I'm self-studying the classification of finite subgroups of $SO(3)$ with this paper.
On page 13 of the paper, there is a paragraph (below the equation (9.3)) that I'm having trouble understanding.
The paragraph:
If two poles $p$ and $p'$ are in the same orbit, their orbits are equal, so $n_p = n_{p'}$ , and therefore $r_p = r_{p'}$ . We label the various orbits arbitrarily, say as $O_1, O_2, ..., O_k$, and we let $n_i = n_p$ and $r_i = r_p$ for $p$ in $O_i$ , so that $n_ir_i = N$. Since the orbit $O_i$ contains $n_i$ elements, there are $n_i$ terms equal to $r_i −1$ on the left side of (9.3) ((9.3) is the counting formula that is derived from the orbit-stabilizer theorem.) We collect those terms together. This gives us the equation $\sum_{i=1}^{k}n_i(r_i-1)=2N-2$
I could understand everything before this paragraph, but the sentence "Since the orbit $O_i$ contains $n_i$ elements, there are $n_i$ terms equal to $r_i −1$ on the left side of (9.3) " is my main problem.
I listed how I understood each sentence of this paragraph.
- If $ap=p'$ and $bp'=p$, then $Gp=Gp'$. (It is clear.)
- I get why $n_ir_i=N$ thanks to the orbit-stabilizer thm, but, I don't understand why we also need "$n_i=n_p$ and $r_i=r_p$" to get "$n_ir_i=N$," which is already clear. I also wonder if $O_1,...,O_k$ are all possible orbits that have some poles in them.
- $O_i$ must have $n_i$ elements by the previous definition in the proof, but why are there $n_i$ terms equal to $r_i-1$? What does $r_i -1 $ mean?
In (9.2), there is $r_p-1$, and I could understand why we get $\sum_{p \in P}(r_p-1)=2N-2$. This is because identity can't have a pole, but except for identity, there are two poles. It looks like the last question is closely related to this, but I can't clearly link them due to the lack of knowledge.
Thus, I wish I could get some help here so that later I can fully understand why there are at most three orbits.
EDIT
I finally found a reason, but I'd like to get verification.
Let $O_1,...,O_k$ be orbits of poles.
But, as the group is $SO(3)$, $O_1,...,O_k$ only contain poles.
Previously, we showed that $Gp=Gp'$ for any arbitrary poles $p,p'$ in some orbit.
And, we also have $\sum_{p\in P} r_p -1 =2N-2$ where $P$ is the set of all poles.
So, if $O_i = \{p_1,p_2,p_3,...,p_{n_i}\}$ for some $i = 1,...,k$ and where $p_1$ is a pole, then $r_{p_1}=...=r_{p_{n_i}}$ and $p_1,...,p_{n_i}$ are all poles.
Therefore, $r_{p_1}-1=...=r_{p_{n_i}}-1$ and there are $n_i$ many terms.
Moreover, $O_i \cap O_j = \phi $ for $i \neq j$.
Hence, $\sum_{p\in P} r_p -1=\sum_{i=1}^{k}n_i(r_i-1)=2N-2$
Did I understand it correctly?