Question

Hello,
One of the subgrouops of $SO(n)$ which acts transitively on the sphere $S^{n-1}$ is the (compact) symplectic group $Sp(n/4)$. The center of $Sp(m)$ is isomorphic to $\mathbb{Z}_2$. Can we embed $Sp(n/4)/\mathbb{Z}_2$ in $SO(n)$ (as a Lie subgroup)? If the answer is yes, what happens to its orbits? Precisely, is $S^{n-1}$ one of its orbits?
Thanks

Answer 1

(I add details to my comments.) The answer depends on $n=4r$. Write $G=Sp(r)/\mu_2$. If $r=1$, then $G\simeq SO_3$, so $G$ admits a faithful 4-dimensional representation into $SO_4$. Similarly, if $r=2$, then $G\simeq SO_5$, hence $G$ admits a faithful 8-dimensional representation into $SO_8$. (Of course, in these cases $S^{4r-1}$ is not an orbit.) For $r\ge 3$ the group $G$ has no nontrivial representations of dimension $4r$, see below, hence it cannot be embedded into $SO_{4r}$.

An irreducible real $n$-dimensional representation of the real algebraic group $G$ induces an irreducible complex $n$-dimensional representation of $G_{\mathbb C}=Sp_{r,{\mathbb C}}/\mu_2$. The irreducible complex representations of the simple group $\widetilde G_{\mathbb C} =Sp_{r,{\mathbb C}}$ of type $C_r$ for $r>1$ of dimension $n<{\rm dim}\ \widetilde{G}_{\mathbb C}$ are listed in the paper of Andreev, Vinberg, and Elashvili, Table 1. They are the fundamental irreducible representations $R(\pi_1)$ of dimension $2r$, $R(\pi_2)$ of dimension $2r^2-r-1$, and, for $r=3$, $R(\pi_3)$ of dimension 14. Since the representations $R(\pi_1)$ and $R(\pi_3)$ are nontrivial on the center $Z(\widetilde G_{\mathbb C})\simeq\mu_2$, we see that the only nontrivial irreducible representation of $G_{\mathbb C}$ of dimension $n<{\rm dim}\ G_{\mathbb C}$ is the representation $R(\pi_2)$ of dimension $2r^2-r-1$, hence $R(\pi_2)$ is the irreducible representation of $G_{\mathbb C}$ of the smallest dimension. For $r\ge 3$ we have $2r^2-r-1>4r$, hence $G_{\mathbb C}$ has no nontrivial representations of dimension $4r$.