No, it is not wrong to write that, you're spot on the mark; therefore, your conclusion is right. Your example has an interesting generalisation beyond $O(3)$ and indeed beyone Lie groups as the following is true for all topological groups. For a topological group $\mathfrak{G}$, the "identity component" $\mathfrak{G}_\mathrm{id}$ (i.e. the connected component of the group with the identity in it) is always:
- Clopen (closed and open at once), therefore, by the connectedness argument, the whole group $\mathfrak{G}$ is connected if and only if $\mathfrak{G}_\mathrm{id}$ is a proper subset of $\mathfrak{G}$;
- A normal subgroup of $\mathfrak{G}$, therefore the connected components of $\mathfrak{G}$ are precisely $\mathfrak{G}_\mathrm{id}$ and its cosets.
See the discussion as Theorem 9.31 on my website here of this situation. Mine is a retelling of this slick little proof that I found a long time ago in
Sagle, A. A. and Walde, R. E., “Introduction to Lie Groups and Lie Algebras“, Academic Press, New York, 1973. §3.3
This is widely known, BTW, but it's a little pearl of a proof, showing how powerful the connectedness argument is. I still enjoy reading it, just like I still like listening to "Walk Like an Egyptian" (for me, they come from about the same time of life!)
In your example, $\mathfrak{G}=O(3)$, the identity component is the smallest Lie group containing $\exp(\mathfrak{g})$, where $\mathfrak{g}=\operatorname{Lie}(\mathfrak{G}) = \mathfrak{so}(3)$; indeed, in this compact case, $\mathfrak{G}_\mathrm{id}=\exp(\mathfrak{g})$ is the whole connected component (in noncompact groups, e.g. $SL(2,\mathbb{C})$, the identity component is strictly bigger than $\exp(\mathfrak{g})$). $\mathfrak{so}(3)$ is of course the Lie algebra of skew-symmetric, real $3\times 3$ matrices and, since $\det(\exp(H)) = \exp(\mathrm{tr}(H))=1$ for any such matrix $H$, we see that $SO(3)$ is the whole of $\mathfrak{G}_\mathrm{id}$. It is a normal subgroup of $O(3)$, and the group of cosets is simply $O(3)/SO(3)\cong\{+1,\,-1\}\cong\mathbb{Z}_2$, which is another way of writing your decomposition.
Answer to edit 1: For manifolds, which are locally Euclidean (locally homeomorphic to $\mathbb{R}^N$) path connectedness and connectedness are the same notion. Moreover, path connectedness always implies connectedness (to prove this, assume otherwise and let $\alpha,\,\beta\in\mathbb{X}$ belong to separate connected components $\mathbb{U},\,\mathbb{U}^\sim$ linked by path $\sigma:[0,\,1]\to\mathbb{X}$ where $\mathbb{X}$ is the topological space in question and $\mathbb{X}=\mathbb{U}\bigcup\mathbb{U}^\sim$. By assumption (path connectedness), $\sigma$ is continuous when $[0,\,1]$ has its wonted topology. But $\mathbb{U},\,\mathbb{U}^\sim$ are disjoint, therefore so is $\mathbb{V}=\sigma([0,\,1])\bigcap\mathbb{X}$, so the inverse image $\sigma^{-1}(\mathbb{V})$, namely $[0,\,1]$, must also be the union of disjoint open sets, contradicting the known connectedness of $[0,\,1]$). However, not all connected topological spaces are path connected (look up the weird "topologist's Sine Curve" as a counterexample).
Answer to Edit 2. We have actually already done this above, because every matrix in $SO(3)$ can be written as a $\exp(H)$, where $H\in\mathfrak{so}(3)$, so you can take your path to be $\sigma:[0,\,1]\to SO(3);\;\sigma(\tau) = e^{\tau\,H}$. To prove that every $SO(3)$ matrix can be written in this was from first principles, simply witness that $\gamma\in SO(3)$ is normal, i.e. commutes with its Hermitian transpose and thus always has a diagonalisation with orthonormal eigenvectors, so $\exists U\ni\,\gamma=U\,\Lambda\,U^\dagger$, where $\Lambda$ is the diagonal matrix of eigenvalues, none of which are nought ($SO(3)$ is a group). Thus, you can always define $\log\gamma = H = U\,\log\Lambda\,U^\dagger$, and you are done.
Note that this does not work for $O(3)$, because now $H$, even though definable as $H = \log \gamma$, $H$ now has diagonal elements if $\gamma\not\in SO(3)$, so $H\not\in\mathfrak{so}(3)$ and you can't find a path through $O(3)$'s charts, because $\mathfrak{so}(3)$ is the Lie algebra of $O(3)$ as well.
This post imported from StackExchange Mathematics at 2014-10-05 10:06 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
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