone like Andy Putman will be able to provide a more detailed explanation, but here is a summary. You build a simplicial complex in which vertices are isotopy classes of closed curves in a surface, and n-simplices are disjoint (n+1)-tuples of curves. If you require that no curve bounds a disc (or an annulus if your surface has boundary) then it is fairly easy to see that the complex is finite dimensional. The mapping class group of the underlying surface acts on this complex.

Now, the homotopy types of this complex and its variations, and of the quotient by the mapping class group, are very interesting objects. In particular, studying the low dimensional homotopy groups allow you to do things like construct presentations for the mapping class group.

This sort of construction is really a small industry in low dim topology/geometric group theory. Braid groups, automorphism groups of free groups, surface mapping class groups, 3-manifold mapping class groups, and various subgroups of these, can all be studied via complexes of this type, and often we only understand the complex as far as its 2-type.

  • $\begingroup$ Doesn't the curve complex have the homotopy type of a wedge of (typically higher dimensional) spheres? I think Harer showed this and computed the dimension. In particular, isn't $\pi_2$ usually trivial? As you say, Andy will know. $\endgroup$
    – HJRW
    Jan 2, 2011 at 5:17
  • $\begingroup$ you're right, it is a wedge of spheres, but then the quotient by the mapping class group is the space with an interesting 2-type, since it is an approximation to the classifying space of the mapping class group. So $\pi_2$ would be trivial, but having an interesting fundamental group makes a 2-type eligible for being considered interesting in my book. $\endgroup$ Jan 2, 2011 at 9:19
  • $\begingroup$ In that case, I don't think I understand the question. Why isn't 'any 4-manifold' a good answer? $\endgroup$
    – HJRW
    Jan 2, 2011 at 19:09
  • $\begingroup$ As I would use the terminology, an $n$-type is an equivalence class of spaces under the relation of being $n$-equivalent, so yes, 'any 4-manifold' would be a perfectly good answer as far as I am concerned. $\endgroup$ Jan 3, 2011 at 10:59

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