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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

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I think it depends on $n$ (divisible by $4$). Write $G=Sp(n/4)/\mu_2$. If $n=4$, then $G\simeq SO_3$, so $G$ admits a faithful 4-dimensional representation. Similarly, if $n=8$, then $G\simeq SO_5$, hence $G$ admits a faithful 8-dimensional representation. (Of course, in these cases $S^{n-1}$ is not an orbit.) It seems that for $n\ge 12$ the group $G$ has no nontrivial representations of dimension $\leq n$, hence cannot be embedded into $SO_n$. –  Mikhail Borovoi Jan 9 '13 at 18:55
    
Thank so much for your helpful comments. I need to be sure about the last part of your comment. Could you introduce me a reference to check that $G$ has no nontrivial representations of dimension $\leq n$. –  purelymath Jan 10 '13 at 10:49
    
The complex simple algebraic group $Sp_{m,\mathbb{C}}$ of $2m$-dimensional space $V$ has, for $m\ge 2$, an irreducible representation of dimension $m(2m-1)-1$ in a subspace of codimension 1 of the space $\Lambda^2 V$. I think that it is the irreducible representation of smallest dimension after $V$ itself - ask this in Math Overflow! On the other hand, for $m\ge 3$ we have $m(2m-1)-1>4m$. –  Mikhail Borovoi Jan 10 '13 at 11:51
    
OK, I'll ask it in Math Overflow. Excuse me for the following question if it is easy, I'm not well in representation theory. Let's think your guess is true about representation of $Sp_{m,\mathbb{C}}$, then how can we get your assertion about representation of $G=Sp(m)/\mathbb{Z}_2$? –  purelymath Jan 10 '13 at 13:10
    
A real $d$-dimensional representation of the real algebraic group $Sp(m)/\mu_2$ gives, by complexification, a complex $d$-dimensional representation of the complex algebraic group $Sp_{m,\mathbb{C}}/\mu_2$, which in turn gives a complex $d$-dimensional representation of $Sp_{m,\mathbb{C}}$. –  Mikhail Borovoi Jan 10 '13 at 13:56

1 Answer 1

up vote 3 down vote accepted

(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 $4r$.

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@Mikhail: In the proof, irreducibility of the representation of $G$ as a subgroup of $SO(n)$ comes from the hypothesis of transitivity of its action on the sphere $S^{n−1}$, true? if it is so, what happens if we drop the transitivity hypothesis? can we still deduce that $G$ can't be a subgroup of $SO(n)$? –  Nrd-Math Jan 17 '13 at 11:38
    
@Nerd-Math: Not true. Any complex representation $\rho$ of dimension $n=4r$ of the compact group $G$ is a direct sum of irreducible representations $\rho_i$. Clearly ${\rm dim}\ \rho_i\le n=4r$. We have seen that for $r\ge 3$ our group $G$ has no nontrivial irreducible representations of dimension $\le 4r$. Thus each $\rho_i$ is trivial, hence $\rho$ is trivial. –  Mikhail Borovoi Jan 17 '13 at 14:12

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