$M^n$. Let $$\begin{aligned}
 X=\bn^D\phi=V^{\g-\a}\nb \phi.\label{vf}\end{aligned}$$ Then $$DX= V^{\g-\a} \left[\bn^2 \phi+\g du\otimes d\phi+\g d\phi \otimes du+ \a \bar g\left(\bn u, \bn \phi\right)\bar g\right]=\bn^{2,D}\phi$$ is symmetric. Moreover, $$\div^D X:=D_i X^i=V^{\g-\a} \left(\bar\De \phi+(2\g+n\a)\<\nb u, \nb \phi\>\right)=\bd^D \phi.$$

Recall under local coordinates, $$\begin{aligned}
   % \nb_i\nb^j\phi=&\<\nb_i\nb \phi,\P_j\>\\
    D_iX^j=&\<D_i X,\bar g^{jk}\P_k\>.
    \label{hessian}\end{aligned}$$ Under normal coordinates, we have $$\begin{aligned}
    D_iX_j=&\<D_i(V^{\g-\a}\bar\nabla\phi), \P_j\>\nonumber\\
    =&\nb_i(V^{\g-\a}\nb_j\phi)+\a V^{\g-\a}u_i\phi_j+\a V^{\g-\a}\nb u\nb \phi \delta_{ij}+\g V^{\g-\a}\phi_iu_j\nonumber\\
    =&V^{\g-\a}\left(\nb_i\nb_j\phi+\g u_i\phi_j+\g \phi_iu_j+\a\<\nb u, \nb \phi\> \delta_{ij}\right).
    \label{Bochner equ 2}\end{aligned}$$ Clearly, $DX$ is symmetric.

\[weight\] Let $W$ be any smooth vector field on $M$. Then $$\begin{aligned}
        V^\tau D_iW^i=\nb_i(V^\tau W^i)
        \label{div}
    \end{aligned}$$ where $\tau=(n+1)\a+\g$, is a divergent form with respect to the Riemannian volume form $d\O$. ($d\O$ denotes the volume form induced by Riemannian metric $\bar g$ throughout this paper)

By definition of $D$, we have $$\begin{aligned}
       D_iW^i=&\nb_iW^i+\a u_i W^i+\a u_k W^k \delta^i_i+\g W^i u_i\nonumber\\
       =&\nb_iW^i+[(n+1)\a+\g] u_i W^i.
       \label{lemm div equ 1}
   \end{aligned}$$ Thus, $$\begin{aligned}
       V^\tau D_iW^i=& V^\tau\nb_iW^i+[(n+1)\a+\g] V^{\tau-1}V_i W^i\nonumber\\
= & V^\tau\nb_iW^i+\tau V^{\tau-1}V_i W^i\nonumber\\
= & \nb_i(V^\tau W^i).
       \label{lemm div equ 2}
   \end{aligned}$$

As an immediate corollary, we can show that the volume form $V^\tau d\O$ is parallel under the new affine connection $D^{\a,\g}$. We thank the referee for pointing out this fact to us. The special case of 