Confusion about quotient of the Lie group $\mathbb{S}^1$

Question

I have read that given a Lie group $G$ and a closed subgroup $H$ then $G/H$ is a smooth manifold. I cannot explain though the following example: take as $G = \mathbb{S}^1$ and as $H =\{\pm 1\}$, $H$ is a closed subgroup but the quotient should look like figure $8$, and thus is not even a topological manifold.

I am surely making some mistake, can you help me in finding it?

Answer 1

The quotient $G/H$ here does not identify just the points in $H$. Like an ordinary quotient of groups, it identifies all the points in the cosets $gH$ for $g \in G$. In this case, it means that for each $z \in G$, the set $\{z, -z\}$ is identified to a point. Can you figure out what the resulting figure is?

Answer 2

You are misunderstanding the meaning of quotient. The quotient $S^1/H$ is the set of all sets $\{z,-z\}$, with $z\in S^1$. It turns out to be homeomorphic to $S^1$ itself. Just consider the map$$\begin{array}{ccc}S^1/H&\longrightarrow&S^1\\\{z,-z\}&\mapsto&z^2.\end{array}$$