We know that $$ O(n+1)/O(n) \simeq SO(n+1)/SO(n) \simeq S^n, $$ based on the result of homogeneous space.
These are in some sense spheres.
If we embed the spin group $Spin(n)$ into $Spin(n+1)$, we may be able to define the quotient space
$$ {Spin(n+1)}/{Spin(n)} \simeq ? $$
- Do we have simpler expressions for the above "manifolds" or "quotient space"?
My Trial/Attempt: We know that $$ Spin(2)\simeq U(1)\simeq SO(2) \simeq S^1 $$ $$ Spin(3)\simeq SU(2)\simeq S^3 $$ $$ Spin(4)\simeq SU(2) \times SU(2) $$ $$ Spin(5)\simeq Sp(2) $$ $$ Spin(6)\simeq SU(4), $$ while $$ Spin(3)/Spin(2) \simeq S^3/ S^1 \simeq S^2 $$ $$ Spin(4)/Spin(3) \simeq (SU(2) \times SU(2))/ S^3 \simeq (S^3 \times S^3 )/ S^3 \simeq S^3 $$ Could we obtain the generic formulas for ${Spin(n+1)}/{Spin(n)} \simeq ? $
Recall that $SO(n+1)$ acts smoothly and transitively on $S^n$ with isotropy subgroup $SO(n)$, so $SO(n+1)/SO(n)$ is diffeomorphic to $S^n$. Now note that $\operatorname{Spin}(n+1)$ also acts smoothly and transitively on $S^n$ by first mapping to $SO(n+1)$. The isotropy subgroup is the subgroup of $\operatorname{Spin}(n+1)$ which projects to $SO(n) \subset SO(n+1)$, this is precisely $\operatorname{Spin}(n)$. Therefore $\operatorname{Spin}(n+1)/\operatorname{Spin}(n)$ is diffeomorphic to $S^n$.
More generally, suppose $G$ is a compact Lie group which acts smoothly and transitively on a smooth manifold $M$ with isotropy subgroup $H$. If $\hat{G}$ is a compact Lie group and $\pi : \hat{G} \to G$ is a surjective Lie group homomorphism, then $\hat{G}$ acts smoothly and transitively on $M$ with isotropy subgroup $\pi^{-1}(H)$, so both $G/H$ and $\hat{G}/\pi^{-1}(H)$ are diffeomorphic to $M$.