I'm trying to compute ranks of the following homotopy groups: $\pi_{n}(S^4 \vee S^5) \otimes \Bbb Q$ for $n\leq 15$.

My thoughts: As I know that there are two ways of computation:

  1. Using the Sullivan model
  2. Using the Hilton theorem

Hilton theorem from his paper "ON THE HOMOTOPY GROUPS OF THE UNION OF SPHERES" allows to compute homotopy groups of wedge sum of spheres $\pi_n(S^{r_1+1}\vee \cdots \vee S^{r_m+1})$ for $r_i\geq 1$

Theorem states that $\pi_n(T)=\sum_{i=1}^{\infty} \pi_n(S^{q_{i}+1})$ where $T=S^{r_1+1}\vee \cdots \vee S^{r_m+1}$. I have the problem with the heights $q_s$ in this equation. I don't understand how to describe homotopy groups using this equation via defining heights for spheres of various dimensions.

I know how to use Hilton theorem for the wedge sum of two spheres of the same dimension for computation of ranks. I don't understand how to apply Hilton's theorem for the wedge sum of spheres with various dimensions.

Can you explain please explicitly. It would be great if someone would show these two ways of calculating ranks.

  • 1
    $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Dec 26, 2022 at 11:32
  • 1
    $\begingroup$ Why do you need to describe $\pi_n(T)$ when $T$ is a wedge of three or more spheres? You state in the first line that you are interested in computing $\pi_n(S^4\vee S^5)\otimes\mathbb{Q}$. Also, what is the problem with the $q_i$'s? Is it that you are looking for exact descriptions of these integers? $\endgroup$
    – Tyrone
    Dec 28, 2022 at 9:03

1 Answer 1


In case it's still relevant to you, here is an answer to your problem: We have $T = S^4 \vee S^5 = S^{r_1+1} \vee S^{r_2+1}$, so $r_1=3$ and $r_2 = 4$. In order to compute the numbers $q_i$, we need to compute the first couple of basic products $p_i$. To this end, consider the positive generators $\iota_1$ and $\iota_2$ of $\pi_3(S^3)$ and $\pi_4(S^4)$ respectively. Then by definition of basic products (see Hilton's paper), we get the following table: \begin{array}{c|c|c|c|c} i& p_i & w_{i1} & w_{i2} & q_i \\ \hline 1 & \iota_1 & 1 & 0 & 3 \\ 2 & \iota_2 & 0 & 1 & 4 \\ 3 & [\iota_1,\iota_2] & 1 & 1 & 7 \\ 4 & [\iota_1, [\iota_1,\iota_2]] & 2 & 1 & 10 \\ 5 & [\iota_2, [\iota_1,\iota_2]] & 1 & 2 & 11 \\ 6 & [\iota_1, [\iota_1, [\iota_1,\iota_2]]] & 3 & 1 & 13 \\ 7 & [\iota_2, [\iota_1, [\iota_1,\iota_2]]] & 2 & 2 & 14 \\ 8 & [\iota_2, [\iota_2, [\iota_1,\iota_2]]] & 1 & 3 & 15 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ \end{array} Note that we computed the numbers $q_i$ by the formula that is stated in the paper, namely $q_i = \sum_{j=1}^kr_jw_{ij}$, where $w_{ij}$ denotes the number of occurences of $\iota_j$ in the $i$-th basic product. Thus, in our case we just have $q_i = 3w_{i1}+4w_{i2}$. Since you're only interested in the cases $n \leq 15$, we don't need to compute any more heights $q_i$, since all further summands $\pi_n(S^{q_i+1})$ will be trivial. This follows for instance from the following lemma, giving a lower bound on the heights:

Lemma: Let $T = S^{r_1+1} \vee S^{r_2+1}$ and let $q$ be the height of a basic product of weight $w$. Then $q \geq r_1(w-1)+r_2$.

For us, the next basic products in the above table are those of weight $w \geq 5$, so indeed we see that $q \geq 3(w-1)+4 \geq 16$, so we don't need to compute any further.

Next, it is just a matter of plugging the values for $q_i$ into Hilton's main theorem $\pi_n(S^4 \vee S^5) \cong \bigoplus_{i=1}^\infty \pi_n(S^{q_i+1})$. This is a bit tedious, but you can look up tables of homotopy groups of spheres for these calculations. Since most summands are torsion, you will find that for $n \leq 15$ we have $$ \pi_n(S^4 \vee S^5) \otimes \mathbb{Q} \cong \begin{cases} \mathbb{Q} & n=4,5,8,11,12,14,\\ \mathbb{Q} \oplus \mathbb{Q} & n=15,\\ 0 & \text{else}, \end{cases} $$ and hence we have also computed the ranks of these groups. I hope this is helpful!


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