Is there a Killing vector field on a complete Riemannian manifold $M$ with finite volume that satisfies the condition

$$\displaystyle\liminf_{r\rightarrow +\infty} \displaystyle\frac{1}{r} \displaystyle\int_{ B(2r)/B(r) } |X| d\nu_g > 0, $$

where $B(r)$ denotes the geodesic ball of radius $r$ and centre $p$, where $p$ is any point in $M$?

  • $\begingroup$ Do you mean "Is there [...]?"? $\endgroup$ – Silvia Ghinassi Jan 25 '16 at 3:20
  • $\begingroup$ Sorry, I made mistakes, the correct is Is there ..... ? Thank you very much. $\endgroup$ – italo lira Jan 25 '16 at 16:30
  • $\begingroup$ @italo lira: Do you know the answer when $M$ is a surface i.e. $dim(M)=2$ ? $\endgroup$ – Holonomia Jan 26 '16 at 14:07
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    $\begingroup$ I tried to find Killing vector fields with this property on surfaces of revolution but in the examples that build the liminf was zero. The motivation of this question is to find a vector field X on a complete manifold with finite volume such that $\rm{div} X$ is integrable and $$ \displaystyle\int_{M} (\rm{div} X) d\nu_{g} =0 $$ but the vector field X does not satisfy the hypothesis of the Karp's theorem in On Stokes’ Theorem for noncompact manifolds , 1981. $\endgroup$ – italo lira Jan 27 '16 at 0:57

Consider a warped product of the hyperbolic plane $H$ with a circle $S^1$, where the length of the circle is rescaled by a function $f\colon H\to(0,1]$ that is radially symmetric around $o\in H$. Write $f(x)=f(r)$ where $d(o,x)=r$ by abuse of notation. A sphere at distance $r$ in $H$ has volume $2\pi\sinh r$, so $f$ needs to satisfy $$\int_0^\infty f(r)\,\sinh r\,dr<\infty\;.$$ For example, take $f(r)=\frac1{1+r^2\sinh r}$ and let $(M,g)$ be the resulting Riemannian manifold with $g=g^{\mathrm{hyp}}\oplus (f\,d\varphi)^2$, with $\varphi$ the coordinate on $S^1$.

Then $(M,g)$ has a rotational symmetry. The corresponding Killing $X$ field has length $\sinh r$, where $r$ is now the pullback of $r$ to $M$. Hence for large $R$, approximately $$\int_{B(2R)\setminus B(R)}|X|\,d\nu_g\sim\int_R^{2R}f(r)\,(\sinh r)^2\,dr \sim\int_R^{2R}\frac{\sinh r}{r^2}\,dr\;.$$

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    $\begingroup$ You can probably also produce surface examples. Here, one would take a warped product of a line and a circle. The warping function $f$ needs to have very slender and high peeks, say centered at $2^k$ of hight $2^k$ and width $2^k/k^2$, outside of these peeks, it has to decay sufficiently fast to produce finite volume. The peeks are so steep that you cannot realise these surfaces as surfaces of revolution in \mathbb R^3$, though. $\endgroup$ – Sebastian Goette Feb 3 '16 at 8:02
  • $\begingroup$ Your solution is very creative and provides very interesting examples. $\endgroup$ – italo lira Feb 3 '16 at 12:06

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