Definition of Lie group

Question

Why Lie group can be saw as a disjoint union of finitely many differentiable manifolds ?

Answer 1

It looks like whatever textbook you found this definition in assumes that manifolds are connected by default. However, this overly rigid definition of a manifold would rule out plenty of Lie groups that we are interested in. For example, the Lie group $O(n)$ of orthogonal matrices has two connected components.

To see why there must be two components, note that the determinant is a continuous map on matrices. Thus, the restriction to orthogonal matrices is a continuous map $\det: O(n) \to \{-1, 1\}$, where both spaces have their usual topologies. Since this map is not constant (since there are orthogonal matrices with determinant $1$ and others with determinant $-1$) then $O(n)$ cannot be connected; it must be a disjoint union of (at least two) manifolds. In fact, there are only two connected components, and they are the preimages $\det^{-1}\{-1\}$ and $SO(n) := \det^{-1}\{1\}$.

To see why there are only two components is a bit more difficult if you don't know much about Lie groups, but one thing you can do is show that any two orthogonal matrices with the same determinant can be joined by a continuous path in $O(n)$. Alternatively, and more generally, if we denote by $G_0$ the connected component containing the identity of a Lie group $G$, then the group $G/G_0$ is the group of path components of $G$. In our case, we have the short exact sequence of Lie group homomorphisms: $$0 \to SO(n) \hookrightarrow O(n) \xrightarrow{\det} \{-1, 1\} \to 0.$$ So it immediately follows that $O(n)/SO(n) \cong \{-1, 1\}$, and hence there are precisely two connected components.

This is explained in any text on Lie groups, but in particular you can find it in section 9.3 of Andrew Baker's Matrix Groups: An Introduction to Lie Group Theory.