Conformal transformation preserve convexity [closed]

Suppose $\Sigma$ a hypersurface in the eucildean space $\mathrm{R}^n$ and equip the inherited metric from the standard metric of $\mathrm{R}^n$.

If $\Sigma$ is $k-$convex, i.e. it has exactly $k$ nonnegative principal curvature.How to show that the $k-$ convex property is preserved under conformal transformation?Thanks!

Solution 1:

You have to see how the extrinsic geometry of $\Sigma$ changes. Let $\rho\colon \mathbb{R}^n\to \mathbb{R}$ be smooth, let $D$ and $\nabla$ denote the Levi-Civita connections of the original metrics $g^\circ$ and $g$ on $\mathbb{R}^n$ and $\Sigma$, and write $U = {\rm grad}_{g^\circ}\rho$ for the gradient field of $\rho$ on $\mathbb{R}^n$. Write $\widetilde{g}^\circ = {\rm e}^{2\rho}g^\circ$ for the conformal metric on $\mathbb{R}^n$, and $\widetilde{g}$ for the restriction of $\widetilde{g}^\circ$ to $\Sigma$. Observe that $\widetilde{g}$ agrees with ${\rm e}^{2\rho|_\Sigma}g$. Let $\widetilde{D}$ and $\widetilde{\nabla}$ be the Levi-Civita connections of $\widetilde{g}^\circ$ and $\widetilde{g}$.

Conformal changes in metric do not affect orthogonality, so the decomposition $\mathbb{R}^n = T_x\Sigma \oplus (T_x\Sigma)^\perp$ is independent of the metric. Let $\sigma$ be the second fundamental form of $(\Sigma, g)$ in $(\mathbb{R}^n, g^\circ)$, and $\widetilde{\sigma}$ be the second fundamental form of $(\Sigma, \widetilde{g})$ in $(\mathbb{R}^n, \widetilde{g}^\circ)$. The big question is: how does $\widetilde{\sigma}$ relate to $\sigma$?

Fix $X,Y\in\mathfrak{X}(\Sigma)$. Let's summarize all the information we know:

$$D_XY = \nabla_XY + \sigma(X,Y), \qquad \widetilde{D}_XY = \widetilde{\nabla}_XY + \widetilde{\sigma}(X,Y),$$but $$\widetilde{D}_XY = D_XY + X(\rho)Y+Y(\rho)X - g^\circ(X,Y)U$$and$$\widetilde{\nabla}_XY = \nabla_XY + X(\rho)Y+Y(\rho)X - g(X,Y)U^\top,$$where we write $U = U^\top + U^\perp$. Subtract both relations to get $$\widetilde{\sigma}(X,Y) = \sigma(X,Y) - g(X,Y)U^\perp.$$This "transformation law" is very benign and holds for any codimension. Now, let $N$ be a $g^\circ$-unit normal field along $\Sigma$, and consider the shape operator $A\colon T\Sigma\to T\Sigma$ characterized by the relation $g^\circ(\sigma(X,Y), N) = g(AX,Y)$. We have that $\widetilde{N} = {\rm e}^{-\rho}N$ is $\widetilde{g}^\circ$-unit and normal to $\Sigma$, and it has its own associated shape operator $\widetilde{A}$, characterized by $\widetilde{g}^\circ(\widetilde{\sigma}(X,Y), \widetilde{N}) = \widetilde{g}(\widetilde{A}X,Y)$.

What is the relation between $A$ and $\widetilde{A}$? Combining the above relations, we have that $$\widetilde{A}X = {\rm e}^{-\rho}(AX - g^\circ(U^\perp, N)X).$$However, the sign of the principal curvatures changes if you change the sign of the normal field. So, if we start by choosing $N$ pointing in the same direction of $U^\perp$, so that $g^\circ(U^\perp, N) > 0$, we conclude that if $\{\kappa_i\}_{i=1}^n$ are the eigenvalues of $A$, then the eigenvalues of $\widetilde{A}$ are $\{{\rm e}^{-\rho}(\kappa_i - g^\circ(U^\perp, N))\}_{i=1}^n$. Clearly the factor ${\rm e}^{-\rho}$ makes no difference for the signs of these curvatures. Now, it should not be difficult to compare $g^\circ(U^\perp, N)$ and the $\kappa_i$'s to get the desired conclusion.