On conformal qc geometry, spherical qc manifolds and convex cocompact subgroups of $\mathrm{Sp}{(n+1,1)}$

Abstract

Conformal qc geometry of spherical qc manifolds is investigated. We construct the qc Yamabe operators on qc manifolds, which are covariant under the conformal qc transformations. A qc manifold is scalar positive, negative or vanishing if and only if its qc Yamabe invariant is positive, negative or zero, respectively. On a scalar positive spherical qc manifold, we can construct the Green function of the qc Yamabe operator, which can be applied to construct a conformally invariant tensor. It becomes a spherical qc metric if the qc positive mass conjecture is true. Conformal qc geometry of spherical qc manifolds can be applied to study convex cocompact subgroups of $\mathrm{Sp}(n+1,1).$ On a spherical qc manifold constructed from such a discrete subgroup, we construct a spherical qc metric of Nayatani type. As a corollary, we prove that such a spherical qc manifold is scalar positive, negative or vanishing if and only if the Poincaré critical exponent of the discrete subgroup is less than, greater than or equal to $2n+2$, respectively.

Appendix: The Green function of the qc Yamabe operator on the quaternionic Heisenberg group

See also §2.4 in [15] for a similar calculation for regular solutions of the qc Yamabe equation.

Proof of Proposition 3.4. Without loss of generality, we assume $\xi =0.$ Denote $\eta =(y,t).$ Recall that $\Delta _{0}=-\frac{1}{2}\sum _{j=1}^{4n}Y_{j}Y_{j}.$ We have

$$\begin{aligned} Y_{4l+j}\left( \frac{1}{(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q-2}{4}}}\right)&= -\frac{Q-2}{4}\frac{Y_{4l+j}\Vert \eta \Vert ^{4}}{(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q+2}{4}}}, \end{aligned}$$

(6.8)

with

$$\begin{aligned} Y_{4l+j}\Vert \eta \Vert ^{4}=4|y|^{2}y_{4l+j}+4\sum _{s=1}^{3}\sum _{k=1}^{4} b_{kj}^{s}y_{4l+k}t_{s}, \end{aligned}$$

(6.9)

using the expression of the vector field $Y_{4l+j}$ in (2.17). Note that $\frac{1}{\sqrt{2}}Y_{j}$ is an orthonormal basis. Then, we get

$$\begin{aligned} \Delta _{0}\left( \frac{1}{(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q-2}{4}}}\right) =-\frac{Q-2}{4}\left[ \frac{\Delta _{0}\Vert \eta \Vert ^{4}}{(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q+2}{4}}} +\frac{(Q+2)}{4}\frac{\left| \nabla _{0}\Vert \eta \Vert ^{4}\right| ^{2}}{(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q+6}{4}}}\right] , \end{aligned}$$

(6.10)

where

$$\begin{aligned} 2\left| \nabla _{0}\Vert \eta \Vert ^{4}\right| ^{2}&=\sum _{l=0}^{n-1}\sum _{j=1}^{4} \left( Y_{4l+j}\Vert \eta \Vert ^{4}\right) ^{2}\nonumber \\&=16\sum _{l=0}^{n-1}\sum _{j=1}^{4}\left( |y|^{4}y_{4l+j}^{2} +\sum _{s,s'=1}^{3}\sum _{k,k'=1}^{4}b_{k'j}^{s'} b_{kj}^{s}y_{4l+k'}y_{4l+k}t_{s}t_{s'}\right) =16\Vert \eta \Vert ^{4}|y|^{2}, \end{aligned}$$

(6.11)

using (6.9) and $\sum _{j=1}^{4}b_{kj}^{s}b_{jk'}^{s'}=(b^{s}b^{s'})_{kk'},$ antisymmetry for $b^{s}b^{s'}$ of $s\ne s'$ and $\left( b^{s}\right) ^{2}=-\mathrm{id}.$ Similarly, by (6.9), we get

$$\begin{aligned} \Delta _{0}\Vert \eta \Vert ^{4}=-\frac{1}{2} \sum _{l=0}^{n-1}\sum _{j=1}^{4}\left( {8y_{4l+j}^{2}+4|y|^{2}+8\sum _{s=1}^{3} \sum _{k,k'=1}^{4}b_{k'j}^{s} b_{kj}^{s}y_{4l+k'}y_{4l+k}}\right) =-2(Q+2)|y|^{2}. \end{aligned}$$

(6.12)

Then apply (6.11) and (6.12) to (6.10) to get

$$\begin{aligned} \Delta _{0}\left( \frac{1}{(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q-2}{4}}}\right) =\frac{(Q-2)(Q+2)|y|^{2}\epsilon ^{2}}{2(\Vert \eta \Vert ^{4}+\epsilon ^{2})^{\frac{Q+6}{4}}}. \end{aligned}$$

(6.13)

Now letting $\epsilon \rightarrow 0$, we see that for any $u\in C^{\infty }_{0}(\mathscr {H})$,

$$\begin{aligned} \int _{R^{4n+3}} L_{0}u\cdot \frac{C_{Q}}{(|y|^{4}+|{t}|^{2})^{\frac{Q-2}{4}}} =&\lim _{\epsilon \rightarrow 0}\int _{R^{4n+3}} L_{0}u\cdot \frac{C_{Q}}{(|y|^{4}+|{t}|^{2}+\epsilon ^{2})^{\frac{Q-2}{4}}}\\ =&\lim _{\epsilon \rightarrow 0}\int _{R^{4n+3}} u\cdot C_{Q}b_{n} \Delta _{0}\left( \frac{1}{(|y|^{4}+|{t}|^{2}+\epsilon ^{2} )^{\frac{Q-2}{4}}}\right) =u(0) \end{aligned}$$

by (6.13) and the formula (3.2) for $C_{Q}^{-1}.$ The proposition is proved.                          $\square$

For $g|_{\xi }=\frac{g_{0}|_{\xi }}{\Vert \xi \Vert ^{2}},$ we can write $g=\phi ^{\frac{4}{Q-2}}g_{0}$ with $\phi ={\Vert \xi \Vert ^{-\frac{Q-2}{2}}}.$ It follows from the transformation law of the scalar curvature (3.6) that

$$\begin{aligned} s_{g,\mathbb {Q}}&=\phi ^{-\frac{Q+2}{Q-2}}b_{n}\Delta _{0}\phi =\Vert \xi \Vert ^{\frac{Q+2}{2}}b_{n}\Delta _{0} \Vert \xi \Vert ^{-\frac{Q-2}{2}}\nonumber \\&=-\frac{Q-2}{8}b_{n}\Vert \xi \Vert ^{\frac{Q+2}{2}}\left( \frac{Q+6}{8} \Vert \xi \Vert ^{-\frac{Q+14}{2}}\left| \nabla _{0}\Vert \xi \Vert ^{4}\right| ^{2}+\Vert \xi \Vert ^{-\frac{Q+6}{2}} \Delta _{0}\Vert \xi \Vert ^{4}\right) \nonumber \\&=\frac{(Q-2)(Q+2)}{2}\frac{|y|^{2}}{\Vert \xi \Vert ^{2}} \end{aligned}$$

(6.14)

by (6.11) and (6.12), where $\eta =(y,t)$. Similarly, we have

$$\begin{aligned} \left| \nabla _{g}\left( \ln \frac{1}{\Vert \xi \Vert }\right) \right| ^{2}=\Vert \xi \Vert ^{2} \left| \nabla _{0}\left( \ln \frac{1}{\Vert \xi \Vert }\right) \right| ^{2} =\frac{1}{16\Vert \xi \Vert ^{6}}\left| \nabla _{0}\Vert \xi \Vert ^{4}\right| ^{2}=\frac{1}{2} \frac{|y|^{2}}{||\xi ||^{2}}. \end{aligned}$$

(6.15)