Suppose a Lie group $G$ acts properly on a manifold $M$. Let $\pi: M\rightarrow M/G$ be the projection.

One can construct a bounded, smooth "cut-off" function $$c:M\rightarrow [0,\infty),$$ with two properties:

  1. For any subset $L\subseteq M$ such that $\pi(L)$ is compact in $M/G$, the intersection of $L$ with the support of $c$ is compact (or its closure is).

  2. $\int_G c(g^{-1}x) dg = 1$ for all $x\in M$.

For example, a construction is given on page 8 of https://arxiv.org/pdf/1608.06375.pdf.

I'm interested in whether one can control the gradient of such a cut-off function at infinity. That is, might it be possible to construct $c$ so that the norm of its differential, $|dc|$, fades to $0$ at infinity? (Or to construct $c$ so that it is a limit of such functions in the space of bounded continuous functions $C_b(M)$ with the sup-norm?)

(By "fading to $0$ at infinity", I mean that for any $\epsilon > 0$, there exists a compact $K\subseteq M$ such that on $M\backslash K$, $|dc|$ is uniformly less than $\epsilon$.)


1 Answer 1


Probably not: $M=R^2$, $G=R$. Action $\alpha_g(x,y) \mapsto (x+g,y)$ ($x,y,g \in R$). (The action seems to be proper?) How could you integrate the function $c$ over an orbit outside a compact ball to get a nonzero integral with a nonvarying function $c$? Not possible.


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