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I see Arnold in his article sets the $S^4$ result in a different series, in terms of a trinity
$\mathbb{R}P^1 \simeq S^1; \mathbb{C}P^2/Conj \simeq S^4; (\mathbb{H}P^4/Aut(\mathbb{H}))/Conj \simeq S^13.$1 to 3 of 3