## A Note on Generalized Douglas-Weyl Spray

Zheng Daxiao,

Department of Mathematics and Statistics Science, Anhui Normal University, Anhui Wuhu 241002

 基金资助: 安徽省自然科学青年基金.  2008085QA05

 Fund supported: The AHNSF.  2008085QA05

Abstract

In this paper, we study Generalized Douglas-Weyl spray. We show that a spray $G$ is a Generalized Douglas-Weyl spray if and only if it is $W$-quadratic. As a corollary, we show that a Finsler metric $F$ is a Generalized Douglas-Weyl metric if and only if it is $W$-quadratic. Furthermore, we consider $R$-quadratic spray and prove that a spray $G$ is $R$-quadratic if and only if $\dot{B}^{~i}_{j~kl}=0$.

Keywords： Spray ; Weyl curvature ; Douglas Tensor

Zheng Daxiao. A Note on Generalized Douglas-Weyl Spray. Acta Mathematica Scientia[J], 2023, 43(1): 43-52 doi:

## 1 引言

$\begin{matrix}G=y^i\frac{\partial}{\partial x^i}-G^i\frac{\partial}{\partial y^i}, \end{matrix}$

Douglas 张量 $D^{~i}_{j~kl}$ 和 Weyl 张量 $W^i_{~k}$ 是两个非常重要的射影不变量. 如果一个喷射的 Douglas 张量为 0, 则称其为 Douglas 喷射. 如果一个喷射的 Weyl 张量为 0, 则称其为 Weyl 喷射. 众所周知, 一个喷射是 Weyl 喷射当且仅当它具有标量旗曲率 [11].

$\begin{matrix} D^{~i}_{j~kl;m}y^m=T_{jkl}y^i, \end{matrix}$

$G^i=\tilde G^i+Py^i,$

$W^i_k=\tilde W^i_k,$

## 2 准备工作

$\begin{matrix}G=y^i\frac{\partial}{\partial x^i}-G^i\frac{\partial}{\partial y^i}, \end{matrix}$

$N^i{}_j=\frac{\partial G^i}{\partial y^j},~~~~\Gamma^i_{jk}=\frac{\partial^2 G^i}{\partial y^j\partial y^k},$
$\omega^i=dx^i,~~~~\omega^{n+i}=dy^i+N^i{}_jdx^j,~~~~\omega^i_j=~\Gamma^i_{jk}dx^k.$

$\Omega^i_j=d\omega^i_j-\omega^k_j\wedge \omega^i_k,$

$\Omega^i_j=\frac12R^{~i}_{j~kl}\omega^k\wedge \omega^l-B^{~i}_{j~kl}\omega^k\wedge\omega^{n+l},$

Douglas 张量的定义如下[2]

$\begin{matrix} D^{~i}_{j~kl}:=\frac{\partial^3}{\partial{y^j}\partial{y^k}\partial{y^l}}(G^i-\frac{1}{n+1}\frac{\partial{G^m}}{\partial{y^m}}y^i)~. \end{matrix}$

$\begin{matrix}D^{~i}_{j~kl}=B^{~i}_{j~kl}-\frac{2}{n+1}(E_{jk}\delta^i_{~l}+E_{jl}\delta^i_{~k}+E_{kl}\delta^i_{~j}+E_{jk.l}y^i), \end{matrix}$

$\begin{matrix}A^i_{~k}=R^i_{~k}-R\delta^i_{~k}, \end{matrix}$

$\begin{matrix}W^i_{~k}=A^i_{~k}+\tau_ky^i, \end{matrix}$

$\begin{matrix} \tau_k=-\frac{1}{n+1}A^m_{~k.m}=-\frac{1}{n+1}\sum^n_{m=1}\frac{\partial A^m_{~k}}{\partial y^m}. \end{matrix}$

$\begin{matrix}\chi_k=-\frac 1 6[2R^m_{~k.m}+R^m_{~m.k}], \end{matrix}$

$\begin{matrix}\tau_k=\frac{3}{n+1}\chi_k+\frac{R_{.k}}{2},~~~~\chi_ky^k=0. \end{matrix}$

## 3 $\boldsymbol GDW$ 喷射和$\boldsymbol W$ -二次型喷射

$G=y^i\frac{\partial}{\partial x^i}-G^i\frac{\partial}{\partial y^i}$ 为流形 $M$ 上的一个喷射. 由(2.10)式, 可得

$\begin{matrix}A^m_{~m}&=R^m_{~m}-nR =(n-1)R-nR=-R, \end{matrix}$

$\begin{matrix}A^i_{~m}y^m&=R^i_{~m}y^m-R\delta^i_{~m}y^m=0-Ry^i=-Ry^i.\end{matrix}$

(2.14)式作用 $y^k$, 我们有

$\begin{matrix}\tau_my^m=\frac{3}{n+1}\chi_my^m+\frac{R_{.m}}{2}y^m=0+R=R. \end{matrix}$

$\begin{matrix} W^m_{~m}=A^m_{~m}+\tau_my^m=-R+R=0, \end{matrix}$
$\begin{matrix} W^i_{~m}y^m=A^i_{~m}y^m+\tau_my^my^i=-Ry^i+Ry^i=0, \end{matrix}$
$\begin{matrix} W^m_{~j.m}=A^m_{~j.m}+(\tau_jy^m)_{.m}=-(n+1)\tau_j+(\tau_{j.m}y^m+n\tau_j)=0. \end{matrix}$

$\begin{matrix} W^i_{~m.j}y^m=-W^i_{~j}. \end{matrix}$

$\begin{matrix} W^i_{~m.j.k}y^m=-(W^i_{~j.k}+W^i_{~k.j}). \end{matrix}$

$\begin{matrix} W^i_{~m.j.k.l}y^m=-(W^i_{~j.k.l}+W^i_{~k.j.l}+W^i_{~l.j.k}). \end{matrix}$

$\begin{matrix}R^i_{~k}=R\delta^i_{~k}+A^i_{~k}=R\delta^i_{~k}-\tau_ky^i+W^i_{~k}. \end{matrix}$

$\begin{matrix} \dot{D}^{~i}_{j~kl}=T_{jkl}y^i+\frac13W^i_{~m.j.k.l}y^m, \end{matrix}$

$\begin{matrix} T_{jkl}=\frac13(R_{.j.k.l}-\tau_{m.j.k.l}y^m)-\frac{2}{n+1}{E}_{jk.l;m}y^m. \end{matrix}$

$\begin{matrix} \dot{D}^{~i}_{j~kl}=\dot{B}^{~i}_{j~kl}-\frac{2}{n+1}(H_{jk}\delta^i_{~l}+H_{jl}\delta^i_{~k}+H_{kl}\delta^i_{~j}+{E}_{jk.l;m}y^my^i), \end{matrix}$

$\begin{matrix} H_{jk}=\dot{E}_{jk} \end{matrix}$

$\begin{matrix} \dot{D}^i_{j~kl}=\tilde T_{jkl}y^i. \end{matrix}$

$\begin{matrix} W^i_{~m.j.k.l}y^m=3(\tilde T_{jkl}-T_{jkl})y^i. \end{matrix}$

(3.22)式两边对 $y^n$ 求导, 可得

$\begin{matrix} W^i_{~m.j.k.l.n}y^m+ W^i_{~n.j.k.l}=3(\tilde T_{jkl}-T_{jkl})_{.n}y^i+3(\tilde T_{jkl}-T_{jkl})\delta_n^i. \end{matrix}$

$\begin{matrix} T_{jkl.n}y^n=-T_{jkl},~~~~~~ \tilde T_{jkl.n}y^n=-\tilde T_{jkl}. \end{matrix}$

$\begin{matrix} 3(n-1)(\tilde T_{jkl}-T_{jkl})=0, \end{matrix}$

$T^i_{~m.j.k.l}y^m=(F^2y^i)_{.m.j.k.l}y^m=0.$

$T^i_{~m.j.k.l}=(F^2y^i)_{.m.j.k.l}=0$

$W^i_{~k}y^k=0.$

$T^i_{~k}y^k=0,$

$T^i_{~m.j.k.l}=0$ 当且仅当 $T^i_{~m.j.k.l}y^m=0$.

## 4 $\boldsymbol R$ -二次型喷射

(3) $H_{jk}=0.$

($\Rightarrow$) 因为 $G$$R -二次型喷射, 通过(2.6)式,(2.10)和(2.11)式, 可得 Ric, W^i_{~k} 是二次型, \tau_j 是一形式且 \tau_{j.k.l}=0. 将它代入(3.19)式, 可得 H_{jk}=0. (\Leftarrow) 由(3.10)式, 我们只需证明 \tau_j 是一形式. 因为 H_{jk}=0, 由(3.19)式, 可得 \begin{matrix} \tau_{m.j.k}y^m=0. \end{matrix} (3.3)式两边对 y^j$$y^k$求导, 可得

$\begin{matrix} \tau_{m.j.k}y^m=R_{.j.k}-\tau_{j.k}-\tau_{k.j}. \end{matrix}$

$\begin{matrix} R_{.j.k}-\tau_{j.k}-\tau_{k.j}=0. \end{matrix}$

(4.3)式两边对 $y^l$求导, 可得

$\begin{matrix} R_{.j.k.l}-\tau_{j.k.l}-\tau_{k.j.l}=0. \end{matrix}$

$\begin{matrix} \tau_{j.k.l}=-\tau_{k.j.l}. \end{matrix}$

$\tau_{j.k.l}=-\tau_{k.j.l}=-\tau_{k.l.j}=\tau_{l.k.j},$

$\tau_{j.k.l}$ 关于三个下指标全对称. 由(4.5)式, 可得

$\begin{matrix} \tau_{j.k.l}=0, \end{matrix}$

(3) $H_{jk}=0.$

$\begin{matrix} \tau_{m.j.k}y^m=0. \end{matrix}$

$\begin{matrix} (R_{.j.k.l}-\tau_{m.j.k.l}y^m)y^i+W^i_{~m.j.l.k}y^m=0, \end{matrix}$

(4.8)式两边对 $y^n$求导, 可得

$\begin{matrix} [R_{.j.k.l.n}-(\tau_{m.j.k.l}y^m)_{.n}]y^i+(R_{.j.k.l}-\tau_{m.j.k.l}y^m)\delta^i_n+W^i_{~m.j.k.l.n}y^m+ W^i_{~n.j.k.l}=0. \end{matrix}$

(3.3)式对 $y^j$, $y^k$$y^l求导, 可得 $$\tau_{m.j.k}y^m=R_{.j.k}-\tau_{j.k}-\tau_{k.j},$$ $$\tau_{m.j.k.l}y^m=R_{.j.k.l}-\tau_{j.k.l}-\tau_{k.l.j}-\tau_{l.j.k},$$ 将(4.7)式代入(4.13)式, 可得 \begin{matrix} R_{.j.k}-\tau_{j.k}-\tau_{k.j}=0. \end{matrix} 将(4.12)式代入(4.14)式, 可得 \begin{matrix} \tau_{j.k.l}+\tau_{k.l.j}+\tau_{l.j.k}=0. \end{matrix} (4.15)式对 y^l 求导, 可得 \begin{matrix} R_{.j.k.l}=\tau_{j.k.l}+\tau_{k.j.l}. \end{matrix} 将(4.16)式代入(4.17)式, 可得 \begin{matrix} R_{.j.k.l}=-\tau_{l.j.k}, \end{matrix} \tau_{j.k.l} 关于三个下指标全对称. 由(4.16)式, 可得 \begin{matrix} \tau_{j.k.l}=0, \end{matrix} 代入到(4.18)式, 可得 \begin{matrix} R_{.j.k.l}=0, \end{matrix} G 是 Ricci 二次型喷射. 证毕. 由引理4.1和引理4.2, 可推出定理 1.2. 注4.1 由 Bianchi 恒等式(3.15), 可得 G$$R$ -二次型喷射当且仅当

$\begin{matrix} {B}^{~i}_{j~kl;m}={B}^{~i}_{j~mk;l}, \end{matrix}$

$B^{~i}_{j~kl;m}$ 关于三个下指标全对称. 由定理 1.2, 可得(4.21)式成立当且仅当(1.4)式成立.

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