I'm ready right now in a paper, that $S^1$ acts on $SO(n+1)/SO(n-1)$ by right translations.
I thought that a Liegroup $G$ acting by right translations, means that we have a right action $\varphi \colon G \times G \to G, \ (g,h) \mapsto \varphi_g(h)$ with $\varphi_g \circ \varphi_h = \varphi_{hg}$.
But we have clearly, that $SO(n+1)/SO(n-1) \neq S^1$. So what do the author mean by "$S^1$ acts on $SO(n+1)/SO(n-1)$ by right translations"?
Edit: Maybe I should mention, that in this paper we get an action of $SO(n+1) \times S^1$, so that $S^1$ has to commute with the action of $SO(n+1)$.
This should refer to the fact that $SO(2)\times SO(n-1)$ naturally is a subgroup of $SO(n+1)$. Since this means that translations with respect to $SO(2)$ commute with translations with resprect to $SO(n-1)$, it provides a natural action of $SO(2)$ on $SO(n+1)/SO(n-1)$.