Is there any equivalent condition for a metric to have all sectional curvatures less or equal to zero?
Basically, I have done some conformal perturbation on a Riemannian metric of negative sectional curvatures and I was trying to find an argument that implies that the sectional curvatures are still non-negative without brutal force calculations on every pair of vectors.
I know it is quite a general question, but I could not find anything in this direction.
For any Riemannian manifold $(M,\mathtt{g})$ and smooth function $\rho\colon M\to \Bbb R$, the sectional curvature $\widetilde{K}$ of $\widetilde{\mathtt{g}} = {\rm e}^{2\rho}\mathtt{g}$ is given in terms of the sectional curvature $K$ of $\mathtt{g}$ via the formula $$\widetilde{K}(\Pi) = {\rm e}^{-2\rho}\big(K(\Pi) - {\rm tr}_{\mathtt{g}_x|_{\Pi}}(B^\rho_x|_{\Pi})\big),\tag{1}$$for all $x\in M$ and $\Pi \in {\rm Gr}_2(T_xM)$, where the twice-covariant symmetric tensor field $B^\rho$ on $M$ is given by $$B^\rho = {\rm Hess}_{\mathtt{g}}\rho - {\rm d}\rho\otimes {\rm d}\rho + \frac{\|{\rm d}\rho\|_{\mathtt{g}}^2}{2}\mathtt{g}.\tag{2}$$ The proof of this fact is a brute force computation using the well-known formula for the curvature tensor $\widetilde{R}$ of $\widetilde{\mathtt{g}}$ in terms of the curvature tensor $R$ of $\mathtt{g}$, namely, that $$\widetilde{R} = {\rm e}^{2\rho}R - 2\widetilde{\mathtt{g}} \wedge B^\rho.\tag{3}$$ Here, $R$ and $\widetilde{R}$ are fully covariant and $\wedge$ stands for the curvaturelike-tensor valued Kulkarni-Nomizu product between twice-covariant symmetric tensor fields, given in coordinates by $$2(T\wedge S)_{ijk\ell} = T_{jk}S_{i\ell} - T_{ik}S_{j\ell} + S_{jk}T_{i\ell} - S_{ik}T_{j\ell}.\tag{4}$$
You can probably figure out what happens if you focus on traces of $B^\rho$, but $(1)$ suggests that you might need more conditions on $\rho$.