Looking for proof that $SO(3)$ is a submanifold of $\mathbb R^3$ It seems to be taken for granted in all sources that $SO(3)$ is a submanifold of $\mathbb R^9$.  However, the one proof of this that I have been able to find has a step or two that doesn't make alot of sense to me.  (In Frankel, "The Geometry of Physics", if anyone's interested.)  Can anyone point me to -- or even supply right here -- a clear proof of this theorem?
 A: Here's a proof: 


*

*It's homogeneous. In fact, multiplication by $\mathbf A$ takes a neighborhood of $\mathbf I$ to a neighborhood of $\mathbf A$ by an isometry, for any $\mathbf A \in \mathbf SO(3)$. So you need only prove it's manifold-like at $\mathbf I$. 

*It's manifold-like at $\mathbf I$. For this, use the implicit function theorem. That says that if you have a map $F: \mathbb R^9 \to \mathbb R^6$ that's regular at $I$, then the preimage near $\mathbf I$ is diffeomorphic to $\mathbb R^3$. The map we'll use consists of $6$ components: the dot product of the $i$th column with the $j$th column for $(i, j) = (1,1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3)$. The first three make sure that the columns are unit vectors; the second three that the columns are orthogonal to one another. Clearly $F^{-1}(0,0,0,0,0,0) = O(3)$. To finish things up, you have to show that $F$ is regular at $\mathbf I$, i.e., that $DF(\mathbf I)$ has rank 6. I leave this to you, since it's a simple computation. 

*Note that this also proves that $O(3)$ is a manifold, by the same reasoning, more or less. SO(3) is just the connected component of the identity. 
