0
$\begingroup$

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)?

$\endgroup$
6
  • 1
    $\begingroup$ 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. $\endgroup$ Jan 25, 2013 at 19:04
  • $\begingroup$ 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. $\endgroup$
    – Name
    Jan 25, 2013 at 19:39
  • $\begingroup$ (Side question: Can a non-closed subgroup of $SO(n)$ be isomorphic to $SO(r)$?) $\endgroup$ Jan 25, 2013 at 19:54
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 25, 2013 at 19:57
  • $\begingroup$ @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 $SO(4)$. $\endgroup$ Jan 25, 2013 at 21:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.