## Parallel surfaces and Minkowski formula

Suppose ${X:M^n\rightarrow \mathbb{R}^{n+1}}$ is an immersed orientable closed hypersurface. ${N}$ is the inner unit normal for ${X(M^n)}$ and denote by ${\sigma}$ the second fundamental form of the immersion and by ${\kappa_i}$, ${i=1,\cdots,n}$ the principle curvatures at an arbitrary point of ${M}$. The ${r-}$th mean curvature of ${H_r}$ is obtained by applying ${r-}$elementary symmetric function to ${\kappa_i}$. Equivalently, ${H_r}$ can be defined through the identity

$\displaystyle P_n(t)=(1+t\kappa_1)\cdots(1+\kappa_n)=1+\binom{n}{1}H_1 t+\cdots+\binom{n}{n}H_n t^n$

for all real number ${t}$. One can see that ${H_1}$ represents the mean curvature of ${X}$, ${H_n}$ is the gauss-Kronecker curvature. ${H_2}$ can reflect the scalar curvature of ${M}$ on the condition that the ambient manifold is a space form.

We want to study the consequence of moving the hypersurface parallel. Namely, define ${X_t}$ to be

$\displaystyle X_t= X-tN.$

When ${t}$ is small enough, ${X_t}$ is well defined immersed hypersurface. Suppose ${e_1,\cdots, e_n}$ are principle directions at a point ${p}$ of ${M}$, then

$\displaystyle \quad(X_t)_*(e_i)=(1+\kappa_it)e_i$

here we identify ${X_*(e_i)=e_i}$ as abbreviation. This implies that ${N_t= N\circ X_t^{-1}}$ is also an unit normal field of ${X_t}$. The area element ${dA_t}$ will be

$\displaystyle dA_t=(1+t\kappa_1)\cdots(1+t\kappa_n)dA=P_n(t)dA.$

The second fundamental form of ${X_t}$ with respect to ${N}$ will be

$\displaystyle \sigma_t(v,w)=\langle N_t,\nabla^{\mathbb{R}^{n+1}}_vw\rangle=-\langle \nabla^{\mathbb{R}^{n+1}}_vN_t,w\rangle$

for all ${v,w}$ tangent vector fields on ${X_t(M)}$. Plugging in ${v=(X_t)_*(e_i)}$ and ${w=(X_t)_*(e_j)}$, we get

$\displaystyle (\nabla^{\mathbb{R}^{n+1}}_vw)(X_t(p))=(\nabla^{\mathbb{R}^{n+1}}_{e_i}e_j)(X(p))$

$\displaystyle \nabla_{v}^{\mathbb{R}^{n+1}}N_t=-\frac{\kappa_i}{1+t\kappa_i}v$

So ${e_1,\cdots, e_n}$ are also principle directions for ${X_t}$ and principle curvatures are

$\displaystyle \frac{\kappa_i}{1+t\kappa_i}$

Another way to see this is by choosing a geodesic local coordinates such that ${\partial_iX}$ are the principle directions of ${X}$ at ${p}$. Then

$\displaystyle \partial_j\partial_iX=\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij}$

$\displaystyle \partial_iN=-\kappa_i\partial_iX$

$\displaystyle \partial_i X_t=\partial_i X-t\partial_i N=\partial_i X+t\kappa_i\partial_iX$

$\displaystyle \partial_j\partial_iX_t=(1+t\kappa_i)\partial_j\partial_iX=(1+t\kappa_i)(\Gamma_{ij}^k\partial_kX+\kappa_iN\delta_{ij})$

Since ${g^{ij}_t=(1+\kappa_it)^{-2}\delta_{ij}}$ at ${p}$. Therefore we get the principle curvature are ${\frac{\kappa_i}{1+t\kappa_i}}$.

Therefore the mean curvature ${H(t)}$ for ${X_t}$ is

$\displaystyle H(t)=\frac{1}{n}\frac{P_n'(t)}{P_n(t)}$

Since we have identity

$\displaystyle \Delta|X_t|^2=2n(1+H\langle X_t,N\rangle)$

which implies

$\displaystyle \int_M\left(1+H(t)\langle X_t,N\rangle\right)dA_t=0$

Plugging in all the information,

$\displaystyle \int_M\left(nP_n(t)+P_n'(t)\langle X,N\rangle-tP_n'(t)\right)dA=0$

Reorder the terms in the above identity by the order of ${t}$, we get

$\displaystyle \int_M (H_{r-1}+H_r\langle X,N\rangle )dA=0$

One can use this to prove Heintze-Karcher inequality. There are Minkowski formula in Hyperbolic space and $\mathbb{S}^n$ also.

Remark: S. Montiel and Anotnio Ros, compact hypersurfaces: the alexandrov theorem for higher order mean curvatures. Differential Geometry, 52, 279-296