3
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – t.b. Aug 18 '12 at 20:59
  • $\begingroup$ yes, In the question paper it was proper inclusion $\endgroup$ – Marso Aug 18 '12 at 21:03
  • 2
    $\begingroup$ 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. $\endgroup$ – t.b. Aug 18 '12 at 21:04
  • 3
    $\begingroup$ Or simply consider any connected Lie group to see that 2 is wrong. $\endgroup$ – Alex Becker Aug 18 '12 at 21:05
  • $\begingroup$ @t.b. yes it was like that. $\endgroup$ – Marso Aug 18 '12 at 21:06
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.