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Assume the topological group $\mathbb{R}$ acts properly on a space $X$. Does then the projection map $p:X\rightarrow \mathbb{R}\backslash X$ have local sections ? (for every $\mathbb{R}x\in \mathbb{R}\backslash X$, there is a open neighbourhood $U \subset \mathbb{R}\backslash X$) and a section of $p|_{p^{-1}(U)}:p^{-1}(U)\rightarrow U$). Are there any nice conditions for $X$, that imply the existence of local sections ?

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you mean $X/ \mathbb{R}$ right? – Pietro Majer Nov 4 '10 at 11:17
Presumably the action is a left action, so writing $\mathbb{R}$ on the left of $X$ to mean the quotient is common (and reasonable) notation. – Sheikraisinrollbank Nov 4 '10 at 13:56
up vote 10 down vote accepted

Such a theorem is proved for completely regular spaces $X$ in my article:

On the Existence of Slices for Actions of Non-Compact Lie Groups, Richard S. Palais, The Annals of Mathematics, Second Series, Vol. 73, No. 2 (Mar., 1961), pp. 295-323 .

Actually, that paper considers more general groups than just $\mathbb{R}$ (any locally compact group) and for actions a little more general than proper (what I call Cartan G-Spaces). The paper is available from JSTOR. (Note that what I prove under these circumstances is the existence of a slice. But since for a proper action the isotropy group at any point is compact, for the case of $\mathbb{R}$ this means that all isotropy groups are trivial, so the action is automatically free, and a slice is a local section.)

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