# Nontrivial copies of SO(r) in SO(n)

If $G=SO(n)=SO(\mathbb R^n)$ and $r\leq n$, it is easy to find a closed subgroup $H\leq G$ that is isomorphic to $SO(r)$, just let $S\subseteq\mathbb R^n$ be an $r$-dimensional subspace and let $H=\{g\in G:(\forall x\in S^\bot)g(x)=x\}$. These are the trivial examples.

If $r=2$ and $m\geq 4$ we can find nontrivial embeddings $SO(r)\to SO(n)$, for example:

$\begin{pmatrix} \cos t&-\sin t\\ \sin t&\cos t\end{pmatrix} \mapsto \begin{pmatrix} \cos(2t)&-\sin (2t)&0&0\\ \sin(2t)&\cos(2t)&0&0\\ 0&0&\cos(3t)&-\sin(3t)\\ 0&0&\sin(3t)&\cos(3t)\end{pmatrix}$

Are there examples for $r=3$ (or more)?

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You are just asking for injective homomorphisms $SO(r)\into SO(n)$ or, equivalently, faithful orthogonal representations of $SO(r)$. They are built from the irreducible ones. For fixed $r$, the irreducible ones occur for special values of $n$ (but infinitely many of them), though the case $r=2$ is special. This all fits in the framework of representations of compact Lie groups. You need to consult a book. – Claudio Gorodski Jan 25 '13 at 19:04
What do you mean exactly by 'nontrivial' embedding? For every embedding $SO(r)\rightarrow SO(n)$, you can just conjugate the image by an inner automorphism. – Name Jan 25 '13 at 19:39
(Side question: Can a non-closed subgroup of $SO(n)$ be isomorphic to $SO(r)$?) – Mariano Suárez-Alvarez Jan 25 '13 at 19:54
@Yazdegerd III, but there are several classes of embeddings, even up to conjugatio. The classification is, as Claudio notes, the classification of faithful orthogonal reps of $SO(r)$ of a given dimension. – Mariano Suárez-Alvarez Jan 25 '13 at 19:57
@mariano: do you mean topologically isomorphic, or abstractly isomorphic? If a subgroup of $SO(n)$ is topologically isomorphic to $SO(r)$, then it is compact, and therefore closed. But if you are thinking about an abstract isomorphism, then for example $SO(2)$ is abstractly isomorphic to a direct sum of $Q/Z$ and an uncountable number of $Q$s. I think it is not hard to embed this group for example into $S^1\times S^1$ and therefore into $SO(4)$. – Gregory Arone Jan 25 '13 at 21:14