# Off _topic question

I would appreciate if you help me to underestand why is this question counted as off_topic.

Is realy all embedding of $F_{2}$ in $SO3$ an obvious question?

Thank you.

Edit: The MO question has now been deleted by OP.

• Your question is about (infinite, nonabelian) compact subgroups $G$ of $SO(3)$, and one thing that came out in comments was the assertion that the only one is $SO(3)$. I believe that is correct. I think it comes down to checking that the normalizer of an $S^1$ in $SO(3)$ is the same subgroup $S^1$. I think this is something you can check with your bare hands; it might be easier to lift the problem up to the universal cover which is the group of unit quaternions, and check it there. See if that helps.
– Todd Trimble Mod
Jan 31, 2015 at 14:28
• In case my last comment was cryptic: certainly a connected compact subgroup of $SO(3)$ is either trivial, $S^1$, or all of $SO(3)$ (this is essentially a Lie algebra computation). And the connected component of a compact subgroup $G$ will be normal in $G$ and of finite index. So the only nontrivial case to check is where the connected component is an $S^1$; since $G$ is contained in the normalizer, it suffices to show the normalizer is just $S^1$ itself.
– Todd Trimble Mod
Jan 31, 2015 at 15:44
• @ToddTrimble Thank you very much for your interesting comments Jan 31, 2015 at 18:19
• @ToddTrimble what about the pure algebraic version of my question.(Not taking closure): are there two copy of F2 in SO3 which can not be carry to each other via an automorphism of SO3? Jan 31, 2015 at 18:47
• Ali, I suspect that all Lie group automorphisms $\phi$ of $SO(3)$ are inner (e.g., see: mathoverflow.net/questions/40666/…). Thus, given $g, g' \in SO(3)$, the angle between the axes of rotation for $g$ and $g'$ should equal the angle between the axes of rotation for $\phi(g), \phi(g')$. On the other hand, "most" homomorphisms $F_2 \to SO(3)$ should be embeddings, and by a cardinality argument the countable sets of angles you get from two such embeddings will usually differ. So $Aut(SO(3))$ should not act transitively on images of embeddings.
– Todd Trimble Mod
Jan 31, 2015 at 19:33

To clarify: the OP actually posed several questions in a barrage within this one, and my reaction in the comments to the original post was to this part:

We fix an embedding of $F_2$, the free group on two generators, in $SO(3)$ and we put $G=\overline{F_2}$. Then we construct the following $C^∗$ algebra: $A=C^∗(G)\cong C^*_r(G)$. What can be said about $A$? Is it simple? Is it idempotentless? Is its $K$-theory computed already?

The OP has asked many questions which involve $C^*$-algebras, clearly coming from a background which is not familiar with them. In itself that is not a problem, but in my personal view, an experienced user of MO should start to learn from what he or she gains from answers. I don't think MO should be used as a substitute for thinking about one's own questions carefully; for me it is somewhere to turn to when one gets stuck, not somewhere one tries as soon as one has thought of a question. Hence my vote to close.

The $C^*$-algebra of a compact group is isomorphic to a $c_0$-direct sum of full matrix algebras. Even if one has not come across this result in one's own background reading about $C^*$-algebras, surely as soon as one defines the group $G$ in the original question, one should be thinking

"I don't know what $G$ looks like but it is compact, so my question is about the $C^*$-algebra of a compact group, let me look online for ideas or consult a textbook or set of notes that deal with $C^*$-algebras of non-discrete groups."

It is the apparent absence of this first level of self-reflection, which I feel makes the question not really MO-quality. We are all meant to be researchers or potential researchers; we must surely aspire to think through some things on our own sometimes.

• +1 thank you very much for you interesting answer. Jan 31, 2015 at 18:18
• May be I should read the MO aims more carefully. But flexibility of participants could help the beginners in an area. Jan 31, 2015 at 18:42