I'm currently reading "Groups acting on the circle" by E. Ghys and came along the group $Homeo_+(\mathbb{S}^1)$ of orientation-preserving circle homeomorphisms and read that it doesn't form a Lie group. The argument was that not every element $f\in Homeo_+(\mathbb{S}^1)$ close to the identity lies on a one-parameter-subgroup. If I understand it correctly, that means that there's no group homomorphism $\phi: \mathbb{R}\to Homeo_+(\mathbb{S}^1)$, s.t. $\phi(t) = f$ for some $t$. To prove this claim, the following proposition was used:
(We get a description of those elements of $Homeo_+(\mathbb{S}^1)$ which have the form $\phi^1$ for some topological flow $\phi^t$ on the circle.)
Proposition 5.10. An element $f$ of $Homeo_+(\mathbb{S}^1)$ can be included in a topological flow iff the rotation number of $f$ is $0$, or $f$ is conjugate to a rotation.
My question now is the following: Is the "flow" meant as map $\phi: \mathbb{S}^1 \times \mathbb{R}\to \mathbb{S}^1$ or $\phi: Homeo_+(\mathbb{S}^1)\times \mathbb{R} \to Homeo_+(\mathbb{S}^1)$? Because the wording makes me think it's the first version but then I don't understand how this proposition can be used in proving that it's not a Lie group.
Thanks for any help!
I'm not engaging with the paper closely, but:
It seems like the first version is exactly what you want here. A map $\phi: \mathbb{S}^1 \times \mathbb{R} \to \mathbb{S}^1$, is (by currying) a map $\widetilde{\phi}: \mathbb{R} \to \text{Homeo}_+(\mathbb{S}^1, \mathbb{S}^1)$ (specifically, $\widetilde{\phi}(t)(x) = \phi(x, t)$). (In practice everyone would use the same symbol $\phi$ for both maps but I'll keep the distinction here for clarity.)
If $\phi$ is a "flow" that ought to mean this map gives a one-parameter subgroup of $\text{Hom}(\mathbb{S}^1, \mathbb{S}^1)$---and vice versa, one-parameter subgroups should be given by such flows. (Recall that a one-parameter subgroup of a Lie group $G$ is a smooth homomorphism $(\mathbb{R}, +) \to G$.)
So the existence of elements $\psi \in \text{Homeo}_+(\mathbb{S}^1, \mathbb{S}^1)$ arbitrarily close to the identity that cannot be written as $\widetilde{\phi}(t)$ for any one-parameter subgroup $\widetilde{\phi}: \mathbb{R} \to \text{Homeo}_+(\mathbb{S}^1, \mathbb{S}^1)$ at least tells you that "one-parameter subgroups" don't form a reasonable notion of "tangent space" at the identity of $\text{Homeo}_+(\mathbb{S}^1, \mathbb{S}^1)$.
(Just to be concrete: As a simple example in classical Lie groups, an example of a one-parameter subgroup is given by $$ \widetilde{\phi}^t = \begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix} \in SL(2, \mathbb{R}). $$ In this analogy, $SL(2, \mathbb{R})$ is playing the role of $\text{Homeo}_+(\mathbb{S}^1, \mathbb{S}^1)$; then $\mathbb{R}^2$ is playing the role of $\mathbb{S}^1$, in the sense that it's a space on which the group acts by some form of automorphism.)