# What is the relation between the usual topology of $S^1$ and its subspace topology in Homeo$(S^{n+1})$?

Let the set of all self homeomorphisms of $S^{2n+1}$ - $\operatorname{Homeo}(S^{2n+1})$, be given the compact open topology. Fix $a_0,\cdots,a_n\in\mathbb Z$ to be $n+1$ coprime integers. Let $S^1$ act on $S^{2n+1}$ as follows - $$\lambda\cdot(z_0,\cdots,z_n)=(\lambda^{a_0}z_0,\cdots,\lambda^{a_n}z_n)$$

$($The resulting quotient space is what is known as the weighted projective space $W\mathbb P(a_0,\cdots,a_n))$

The action is clearly faithful and hence we can think of $S^1$ as a sub-group of $\operatorname{Homeo}(S^{2n+1})$. We know that $S^1$ with its usual topology is a compact Lie group.

1. What can be said about $S^1$ as a subspace of $\operatorname{Homeo}(S^{2n+1})$? Will the subspace topology on $S^1$ be the same as its usual topology?

2. Will the topology change depending on the choice of $a_i$?

Thank you.

• I think there's essentially only one compact group structure on $S^1$, so probably yes. – Henno Brandsma Feb 20 '16 at 7:15
• @HennoBrandsma, Thank you for your comment. It gives me a direction to think in. – R_D Feb 20 '16 at 7:18

The induced map $S^1\to \operatorname{Homeo}(S^{2n+1})$ is a continuous injection from a compact space to a Hausdorff space, so it is a homeomorphism onto its image. That is, the subspace topology is the same as the usual topology of $S^1$.
• The map $S^1\times S^{2n+1}\to S^{2n+1}$ is continuous, and so the adjoint map $S^1\to \operatorname{Maps}(S^{2n+1},S^{2n+1})$ is continuous with respect to the compact-open topology. – Eric Wofsey Feb 20 '16 at 7:37
• This works for any topological group $G$ acting continuously on a topological space $X$ right? The map $G\times X\to X$ is continuous $\Rightarrow$ the map $G\to\operatorname{Maps}(X,X)$ is continuous wrt compact-open topology? – R_D Feb 20 '16 at 8:19