# Is there an example of a Killing vector field on a complete Riemannian manifold with finite volume?

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 $M$?

• Do you mean "Is there [...]?"? Jan 25, 2016 at 3:20
• Sorry, I made mistakes, the correct is Is there ..... ? Thank you very much. Jan 25, 2016 at 16:30
• @italo lira: Do you know the answer when $M$ is a surface i.e. $dim(M)=2$ ? Jan 26, 2016 at 14:07
• 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. Jan 27, 2016 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 $$\int_{B(2R)\setminus B(R)}|X|\,d\nu_g\sim\int_R^{2R}f(r)\,(\sinh r)^2\,dr \sim\int_R^{2R}\frac{\sinh r}{r^2}\,dr\;.$$
• 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. Feb 3, 2016 at 8:02