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Suppose we have a smooth action of a compact Lie group $G$ on a non-compact smooth manifold $M$, denoted by $$\phi:G\times M\rightarrow M.$$ Differentiating $\phi$ at a point $x\in M$ gives a map that I'll call $$\psi_x:\mathfrak{g}\rightarrow T_x M.$$ Suppose we fix an Ad-invariant inner product on $\mathfrak{g}$.

Question: Is it always possible to choose a Riemannian metric $g$ on $M$ so that the norm of the map $\psi_x$ is uniformly bounded across $M$?

(In other words, can we always choose $g$ so that the norm of $d\phi$ is uniformly bounded across $M$?)

Remark: I'm motivated by considering the action of $S^1$ on $\mathbb{R}^2$ by rotation around the origin. Here the norm of $\psi_x$ goes to infinity as $|x|\rightarrow\infty$, but one can deform $\mathbb{R}^2$ diffeomorphically so that this is no longer the case. The question above asks whether this is true generally of compact Lie group actions.

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This is always possible. Namely, fix a metric $\langle\cdot,\cdot\rangle$ on $M$ and an orthonormal basis $X_1,...,X_n$ of $\mathfrak{g}$. Set now $\rho(x)=\sqrt{\sum_i \|d\psi_x(X_i)\|^2+1}$ and consider the conformally equivalent metric $\langle\cdot,\cdot\rangle'=\frac{1}{\rho(x)}\langle\cdot,\cdot\rangle$. Then if $v=\sum_i a_i X_i$ we get $$ \|d\psi_x(v)\|^2\le \|(a_1,...,a_n)\|^2_{\ell_2}\left(\sum_i\|d\psi_x(X_i)\|_{\langle\cdot,\cdot\rangle'}^2\right) = $$ $$ \|v\|^2\frac{\sqrt{\rho(x)^2-1}}{\rho(x)}\le\|v\|^2. $$

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  • $\begingroup$ One should be able to improve this answer such that the new metric is complete whenever the old one was. But then a conformal change is not quite enough - one should preserve the metric in directions perpendicular to the orbits. $\endgroup$
    – Sebastian Goette
    Commented Jul 6, 2019 at 8:54
  • $\begingroup$ Right. Actually I’ve been wondering if it is possible to modify the construction so as to preserve the property of “bounded geometry” (curvature and its derivatives are bounded, and injectivity radius is bounded away from 0). It seems, by your suggestion, that such a thing might be possible if there was a single isotropy type at least. $\endgroup$
    – geometricK
    Commented Jul 7, 2019 at 1:58
  • $\begingroup$ @SebastianGoette do you have an easy example in which this change of coordinates does not preserve completeness? It should be easy to find if it exists but I find it a little hard to imagine how it happens when the group is compact, probably due to my poor imagination. $\endgroup$
    – S. carmeli
    Commented Jul 7, 2019 at 21:45
  • $\begingroup$ Let's do the opposite. Start with a product $(-1,1)\times S^1$ with standard metric, which is not complete. The rotation on $S^1$ produces a vector field of constant length. By a conformal change of variables depending on $t\in(-1,1)$, we get a complete metric with an isometric $S^1$ action. I think $f(t)=1/(1-t^2)^2$ does the trick. $\endgroup$
    – Sebastian Goette
    Commented Jul 8, 2019 at 9:42
  • $\begingroup$ @SebastianGoette I wonder if the Cheeger deformation gives a complete metric and also satisfies that $\psi_x$ is uniformly bounded. In the case of $S^1$ acting on $\mathbb{R}^2$ this seems to be true. $\endgroup$
    – geometricK
    Commented Jul 8, 2019 at 15:14

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