I am trying to obtain the relation between the Hessians when I consider the transformation $\tilde{g}=\sigma^{-1}g$, for $(M^n,g)$ semi-Riemannian manifold and $f\in C^\infty(M)$, then if
$$\text{Hess}f(X,Y)=g(\nabla_X\nabla f,Y),\hspace{.5cm}\widetilde{\text{Hess}}f(X,Y)=\text{Hess}f(X,Y)+\text{other things} $$
for this, I initially thought of applying Koszul's formula to get
$$\tilde{g}(\widetilde\nabla_XY,Z)=\sigma^{-2}\big\{g(\nabla_X Y,Z)-\sigma^{-1}\left[X(\sigma)g(Y,Z)+Y(\sigma)g(Z,X)-Z(\sigma)g(X,Y)\right]\big\}$$
and replace in the expression of the Hessian
$$\widetilde{\text{Hess}}f(X,Y)=\tilde{g}\Big(\nabla_X\widetilde\nabla f-\sigma^{-1}\big[g(\nabla\sigma,X)\widetilde\nabla f+g(\nabla\sigma,\widetilde\nabla f)X-g(X,\widetilde\nabla f)\nabla \sigma\big],Y\Big) $$
but I don't know how to continue. If you could guide me I would be very grateful, also with recommended bibliography.
Best Answer
Setting $\sigma^{-1} = {\rm e}^{2\varphi}$, a standard computation gives $$\widetilde{\nabla}_XY = \nabla_XY + X(\varphi)Y+Y(\varphi)X - g(X,Y){\rm grad}_g\varphi.$$Also note that $${\rm d}f(Z) = g({\rm grad}_gf,Z) = \widetilde{g}({\rm grad}_{\widetilde{g}}f,Z) = {\rm e}^{2\varphi}g({\rm grad}_{\widetilde{g}}f,Z)$$leads to ${\rm grad}_{\widetilde{g}}f = {\rm e}^{-2\varphi} {\rm grad}_gf$. Hence: $$\begin{align} {\rm Hess}_{\widetilde{g}}f(X,Y) &= \widetilde{g}(\widetilde{\nabla}_X{\rm grad}_{\widetilde{g}}f, Y) \\ &= {\rm e}^{2\varphi} g(\widetilde{\nabla}_X({\rm e}^{-2\varphi}{\rm grad}_gf), Y) \\&= {\rm e}^{2\varphi} g(X({\rm e}^{-2\varphi}){\rm grad}_gf + {\rm e}^{-2\varphi} \widetilde{\nabla}_X{\rm grad}_gf, Y) \\ &= {\rm e}^{2\varphi}X({\rm e}^{-2\varphi}) {\rm d}f(Y) + g(\widetilde{\nabla}_X{\rm grad}_gf,Y) \\ &= {\rm e}^{2\varphi}(-2{\rm e}^{-2\varphi}X(\varphi)) {\rm d}f(Y) + g(\nabla_X{\rm grad}_gf+ X(\varphi){\rm grad}_gf+{\rm d}\varphi({\rm grad}_gf)X - {\rm d}f(X){\rm grad}_g\varphi,Y) \\ &= -2{\rm d}\varphi(X){\rm d}f(Y)+{\rm Hess}_gf(X,Y)+{\rm d}\varphi(X){\rm d}f(Y) + g({\rm d}\varphi,{\rm d}f)g(X,Y) - {\rm d}f(X){\rm d}\varphi(Y) \\ &= -2({\rm d}\varphi\odot {\rm d}f)(X,Y) + {\rm Hess}_gf(X,Y) + g({\rm d}\varphi,{\rm d}f)g(X,Y).\end{align}$$Thus $${\rm Hess}_{\widetilde{g}}f = {\rm Hess}_gf + g({\rm d}\varphi,{\rm d}f)g - 2{\rm d}\varphi\odot{\rm d}f.$$Now, to translate this back to $\sigma$, note that $\sigma = {\rm e}^{-2\varphi}$ implies that ${\rm d}\sigma = -2{\rm e}^{-2\varphi}{\rm d}\varphi = -2\sigma{\rm d}\varphi$, so $${\rm Hess}_{\widetilde{g}}f = {\rm Hess}_gf -\frac{1}{2} g({\rm d}\sigma, {\rm d}f)\widetilde{g}+\frac{1}{\sigma}{\rm d}\sigma \odot {\rm d}f.$$If the above is not convenient for you, ${\rm d}\varphi = -(1/2) {\rm d}(\log \sigma)$ also makes the cut, so $${\rm Hess}_{\widetilde{g}}f = {\rm Hess}_gf -\frac{1}{2} g({\rm d}(\log\sigma),{\rm d}f) + {\rm d}(\log \sigma)\odot {\rm d}f.$$