Connectedness of punctured $G$

Question

If I take a connected topological group $G$ and I remove the indentity from $G$, when I can say that $G-\lbrace 1 \rbrace $ is connected?

Any suggestion or reference is appreciated.

Answer 1

Suppose that the dimension of $G$ is finite $G-\{1\}$ is connected if $dim_G>1$ or if $G$ is a circle.

Answer 2

For a connected topological group $G$ the following are equivalent:

1. $G \setminus \{1\}$ is disconnected.
2. $G$ is algebraically isomorphic to $\mathbb{R}$ and the topology is finer than the order topology.

The implication $2. \implies 1.$ is easy. The converse is the subject of a lengthy exercise in Bourbaki's General topology (1974). [Chapter V, section 3, exercise 4]

A brief overview of the proof: first it is proved that $G \setminus \{1\}$ has only two components, $A$ and $B$, with $B = A^{-1}$. It is then shown that $x \preceq y \iff xy^{-1} \in A$ defines a translation-invariant order on $G$ and that the order topology is coarser than the given topology. The conclusion then follows from the fact that in a connected linear order space every bounded set has a supremum, and that $\mathbb{R}$ is the only non-discrete totally ordered group in which this is true.