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## The second homotopy group of a simple CW-complex

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 $\pi_2(X)\ne 1$?

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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 2011 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.) – Andrew Stacey Dec 15 2011 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 2011 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 2011 at 16:41
What does oPhone call it? – Tom Goodwillie Dec 15 2011 at 16:54
show 15 more comments

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

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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 2011 at 5:47
Well thank you very much, this is a very nice answer! – Julien Marché Dec 16 2011 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 2011 at 13:09

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? ;-)

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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 2011 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 2011 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 2011 at 14:00

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.

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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 2011 at 19:42
@Benjamin: not enough sleep... – Igor Rivin Dec 15 2011 at 20:18
I've been guilty of that... – Benjamin Steinberg Dec 15 2011 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 2011 at 1:16
It is not, because $F/\gamma_4$ has cohomological dimension 6. – Sergey Melikhov Dec 16 2011 at 2:17