I am trying to prove that the isometry group of Riemannian manifold is Lie transformation group. One of the tools used is the following theorem but in place of $M$ we work on $O(M)$ and the parallelism structure are given by the canonical 1-form $\theta=\theta^{i}$ of $L(M)$ restricted to $O(M)$ and by the connection forms $\omega=\omega^{j}_{k},j<k$, and the problem the author is assuming $M$ to be connected in Lemma 1, so my question is this theorem hold for $M$ with finite number connected components and what's the case with infinite connected components?
In general, the bundle of orthonormal frame bundle isn't connected see.
Refrence: Kobayashi transformation group in differential geometry.
The possible answer in case we have finite connected components $\{K_{i}\}_{i}$ of $M$, we simply apply the trick for each connected components with absolute parallelism restricted to $K_{i}$, hence $\mathfrak{U}(K_{i})$ is Lie transformation group, we take the product of $\{\mathfrak{U}(K_{i})\}_{i}$, this a finite product of lie transformation group hence its lie transformation group and consider as normal subroup of $\mathfrak{U}(M)$ hence we transfer the topology and lie group structure from this normal subgroup to the entier group of automorphism of absolute parallelism. Other question I don't really know precisely but I think that because we dealing with compact open topology, that is above groups are locally compact with this topology in case of finite connected components.