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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.

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    $\begingroup$ 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. $\endgroup$ Oct 7, 2020 at 19:37
  • $\begingroup$ @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? $\endgroup$
    – Tim Campion
    Aug 9, 2021 at 16:19
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    $\begingroup$ @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. $\endgroup$ Jan 6 at 22:21
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    $\begingroup$ Chris, but not always with a symmetric braiding. $\endgroup$ Jan 9 at 2:50

2 Answers 2

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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: $\beta(x + y) = \beta(x) + \beta(y)$.

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    $\begingroup$ Are there non-zero beta which satisfy this condition? $\endgroup$ Jan 6 at 22:16
  • $\begingroup$ @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. $\endgroup$ Jan 7 at 14:38
  • $\begingroup$ 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... $\endgroup$ Jan 7 at 14:51
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    $\begingroup$ @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. $\endgroup$ Jan 7 at 19:16
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    $\begingroup$ 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 $H^4(K(\mathbb{Z}/2,2), \mathbb{Z}/4)\cong \mathbb{Z}/4$. $\endgroup$ Jan 8 at 23:58
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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.

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