Lie Groups: Identity Component

Main Problem

Given a Lie group.

The connected component of the identity is a Lie subgroup:

1. It is a subgroup.

2. It is open.

How to check this using topological tools?

Extra Problem

The quotient by the above is the group of connected components: $$G_e\lhd G:\quad (gG_e)(hG_e)=(ghG_e)=G_{gh}$$

How to check this using topological tools?

• I don't understand 2. Are you sure you want $\subsetneq$, not $\subseteq$ (in fact you have equality)? The first two properties immediately follow from connectedness and continuity of the maps involved. The fourth follows directly from 1.,2.,3. It remains to ponder 3. – t.b. Aug 18 '12 at 20:59
• yes, In the question paper it was proper inclusion – Marso Aug 18 '12 at 21:03
• Then it is wrong. Was it $\subset$? Note that many people allow equality when writing $\subset$. Consider some examples, like $O(2)$ and $O(3)$, or whatever Lie group you understand well. – t.b. Aug 18 '12 at 21:04
• Or simply consider any connected Lie group to see that 2 is wrong. – Alex Becker Aug 18 '12 at 21:05
• @t.b. yes it was like that. – Marso Aug 18 '12 at 21:06

It is open because every element of a Lie group (or indeed any manifold) has an open neighbourhood which is homeomorphic to $\Bbb{R}^n$ and is in particular connected.
To see it is a subgroup, denote by $G^\circ$ the connected component of the identity of $G$. The map $$\mu:G^\circ\times G^\circ\to G$$ defined by $$(g,h)\mapsto g\cdot h$$ is continuous, so its image is connected (because $G^\circ\times G^\circ$ is connected). It also contains the identity (because $\mu(1,1)=1$), so it must lie inside the connected component of the identity, i.e. $g\cdot h\in G^\circ$ for every $g,h\in G^\circ$. Similarly (only using the map $\iota:G^\circ\to G$ defined by $g\mapsto g^{-1}$ instead of $\mu$) we can see that $g^{-1}\in G^\circ$ for every $g\in G^\circ$. Thus, $G^\circ$ is a subgroup. (You can also see this way it is a normal subgroup).