Why can't the 3-type of $S^2$ be delooped?

Question

In 2.2 of this paper, the argument runs through the fact that the 3-type of $S^2$ cannot be delooped. Why not?

I understand that this is probably a fairly basic fact of homotopy theory (hence neither that paper nor this paper, which it cites, providing an explanation or source for the fact). All I know of homotopy stems from studying higher categories & groupoids, so I'm weak on the basics. Unfortunately, my google-fu has failed me, and I haven't been able to find a source that discusses why the 3-type of the 2-sphere cannot be delooped. I'd appreciate a pointer to an appropriate text, and I'd appreciate a quick explanation of why this should be so even more.

Answer 1

Let $X$ denote the $3$-type of $S^2$. The only nontrivial homotopy groups of $X$ are $\pi_2(X)\cong\mathbb{Z}$ and $\pi_3(X)\cong\mathbb{Z}$. Suppose $X$ could be delooped, so there is a connected space $Y$ such that $X\simeq \Omega Y$. The only nontrivial homotopy groups of $Y$ would be $\pi_3(Y)\cong\mathbb{Z}$ and $\pi_4(Y)\cong\mathbb{Z}$.

Now a homotopy type with only two nontrivial homotopy groups $\pi_m(X)\cong A$ and $\pi_n(X)\cong B$ ($n>m$) is classified by its $k$-invariant, a cohomology class in $H^{n+1}(K(A,m);B)$. Explicitly, such a cohomology class corresponds to a map $K(A,m)\to K(B,n+1)$ (up to homotopy) and then you take the homotopy fiber of this map to get your space.

So, our delooping $Y$ would be classified by a class in $H^5(K(\mathbb{Z},3);\mathbb{Z})$. Now it happens that $H^5(K(\mathbb{Z},3);\mathbb{Z})$ is trivial (you can compute the cohomology of $K(\mathbb{Z},3)$ from the cohomology of $K(\mathbb{Z},2)=\mathbb{CP}^\infty$ using the Serre spectral sequence, for instance), so the $k$-invariant of $Y$ is trivial, so actually $Y\simeq K(\mathbb{Z},3)\times K(\mathbb{Z},4)$. It then follows that $X\simeq K(\mathbb{Z},2)\times K(\mathbb{Z},3)$. But this is false: it would imply, for instance, that $H_3(X;\mathbb{Z})$ is nontrivial (since $H_3(K(\mathbb{Z},3);\mathbb{Z})\cong\mathbb{Z}$), but $H_3(X;\mathbb{Z})\cong H_3(S^2;\mathbb{Z})$ is trivial.

Answer 2

You know, the $h$-space approach suggested by Tyrone and by the paper may feel more elementary, as it doesn't need Eilenberg-Mac Lane spaces, spectral sequences, or any cohomology.

Let's also let $X$ be the $3$-type of $S^2$. If $X$ were equivalent to a loop space $\Omega Y$, then concatenation of loops would make $X$ into an $h$-space, in fact an $A_\infty$-space, though that's more than we need. An $h$-space is just a space admitting a multiplication map which admits a unit up to homotopy, so this should be easy to verify directly.

Next, you need to know that there is something called a Whitehead product, a bilinear mapping $\pi_{n_1}(X)\times \pi_{n_2}(X)\to \pi_{n_1+n_2-1}(X)$, which is defined using the natural map $S^{n_1+n_2-1}\to S^{n_1}\vee S^{n_2}$ which is the attaching map in the cell structure on $S^{n_1}\times S^{n_2}$ that has a $0$-cell, an $n_1$-cell, an $n_2$-cell, and an $n_1+n_2$-cell. These Whitehead products should in principle be explainable in terms of higher groupoid theory, but of course nobody has a really workable definition of higher groupoid that's not more or less based on topological spaces or simplicial sets, so this is difficult to understand in a dramatically different way. Samelson products are in principle a bit more algebraic.

OK, so once we have Whitehead products, it's an exercise to prove that every $h$-space has trivial Whitehead products (indeed, it's Hatcher 4.2.37.) The idea is to show that every map $(\alpha,\beta):S^{n_1}\vee S^{n_2}\to X$ extends to $S^{n_1}\times S^{n_2}$, which is the cofiber of the attaching map that defines the Whitehead product. This implies the Whitehead products are trivial because cofiber sequences of spaces are weak pushouts in the homotopy category, so mapping them into $X$ gives a weak pullback of pointed sets.

Such an extension $\gamma:S^{n_1}\times S^{n_2}\to X$, for $X$ an $h$-space whose base point is its unit $e$, is given simply by $\gamma(x,y)=\alpha(x)\beta(y)$, where the multiplication uses the $h$-space structure on $X$. $\gamma$ restricts to $\alpha$ and $\beta$ on $S^{n_1}\times \{*\}$ and $\{*\}\times S^{n_2}$ respectively, using the fact that multiplication with $e=\alpha(*)=\beta(*)$ is trivial. (In principle it's only homotopy trivial, but every $h$-space is homotopy equivalent to one with a strict unit.)

Answer 3

This is an answer to a different question which I can't find: when is X an H-space?

Let G = \pi_1(X) and H = \pi_2(X) Interpret the Postnikov invariant beta as a (homotopy class of maps) K(G,1) to K(H,3) Then X is an H-space if and only if beta is primitive i.e. for m the multiplication on K(G,1) if and only if m^* beta = beta \otimes 1 + 1 \otimes beta

In this special case of K(G,1)  K(H,3) This is also sufficient for beta to be the looping of B beta \in H^4(BG,H)

More generally there is a result due to Art Copeland that if X is induced by f :K(G,p) to K(H,q) with q \geq p then X has an H-space structure if and only if f represents a primitive class

Outside that range, things are more subtle

Not sure what you mean by H-group