Take the 2-minute tour ×

Let $X$ be a CW-complex with

  • one 0-cell
  • two 1-cells
  • three 2-cells
  • no cells in dimensions 3 or higher.

Is it always true that $\pi_2(X)\ne 1$?

share|improve this question
Well, it is not homework and is not so simple if one thinks a little. Maybe the simple formulation is misleading, but this is what I find attracting in that question. –  Julien Marché Dec 15 '11 at 12:34
Ditto. I was thinking along the lines of Mikael's answer. (I do think that the question could be fleshed out a little with some background and motivation to make it more focussed.) –  Loop Space Dec 15 '11 at 14:33
You are asking if there is a 2-generator 3-relator group with a 2-dimensional Eilenberg-MacLane space. I suggest retagging with group theory and geometric group theory and you will get a quick answer. Maybe retitle to make clear this is what you want. –  Benjamin Steinberg Dec 15 '11 at 15:32
It is the same as asking if the presentation complex of a 2-generated group with 3-relators can be an Eilenberg-MacLane space. I oversimplified in my haste :) –  Benjamin Steinberg Dec 15 '11 at 16:41
What does oPhone call it? –  Tom Goodwillie Dec 15 '11 at 16:54

3 Answers 3

up vote 11 down vote accepted

There are classic examples, coming from small cancellation theory. See the section of the Wikipedia article on asphericity.

share|improve this answer
Just to make this nice answer explicit, let $w(x,y)$ be the word $xyxy^2\cdots xy^{100}$. Then the presentation $$\langle a,b\mid w(a,b), w(b,a), w(ab,ba)\rangle$$ satisfies $C'(1/6)$ if I am not mistaken and so has a presentation complex with 1 0-cell, 2 1-cells, 3 2-cells and nothing more with trivial $\pi_2$ by the result cited in the Wiki article. –  Benjamin Steinberg Dec 16 '11 at 5:47
Well thank you very much, this is a very nice answer! –  Julien Marché Dec 16 '11 at 12:24
@Julien, if you click the check mark next to the answer it accepts Agol's answer as the official one. –  Benjamin Steinberg Dec 16 '11 at 13:09

I believe that the answer is NO. If you look at

Gutiérrez, Mauricio A.; Ratcliffe, John G. On the second homotopy group. Quart. J. Math. Oxford Ser. (2) 32 (1981), no. 125, 45–55.

Corollary 3 states that a "reduced 2-complex $K(X; R)$ is aspherical if and only if each element of $R$ is independent and not a proper power."

Now, "reduced" means that there is (a) only one 0-cell (true in your case), and the one cells represent distinct nontrivial elements of $\pi_1(K^1),$ where $K^1$ is the one-skeleton. Again seems to be true under your assumptions. $R$ are the relations (given by attaching maps of the 2-cells, I imagine), "independent" is too complicated to explain here (look at the paper), but in any case, the "not a proper power" condition is easy to violate.

EDIT Actually, independent is not too hard to explain. The definition is: a relator $r$ is independent if, setting $M$ to be the normal closure of $r,$ and $N$ the normal closure of $R - r,$ $M \cap N = [ M, N].$

As @Benjamin points out, above I am answering the complementary question, so to get the example that the OP wants, we need three independent elements in the free group on two generators which are not proper powers.

share|improve this answer
Igor, he is asking if aspherical examples do not exist. So I think you need an example of 3 independent relators that are not proper powers. –  Benjamin Steinberg Dec 15 '11 at 19:42
@Benjamin: not enough sleep... –  Igor Rivin Dec 15 '11 at 20:18
I've been guilty of that... –  Benjamin Steinberg Dec 15 '11 at 20:35
One obvious example to look at would be the basic commutators $x,y,z$ of weight 4, i.e. $[[[a,b],a],a]$, $[[[a,b],b],a]$ and $[[[a,b],b],b]$ in $F=\left<a,b\mid\right>$. If $F/[\left<x\right>^F,\left<y,z\right>^F]$ is residually nilpotent, then it is easy to show, using P. Hall's basis theorem, that $x$ is independent (as a relator of $\left<a,b\mid x,y,z\right>$; the group given by this presentation is in fact known to be $F/\gamma_4$). But I don't know if this quotient is residually nilpotent. –  Sergey Melikhov Dec 16 '11 at 1:16
It is not, because $F/\gamma_4$ has cohomological dimension 6. –  Sergey Melikhov Dec 16 '11 at 2:17

So, the one 0-cell forces the 1-skeleton to be a figure-8. And we attach three 2-cells to this figure-8. These cells can be attached to:

  • loop 1, with some winding number n.
  • loop 2, with some winding number m.
  • the 0-cell, and it's degenerate 1-cell.

In the last case, we get a generator for $\pi_2$ from the resulting sphere; and without any 3-cells, any generator that shows up will produce non-trivial homotopy.

Suppose, thus, that the last case does not occur. Then we would be distributing three 2-cells on 2 loops. Regardless of how we do this, at least two 2-cells attach to the same loop, possibly with different winding numbers. Unless all three 2-cells attach to the same loop, the fundamental group will be trivial. If $\pi_1$ is indeed trivial, then because $H_2(X)=Ab \pi_2(X)$, it follows that $\pi_2(X)$ is indeed non-trivial. If all three 2-cells attach to the same loop, then the space is a wedge of a circle and the CW-complex on 1 0-cell, 1 1-cell and 3 2-cells. Being a wedge, if the homotopy on a factor is non-trivial, the entire homotopy will also be, and for the factor of the three attached 2-cells, the above argument with the abelianization also works out.

... or at least, that's how I would approach it. Would those here who know homotopy theory now please tell me why this cannot possibly work? ;-)

share|improve this answer
A 2-cell can be attached to some interlacing of both loops (the fundamental group of the figure-8 is free on the two generators that are the two loops). –  Guillaume Brunerie Dec 15 '11 at 13:14
I wonder about $H_2(X) = Ab \pi_2(X)$: Isn't the second (and higher) homotopy group always abelian ? I only know this formula in degree one, does it also hold in degree 2 ? (I don't know much about topology, so I my be wrong.) –  Ralph Dec 15 '11 at 13:33
I agree with Guillaume, the attaching map from the boundary of the 2-cells to the figure-eight can be anything, so your assumption is too strong. To Ralph, $\pi_2(X)$ is always commutative and by Hurewicz's theorem, if $\pi_1(X)=1$, then the map $\pi_2(X)\to H_2(X)$ is an isomorphism, showing the result. Of course, one cannot suppose that $\pi_1(X)=1$! –  Julien Marché Dec 15 '11 at 14:00

Your Answer


