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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$\begingroup$ you mean $X/ \mathbb{R}$ right? $\endgroup$– Pietro MajerCommented Nov 4, 2010 at 11:17
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2$\begingroup$ 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. $\endgroup$– SheikraisinrollbankCommented Nov 4, 2010 at 13:56
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1 Answer
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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.)
answered Nov 4, 2010 at 16:27
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$\begingroup$ Dear Richard: While your answer is of course correct for topological actions of Lie groups, the statement is false for free proper actions of locally compact (even compact) groups. An example is due to Kolmogorov, see my answer here: math.stackexchange.com/questions/560371/… $\endgroup$ Commented Jul 19, 2016 at 0:57