1 Introduction

The homology groups of a space do not determine its homotopy type, but when the space is a simply connected CW-complex, it is a well-known fact that if the space has the homology of a wedge of spheres of the same dimension, then it must have the homotopy type of that wedge of spheres. Therefore, we can ask if there are other cases where the homology groups of a simply connected CW-complex determine that the space is a wedge of spheres. In this brief note, we show that if a complex has free finitely generated reduced homology groups for two consecutive dimensions and trivial homology for all other dimensions, then it must have the homotopy type of a wedge of spheres of two consecutive dimensions. We also give other pairs of dimensions for which the last result can be generalized.

2 Main result

To fix ideas, we first give a proof of the above-mentioned folklore result; the map constructed in this proof will be use in the proof of our main theorem. Throughout we will take homology with integral coefficients.

Theorem 1

Let X be a simply connected CW-complex, such that the only nonzero reduced homology group is \({\tilde{H}}_d(X)\cong {\mathbb {Z}}^a\). Then

$$\begin{aligned} X\simeq \bigvee _a{\mathbb {S}}^d. \end{aligned}$$

Proof

By the Hurewicz Theorem (see [2, Theorem 4.3.2, p. 199]), \(\pi _d(X)\cong {\tilde{H}}_d(X)\cong {\mathbb {Z}}^a\). Therefore, there are maps \(i_j:{\mathbb {S}}^d\longrightarrow X\) for \(1\le j\le a\), such that the combined map

$$\begin{aligned} i:\bigvee _a{\mathbb {S}}^d\longrightarrow X \end{aligned}$$

is an isomorphism on \(\pi _d\). Thus, i induces an isomorphism on reduced homology groups and, by Whitehead’s Theorem (see [2, Theorem 4.3.5, p. 202]), is a homotopy equivalence. \(\square\)

The case \(k=1\) of the following theorem is a special case of example 4C.2 of [1]. For completeness, we give a different proof.

Theorem 2

Let X be a simply connected CW-complex, such that

$$\begin{aligned} {\tilde{H}}_q(X)\cong \left\{ \begin{array}{cc} {\mathbb {Z}}^a &{} \text{ for } q=d \\ {\mathbb {Z}}^b &{} \text{ for } q=d+k \\ 0 &{} \text{ for } q\ne d,d+k, \end{array}\right. \end{aligned}$$

where a, b, d, k are positive integers, with \(d>k\) and \(k \in \{1,5,6,13,62\}\). Then

$$\begin{aligned} X\simeq \bigvee _{a}{\mathbb {S}}^d\vee \bigvee _{b}{\mathbb {S}}^{d+k}. \end{aligned}$$

Proof

The key step of the proof is to show that \(\pi _{d+k}(X)\cong {\mathbb {Z}}^b\oplus H\) for some H, so we can pick a map

$$\begin{aligned} s:\bigvee _b{\mathbb {S}}^{d+k}\longrightarrow X \end{aligned}$$

with \(\textrm{im}(s_*)={\mathbb {Z}}^b \le \pi _{d+k}(X)\).

First, we do this for the case \(k=1\) using a Serre spectral sequence argument. The dth space \(P_d(X)\) in the Postnikov tower is given by \(K({\mathbb {Z}}^a, d)\). Let \(\phi _d\) be the canonical map \(\phi _d:X\longrightarrow P_d(X)\). We have that \(\phi _d\) is \((d+1)\)-connected and that \(F:=\textsf{hofib}(\phi _d)\) is d-connected. In the Serre spectral sequence for \(F\longrightarrow X\longrightarrow K(Z^a,d)\), we have that

$$\begin{aligned} E_{p,q}^2=H_p\left( K\left( {\mathbb {Z}}^a,d\right) ,H_q(F)\right) . \end{aligned}$$

Thus, \(E_{p,q}^2 = 0\) for any \(1\le q\le d\) and for any \(1\le p<d\). Also, \(E_{0,d+1}^2=H_{d+1}(F)\cong \pi _{d+1}(F)\cong \pi _{d+1}(X)\) and, \(E_{d+1,0}^2 = H_{d+1}(K({\mathbb {Z}}^a, d)) = 0\) (the easiest way to see this last fact is that one can build a \(K({\mathbb {Z}}^a, d)\) by killing the homotopy groups of \(\bigvee _a S^d\) in degrees \(d+1\) and higher and this is done by attaching cells of dimensions \(d+2\) and higher).

Therefore, the second page of the sequence looks like

Then, \({\mathbb {Z}}^b\cong {\tilde{H}}_{d+1}(X)\cong E_{0,d+1}^\infty \cong \pi _{d+1}(X)/H\), where \(H=im\left( d_{d+2}^{^{d+2,0}}\right)\). The short exact sequence \(0 \rightarrow H \rightarrow \pi _{d+1}(X) \rightarrow {\mathbb {Z}}^b \rightarrow 0\) splits, because \({\mathbb {Z}}^b\) is free abelian, and so

$$\begin{aligned} \pi _{d+1}(X)\cong {\mathbb {Z}}^b\oplus H. \end{aligned}$$

For the other cases, namely \(k \in \{5,6,13,62\}\), consider the following diagram:

where the inner square is an homotopy push-out and the outer square is a homotopy pullback, i.e., \(Q = \textsf{hofib}(q)\) and \(C = \textsf{hocofib}(i)\). Then, by the Blakers–Massey Theorem (see [2, Theorem 4.2.1, p. 188]), f is \((2d-1)\)-connected. Next, we will look at the long exact sequence of homotopy groups for the fiber sequence \(Q \longrightarrow X \longrightarrow C\) to compute homotopy groups of X up to dimension \(d+k\). For this, we will need some information about the homotopy groups of Q and C.

By the Seifert–van Kampen theorem, C is simply connected, and the Mayer–Vietoris homology sequence for C tells us that C has reduced homology \({\mathbb {Z}}^b\) in degree \(d+k\) and zero otherwise, so that \(C \simeq \bigvee _b {\mathbb {S}}^{d+k}\). The long exact sequence in homotopy then gives us \(\pi _r(Q) \cong \pi _r(X)\) for \(r \le d+k-2\).

As for Q, the Blakers–Massey theorem gave us that \(\pi _r(Q) \cong \pi _r(\bigvee _a {\mathbb {S}}^d)\) for \(r<2d-1\). Taking \(\bigvee _a{\mathbb {S}}^d\) as the d-skeleton of \(\prod _a{\mathbb {S}}^d\), we have that the pair \(\left( \prod _a{\mathbb {S}}^d,\bigvee _a{\mathbb {S}}^d\right)\) is \((2d-1)\)-connected; therefore

$$\begin{aligned} \pi _r(Q) \cong \pi _r\left( \bigvee _a{\mathbb {S}}^d\right) \cong \bigoplus _a\pi _r\left( {\mathbb {S}}^d\right) \end{aligned}$$

for all \(r<2d-1\). Since \(d>k\), we have \(d+k-1<2d-1\), and therefore

$$\begin{aligned} \pi _{d+k-1}(Q)\cong \bigoplus _a\pi _{d+k-1}\left( {\mathbb {S}}^d\right) . \end{aligned}$$

Since we are assuming that \(d>k\), the homotopy groups of spheres appearing in the above direct sum are in the stable range, that is, by the Freudenthal suspension theorem (see [2, Theorem 4.2.8, p. 190]), for any \(i \ge 0\), we have \(\pi _{d+k-1}({\mathbb {S}}^d) \cong \pi _{i+d+k-1}({\mathbb {S}}^{i+d})\); in other words, we have \(\pi _{d+k-1}({\mathbb {S}}^d) \cong \pi ^s_{k-1}\), where \(\pi ^s_{k-1}\) is the \((k-1)\)-st stable homotopy group of spheres, also called the \((k-1)\)-st stable stem. These groups are zero for \(k=5,6,13\) (see [3, Table I, p. 186 and Table III, p. 188]) and \(k=62\) [4], and are, in fact, all the stable homotopy groups of spheres that are known to be zero.

Now, we see that the long exact sequence in homotopy near degree \(d+k\) is

$$\begin{aligned} \pi _{d+k}(Q) \xrightarrow {g_*} \pi _{d+k}(X) \xrightarrow {q_*} \pi _{d+k}(C) \rightarrow 0 \rightarrow \pi _{d+k-1}(X) \rightarrow 0. \end{aligned}$$

Therefore, \(\pi _{d+k-1}(X) = 0\) and, since \(\pi _{d+k}(C)\) is free abelian, we get the desired splitting

$$\begin{aligned} \pi _{d+k}(X)\cong {\mathbb {Z}}^b\oplus \textrm{im}\left( g_*\right) . \end{aligned}$$

Now, in all cases, we can pick a map

$$\begin{aligned} s:\bigvee _b{\mathbb {S}}^{d+k}\longrightarrow X \end{aligned}$$

with \(\textrm{im}(s_*)={\mathbb {Z}}^b \le \pi _{d+k}(X)\), and we will show that the map

$$\begin{aligned} i\vee s:\bigvee _a{\mathbb {S}}^d\vee \bigvee _b{\mathbb {S}}^{d+k}\longrightarrow X \end{aligned}$$

is a homotopy equivalence. Since its source and target are simply connected, it is enough to show it induces and isomorphism on homology, which it clearly does in degree d.

All that is left to show is that s is an isomorphism on \(H_{d+k}\). In the case, \(k = 1\), this is because we had an isomorphism \(\pi _{d+1}(X)/H \cong H_{d+1}(X)\) coming from the Serre spectral sequence. In the other cases, where \(k>1\), the summand \({\mathbb {Z}}^b\) of \(\pi _{d+k}(X)\) was \(\pi _{d+k}(C)\), so that \(q_*\circ s_*: \pi _{d+k}(\bigvee _b {\mathbb {S}}^{d+k}) \rightarrow \pi _{d+k}(C)\) is an isomorphism. By the Hurewicz theorem, \(q \circ s\) is an isomorphism on \(H_{d+k}\), and since q is also an isomorphism on \(H_{d+k}\), we conclude that s is as well. \(\square\)

The lower bound \(d>k\) in the theorem cannot be reduced. The product \(X={\mathbb {S}}^k\times {\mathbb {S}}^k\) is not homotopy equivalent to a wedge of spheres and is a counterexample to a hypothetical \(d \ge k\) generalization of the theorem. It should be noticed that for any CW-complex X which satisfies the hypothesis of the theorem but with \(d=k\) instead of \(d>k\), the suspension \(\Sigma X\) does satisfy all the hypothesis and thus has the homotopy type of a wedge of spheres.

Question 1

Are there more pairs q and k such that the result holds for dimensions d and \(d+k\) for all \(d\ge q\)?

Certainly, if any other stable stem \(\pi ^s_m\) were 0, then \((q,k)= (m+2,m+1)\) would be an example of such a pair, but notice the theorem also holds for \((q,k) = (2,1)\), even though \(\pi ^s_0 \cong {\mathbb {Z}}\) is not zero.

Also, note that for \(k=2\), there is no such q: consider the space \(\Sigma ^n\mathbb{C}\mathbb{P}^2\) which is simply connected, only has nonzero reduced homology in degrees n and \(n+2\) (where it is \({\mathbb {Z}}\)), but is not homotopy equivalent to a wedge of spheres—because \(\textrm{Sq}^2\) is nonzero, for example. Similarly, suspensions of the quaternionic projective plane show that for \(k=4\), there can be no q. The \(\Sigma ^n\mathbb{C}\mathbb{P}^2\) example also shows that the theorem cannot be generalized to spaces with free abelian homology in three or more consecutive dimensions.