Sectional curvature on conformally equivalent metric Let $(M,g)$ be a Riemannian manifold, $\tilde{g}:=e^{2 \psi} g$ for some $\psi \in C^{\infty}(M)$. 
I want to show, that for $X,Y$ orthonormal
$e^{2 \psi}\tilde{K}(X,Y)= K(X,Y)-Hess_{\psi}(X,X)-Hess_{\psi}(Y,Y)-g(\text{grad} \psi, \text{grad}  \psi)+(X(\psi))^2+(Y(\psi))^2$
So far I computed the left hand side to be:
$K(X,Y)-Y(\psi)g([X,Y],X)-X(\psi)g([X,Y],Y)$
but I don't know what to do with the right hand side. I somehow need to get rid of the Hessian and the gradient but the only thing I know about that is that
1)$Hess_{\psi}(X,X)=g(\nabla_X grad \psi,X)$
and
2)$g(grad_{\psi},X)=X(\psi)$
Using this and the Konszul formula for the right hand side I just always come back to the starting point. 
Does anybody have some hint for me here?
 A: OK, this is a long computation.
First, the Koszul formula gives
\begin{align*}\newcommand{\grad}{\operatorname{grad}_g}\newcommand{\Hess}{\operatorname{Hess}_g}
2g(\nabla_XY,Z)&=Xg(Y,Z)+Yg(X,Z)-Zg(X,Y)\\
&\quad+g([X,Y],Z)-g([X,Z],Y)-g([Y,Z],X)\\
2\tilde{g}(\tilde{\nabla}_XY,Z)&=X\tilde{g}(Y,Z)+Y\tilde{g}(X,Z)-Z\tilde{g}(X,Y)\\
&\quad+\tilde{g}([X,Y],Z)-\tilde{g}([X,Z],Y)-\tilde{g}([Y,Z],X)\\
&=X(e^{2\psi}g(Y,Z))+Y(e^{2\psi}g(X,Z))-Z(e^{2\psi}g(X,Y))\\
&\quad+e^{2\psi}g([X,Y],Z)-e^{2\psi}g([X,Z],Y)-e^{2\psi}g([Y,Z],X)\\
&=2(X\psi)\tilde{g}(Y,Z)+2(Y\psi)\tilde{g}(X,Z)-2(Z\psi)\tilde{g}(X,Y)\\
&\quad+2\tilde{g}(\nabla_XY,Z)\\
&=2(X\psi)\tilde{g}(Y,Z)+2(Y\psi)\tilde{g}(X,Z)-2\tilde{g}(\grad\psi,Z)g(X,Y)\\
&\quad+2\tilde{g}(\nabla_XY,Z)
\end{align*}
So
$$
\tilde{g}(\tilde\nabla_XY-\nabla_XY,Z)=
\tilde{g}((X\psi)Y+(Y\psi)X-g(X,Y)\grad\psi,Z)
$$
and hence
$$
\tilde\nabla_XY=\nabla_XY+(X\psi)Y+(Y\psi)X-g(X,Y)\grad\psi.
$$
Now we repeat this and have
\begin{align*}
\tilde\nabla_X\tilde\nabla_YZ
&=\nabla_X\tilde\nabla_YZ
+(X\psi)\tilde\nabla_YZ
+((\tilde\nabla_YZ)\psi)X-g(X,\tilde\nabla_YZ)\grad\psi\\
&=\nabla_X[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\\
&\quad
+(X\psi)[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\\
&\quad
+([\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\psi)X\\
&\quad-g(X,[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi])\grad\psi\\
&=\nabla_X\nabla_YZ+(X(Y\psi))Z+(Y\psi)\nabla_XZ+(X(Z\psi))Y+(Z\psi)\nabla_XY-(Xg(Y,Z))\grad\psi+g(Y,Z)\nabla_X\grad\psi\\
&\quad
+(X\psi)\nabla_YZ+(X\psi)(Y\psi)Z+(X\psi)(Z\psi)Y-(X\psi)g(Y,Z)\grad\psi\\
&\quad
+((\nabla_YZ)\psi+2(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\
&\quad-g(X,[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi])\grad\psi\\
%
&=\nabla_X\nabla_YZ\\
&\quad
+(Y(Z\psi)-g(Z,\nabla_Y\grad\psi)+2(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\
&\quad+(X(Z\psi)+(X\psi)(Z\psi))Y\\
&\quad+(X(Y\psi)+(X\psi)(Y\psi))Z\\
&\quad+(Z\psi)\nabla_XY\\
&\quad+(Y\psi)\nabla_XZ\\
&\quad+(X\psi)\nabla_YZ\\
&\quad-(Xg(Y,Z)+g(X,[\nabla_YZ+(Y\psi)Z+(Z\psi)Y]))\grad\psi\\
&\quad-g(Y,Z)\nabla_X\grad\psi
\end{align*}
and
\begin{align*}
\tilde\nabla_{\tilde\nabla_XY}Z
&=\tilde\nabla_{\nabla_XY+(X\psi)Y+(Y\psi)X-g(X,Y)\grad\psi}Z\\
&=\tilde\nabla_{\nabla_XY}Z+(X\psi)\tilde\nabla_YZ+(Y\psi)\tilde\nabla_XZ-g(X,Y)\tilde\nabla_{\grad\psi}Z\\
&=\nabla_{\nabla_XY}Z+((\nabla_XY)\psi)Z+(Z\psi)\nabla_XY-g(\nabla_XY,Z)\grad\psi\\
&\quad+(X\psi)[\nabla_YZ+(Y\psi)Z+(Z\psi)Y-g(Y,Z)\grad\psi]\\
&\quad+(Y\psi)[\nabla_XZ+(X\psi)Z+(Z\psi)X-g(X,Z)\grad\psi]\\
&\quad-g(X,Y)[\nabla_{\grad\psi}Z+((\grad\psi)\psi)Z+(Z\psi)\grad\psi-g(\grad\psi,Z)\grad\psi]\\
%
&=\nabla_{\nabla_XY}Z\\
&\quad+(Y\psi)\nabla_XZ\\
&\quad+(X\psi)\nabla_YZ\\
&\quad+(Z\psi)\nabla_XY\\
&\quad+(Y\psi)(Z\psi)X\\
&\quad+(X\psi)(Z\psi)Y\\
&\quad+((\nabla_XY)\psi+2(X\psi)(Y\psi)-g(X,Y)g(\grad\psi,\grad\psi))Z\\
&\quad-[(X\psi)g(Y,Z)+(Y\psi)g(X,Z)+g(\nabla_XY,Z)]\grad\psi\\
&\quad-g(X,Y)\nabla_{\grad\psi}Z
\end{align*}
So
\begin{align*}
\tilde\nabla^2_{X,Y}Z&=\nabla^2_{X,Y}Z\\
&\quad+(Y(Z\psi)+(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\
&\quad+(X(Z\psi))Y\\
&\quad+(X(Y\psi)-(\nabla_XY)\psi-(X\psi)(Y\psi)+g(X,Y)g(\grad\psi,\grad\psi))Z\\
&\quad-(g(Y,\nabla_XZ)+g(X,\nabla_YZ)+(Z\psi)g(X,Y)-(X\psi)g(Y,Z))\grad\psi\\
&\quad-g(Y,Z)\nabla_X\grad\psi\\
&\quad+g(X,Y)\nabla_{\grad\psi}Z
\end{align*}
and hence
\begin{align*}
\tilde{R}(X,Y)Z-R(X,Y)Z
&=(-g(Z,\nabla_Y\grad\psi)+(Y\psi)(Z\psi)-g(Y,Z)g(\grad\psi,\grad\psi))X\\
&\quad-(-g(Z,\nabla_X\grad\psi)+(X\psi)(Z\psi)-g(X,Z)g(\grad\psi,\grad\psi))Y\\
&\quad+((X\psi)g(Y,Z)-(Y\psi)g(X,Z))\grad\psi\\
&\quad-g(Y,Z)\nabla_X\grad\psi\\
&\quad+g(X,Z)\nabla_Y\grad\psi
\end{align*}
Now use
$$
e^{2\psi}\tilde{K}(X,Y)-K(X,Y)=g(R(X,Y)Y,X)
$$
for $X,Y$ $g$-orthonormal to finish.
