0
$\begingroup$

My question is:

How to find out all the finite subgroup of SO(n)? Or just for the simple case SO(4) SO(5)?

With more discribe:

If $S^n\backslash \Gamma$ is a manifold,

I just want to know that how many subgroups of SO(n) can be realized as the $\Gamma$ above?

If for generic n, it may be difficult, can we just figure out the case when n=3,4, I only care about $S^3$ and $S^4$

$\endgroup$
4

1 Answer 1

5
$\begingroup$

There are a pair of double covers $\text{SU}(2) \times \text{SU}(2) \to \text{SO}(4) \to \text{SO}(3) \times \text{SO}(3)$, and the first resp. the second more or less reduces the classification of finite subgroups of $\text{SO}(4)$ to the classification of finite subgroups of $\text{SU}(2)$ resp. $\text{SO}(3)$ (these classifications are in turn more or less equivalent and well-known) using Goursat's lemma. This classification can be found more explicitly in Conway and Smith's On Quaternions and Octonions (Section 4.3).

There is also a double cover $\text{Sp}(2) \to \text{SO}(5)$ which reduces the classification of finite subgroups of $\text{SO}(5)$ to the classification of finite subgroups of $\text{Sp}(2)$. This is probably easier but I don't know if it's well-known (in any case I don't know it).

$\endgroup$
1
  • $\begingroup$ Thank you for helping me! These days I cann't assess the Internet, sorry to accept it in such later time. $\endgroup$
    – Siqi He
    Commented Aug 8, 2012 at 1:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .