# Which homotopy 2-types are H-spaces?

Let $$X$$ be a connected CW-complex with $$\pi_k(X)$$ trivial for $$k >2$$. Is it known under which circumstances $$X$$ is an $$H$$-group?

I have been able only to deduce the necessary condition that $$\pi_1(X)$$ has to be abelian and act trivially on $$\pi_2(X)$$. Furthermore, if the necessary condition holds, vanishing of the Postnikov invariant $$\beta \in H^3( \pi_1(X), \pi_2(X))$$ is a sufficient condition. Thus the interesting case is that of $$\beta$$ nonzero.

• A slightly stronger and more natural condition (I don't know if it's equivalent to being an $H$-group in this case) is that $X$ has a delooping $BX$; this means not only that $\pi_1$ is abelian and acts trivially on $\pi_2$ but that the Postnikov invariant must arise from a Postnikov invariant in $H^4(B^2 \pi_1, \pi_2)$. This is exactly the space of quadratic maps $\pi_1 \to \pi_2$, although I don't know how to describe the map to $H^3(B \pi_1, \pi_2)$ in these terms. Oct 7, 2020 at 19:37
• @Qiaochu Yuan If $X$ has a delooping $BX$, then $X$ admits a grouplike $A_\infty$ structure, which is much stronger than being an $H$-space, right? Aug 9, 2021 at 16:19
• @QiaochuYuan I think that the map to $H^3(B \pi_1, \pi_2)$ is always zero. The $H^4(B^2\pi_1, \pi_2)$ classes correspond to certain braided monoidal categories, and I am pretty sure these can be realized with trivial associators - so the map should be zero. Jan 6 at 22:21
• Chris, but not always with a symmetric braiding. Jan 9 at 2:50

The necessary condition is a "additivity"-type condition on the $$k$$-invariant.

Suppose $$\pi_1 X = G$$ and $$\pi_2 X = A$$. As you correctly point out, $$G$$ must be abelian and act trivially on $$A$$. Under these circumstances, the $$k$$-invariant is expressible as a natural map of pointed spaces $$\beta: K(G,1) \to K(A,3).$$ (Normally $$\beta$$ can only be constructed as a map $$K(G,1) \to K(A_G, 3)$$ in terms of the coinvariants.) The space $$X$$ is the homotopy fiber of $$\beta$$, and the $$k$$-invariant of $$X \times X$$ is $$\beta \times \beta$$.

This means that, in order to get an $$H$$-space, it is necessary and sufficient to get a homotopy commutative diagram of pointed spaces: $$\require{AMScd}$$ $$\begin{CD} K(G,1) \times K(G,1) @>m>> K(G,1)\\ @V \beta \times \beta V V @VV \beta V\\ K(A,3) \times K(A,3) @>>m> K(A,3) \end{CD}$$ Here $$m$$ is the $$H$$-space multiplication on Eilenberg--Mac Lane spaces for abelian groups, which induces addition on cohomology. In terms of the projection maps $$p_i: K(G,1) \times K(G,1) \to K(G,1)$$, homotopy commutativity of this diagram says that $$m^*(\beta) = p_1^*(\beta) + p_2^*(\beta).$$

If there is a sufficiently good Kunneth formula (e.g. if $$A$$ is a ring and $$H_*(K(G,1);A)$$ are finitely generated projective) then $$m^*$$ is a "coproduct" $$m^*: H^*(K(G,1); A) \to H^*(K(G,1); A) \otimes_A H^*(K(G,1); A)$$ and under this identification we are asking that $$\beta$$ is primitive: $$m^*(\beta) = \beta \otimes 1 + 1 \otimes \beta$$.

If we think of $$\beta$$ as representing a cohomology operation $$H^1(-;G) \to H^3(-;A)$$, this condition equivalently asks that $$\beta$$ is additive: $$\beta(x + y) = \beta(x) + \beta(y)$$.

• Are there non-zero beta which satisfy this condition? Jan 6 at 22:16
• @ChrisSchommer-Pries That I don't know. As you noted above, the most natural source of additive cohomology operations does not give any. If $G$ is free then there are also no such cohomology operations. Jan 7 at 14:38
• Using a free resolution $0 \to R \to F \to G \to 0$ to get a fiber sequence $K(F,1) \to K(G,1) \to K(R,2)$ we can use the Serre spectral sequence. There are two potential nonzero contributions to $H^3$, but the first is from $H^3(K(F,1);A)$ and we already said that has no primitives. That reduces to checking if there are any primitive elements in the cokernel of $$H^2(K(F,1); A) \to H^2(K(R,2); H^1(K(F,1); A))$$ which is isomorphic to a map $$Hom(\Lambda^2 F, A) \to Hom(R \otimes F, A).$$ This is about where I ran out of steam... Jan 7 at 14:51
• @mme If the groups involved (even just G) are finitely generated, yes, absolutely; a direct sum decomposition gives you a matrix decomposition of the primitives, and you can reduce to the case where G is cyclic. I'm worried (perhaps needlessly?) about eg the possibility of lim^i terms if G is not finitely generated. Jan 7 at 19:16
• A non-zero $\beta$ is the non-trivial element of $H^3(K(\mathbb{Z}/2,1), \mathbb{Z}/4)\cong \mathbb{Z}/2$. The corresponding 2-type is the loop space of the Postnikov piece with $k$-invariant the generator of $H^4(K(\mathbb{Z}/2,2), \mathbb{Z}/4)\cong \mathbb{Z}/4$. Jan 8 at 23:58

I just wanted to add to Tyler Lawson's answer that all the maps $$\beta\colon K(G,1)\rightarrow K(A,3)$$ ($$G$$ and $$A$$ abelian and no action of $$G$$ on $$A$$) satisfying his additivity condition are loop maps by Stasheff's Homotopy Associativity of $$H$$-Spaces II (Theorem 5.3). Hence for 2-types being an $$H$$-space is the same as being a loop space. See also Qiaochu Yuan's comment.

In terms of $$k$$-invariants, the condition is that the $$k$$-invariant of the 2-type is in the image of the so-called cohomology 'suspension' morphism

$$H^4(K(G,2),A)\longrightarrow H^3(K(G,1),A),$$

which is induced by taking loops on the corresponding sets of homotopy classes of maps between Eilenberg-MacLane spaces.

The source is very well understood, it coincides with $$\hom(\Gamma(G),A).$$ Here $$\Gamma(G)$$ is the target of the universal quadratic map $$G\rightarrow\Gamma(G)$$. Recall that a map $$\gamma\colon G\to B$$ between abelian groups is quadratic if $$G\times G\to B\colon (x,y)\mapsto\gamma(x+y)-\gamma(x)-\gamma(y)$$ is bilinear.

An example with non-trivial $$k$$-invariant can be constructed as follows. It suffices to show that

$$H^4(K(\mathbb{Z}/2,2),\mathbb{Z}/4)\longrightarrow H^3(K(\mathbb{Z}/2,1),\mathbb{Z}/4)$$

coincides with

$$\mathbb{Z}/4\twoheadrightarrow \mathbb{Z}/2.$$

Indeed, $$H^3(K(\mathbb{Z}/2,1),\mathbb{Z}/4)$$ is well-known to be the elements annihilated by $$2$$ in $$\mathbb{Z}/4$$, which identifies with $$\mathbb{Z}/2$$. A normalised $$3$$-cocycle representing the generator is $$f\colon \mathbb{Z}/2\times \mathbb{Z}/2\times \mathbb{Z}/2\longrightarrow \mathbb{Z}/4,\qquad f(1,1,1)=2.$$ (I will omit notations for classes in quotients of $$\mathbb{Z}$$ since the meaning in each case will be clear from the context.)

The universal quadratic map for $$\mathbb{Z}/2$$ is $$\gamma\colon \mathbb{Z}/2\rightarrow \mathbb{Z}/4$$, $$\gamma(0)=0$$, $$\gamma(1)=1$$. Hence $$\Gamma(\mathbb{Z}/2)=\mathbb{Z}/4$$ and $$\hom(\Gamma(\mathbb{Z}/2),\mathbb{Z}/4)=\hom(\mathbb{Z}/4,\mathbb{Z}/4)=\mathbb{Z}/4.$$

In order to compute the suspension morphisms I'm going to use crossed modules and their semi-stable version called reduced quadratic modules by Baues (see his book on 4-dimensional complexes). It is well known that 3-dimensional group cohomology classifies crossed modules (Eilenberg-MacLane or one of them alone, I don't currently remember). Similarly $$\hom(\Gamma(\mathbb{Z}/2),\mathbb{Z}/4)$$ classifies reduced quadratic modules. The generator is represented by the reduced quadratic module $$\mathbb{Z}\otimes \mathbb{Z} \stackrel{\langle-,-\rangle}\longrightarrow \mathbb{Z}\oplus \mathbb{Z}/4\stackrel{\partial}\longrightarrow \mathbb{Z}$$ where $$\partial(a,b)=2a$$ and $$\langle x,y\rangle =(0,{xy}).$$ This is because $$\ker \partial=\mathbb{Z}/4$$, $$\operatorname{coker}\partial=\mathbb{Z}/2$$ and the map $$\mathbb{Z}/2\mapsto \mathbb{Z}/4$$ defined by $${x}\mapsto{\langle x,x\rangle}$$ coincides with the aforementioned universal quadratic map. The loop crossed module of this reduced quadratic module is $$\mathbb{Z}\oplus \mathbb{Z}/4\stackrel{\partial}\longrightarrow \mathbb{Z},$$ where the source is equipped with an exponential action of the target defined by $$x^a=x+\langle\partial(x),a\rangle.$$ (Crossed modules usually consist of non-abelian groups and reduced quadratic modules too, but in this case everything is abelian because they are very small, this simplifies a lot the computations). A 3-cocycle $$g\colon \mathbb{Z}/2\times \mathbb{Z}/2\times \mathbb{Z}/2\longrightarrow \mathbb{Z}/4$$ representing the cohomology class of this crossed module is defined by the following choices:

We first need a set-theoretic splitting of $$\partial$$, that we define as $$s\colon \mathbb{Z}/2\longrightarrow \mathbb{Z},\qquad s(0)=0,\quad s(1)=1.$$ Now, for each pair of elements $$x,y\in \mathbb{Z}/2$$ we need $$t(x,y)\in \mathbb{Z}\oplus \mathbb{Z}/4$$ such that $$\partial(t(x,y))=-s(y)-s(x)+s(x+y).$$ This measures the failure of $$s$$ to be a morphism. We take $$t(0,-)=(0,0)$$, $$t(-,0)=(0,0)$$, and $$t(1,1)=(1,0)$$. With these choices $$g$$ is $$g(x,y,z)=t(x,y)^{s(z)}+t(x+y,z)-t(x,y+z)-t(y,z)\in\ker\partial=\mathbb{Z}/4.$$ Different choices produce cohomologous cocycles. This cocycle is normalised because $$t(0,-)$$ and $$t(-,0)$$ vanish, so we only have to compute $$\begin{array}{rcl} g(1,1,1)&=&t(1,1)^{s(1)}+t(1+1,1)-t(1,1+1)-t(1,1)\\ &=&t(1,1)+\langle \partial t(1,1), s(1)\rangle +t(0,1)-t(1,0)-t(1,1)\\ &=&\langle \partial t(1,1), s(1)\rangle\\ &=&\langle \partial (1,0), 1\rangle\\ &=&\langle 2, 1\rangle\\ &=&(0,2). \end{array}$$ The second coordinate (the kernel of $$\partial$$) is $$2\in \mathbb{Z}/4$$, therefore, $$g=f$$ above. This concludes the proof of the claim.