Loop space of wedge sum of spheres 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.
 A: The reference for Hilton's theorem is

Hilton, P. J. (1955), "On the homotopy groups of the union of spheres", Journal of the London Mathematical Society. Second Series 30 (2): 154–172, doi:10.1112/jlms/s1-30.2.154, ISSN 0024-6107, MR0068218.

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

Jeffrey Strom. Modern classical homotopy theory. Graduate Studies in Mathematics 127. American Mathematical Society, Providence, RI, 2011, pp. xxii+835. ISBN: 978-0-8218-5286-6. MR2839990.

More precisely Problem 17.68. The Hilton–Milnor–Gray theorem states that:

Theorem (Hilton–Milnor–Gray). If $A$ and $B$ are well-pointed spaces, then $$\Omega(\Sigma A \vee \Sigma B) \simeq \Omega(\Sigma A) \times \Omega\Sigma \left( B \vee \bigvee_{n=1}^\infty B \wedge A^{\wedge n} \right).$$

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 $\Omega\Sigma(S^2 \vee S^2)$:
$$\Omega(\Sigma S^1 \vee \Sigma S^1) \simeq \Omega\Sigma S^1 \times \Omega\Sigma \left( S^1 \vee \bigvee_{n=1}^\infty S^1 \wedge (S^1)^{\wedge n} \right) \\
\simeq \Omega S^2 \times \Omega\Sigma \left(S^1 \vee \bigvee_{n=1}^\infty S^{n+1} \right) \\
\simeq \Omega S^2 \times \Omega \left( S^2 \vee \Sigma \left(\bigvee_{n=1}^\infty S^{n+1} \right) \right) \\
\simeq \Omega S^2 \times \Omega S^2 \times \Omega\Sigma\left( \bigvee_{n=1}^\infty S^{n+1} \vee \bigvee_{k=1}^\infty \bigvee_{n=1}^\infty S^{k+n} \right)$$
and after that it gets complicated.
