What is an explicit formula for $\Omega (S^n \vee S^m)$? I know that it follows from Hilton-Milnor theorem. But I don't quite understand it's formulation.
2026-03-25 23:38:38.1774481918
Loop space of wedge sum of spheres
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The reference for Hilton's theorem is
In fact it's difficult to work out combinatorially which spheres appear in the decomposition of $\Omega(S^n \vee S^m)$. A possible reference is in
More precisely Problem 17.68. The Hilton–Milnor–Gray theorem states that:
If $m \le n$, you can apply this theorem to compute $\Omega(S^m \vee S^n)$ inductively, because the second term in the product will always become more and more connected, and so the map to the product you find will have a contractible homotopy fiber. According to Strom it was done by Hilton and put in bijection with the generators of a free Lie algebra, but I didn't find a reference.
To give an example of computation, here is the beginning for $\Omega(S^2 \vee S^2)$: $$\Omega(\Sigma S^1 \vee \Sigma S^1) \simeq \Omega\Sigma S^1 \times \Omega\Sigma \left( S^1 \vee \bigvee_{n=1}^\infty S^1 \wedge (S^1)^{\wedge n} \right) \\ \simeq \Omega S^2 \times \Omega\Sigma \left(S^1 \vee \bigvee_{n=1}^\infty S^{n+1} \right) \\ \simeq \Omega S^2 \times \Omega \left( S^2 \vee \Sigma \left(\bigvee_{n=1}^\infty S^{n+1} \right) \right) \\ \simeq \Omega S^2 \times \Omega S^2 \times \Omega\Sigma\left( \bigvee_{n=1}^\infty S^{n+1} \vee \bigvee_{k=1}^\infty \bigvee_{n=1}^\infty S^{k+n} \right)$$ and after that it gets complicated.