Category Archives: Riemannian Geo

Surface intersection with a ball

Suppose M is 2-dimensional surface in \mathbb{R}^3. If M is simply connected, then B\cap M may not be simply connected, where B is ball in \mathbb{R}^3. See the following figure.

nose

But if M is minimal surface, then M\cap B must be simply connected. The reason is M has convex haul property.

Learn this example from Jacob Bernstein.

 

 

Hessian of radial functions

Suppose {u=u(r)} is a radial function on {\mathbb{R}^n}, here {r=|x|}.

\displaystyle u_{x_i}=u'\frac{x_i}{r}

\displaystyle u_{x_ix_j}=u''\frac{x_ix_j}{r^2}+u'(\frac{\delta_{ij}}{r}-\frac{x_ix_j}{r^3})=\frac{u'}{r}\delta_{ij}+(\frac{u''}{r^2}-\frac{u'}{r^3})x_ix_j

therefore

\displaystyle \det D^2u = \left(\frac{u'}{r}\right)^{n}\det[ \delta_{ij}+\frac{r}{u'}(u''-\frac{u'}{r})\frac{x_ix_j}{r^2}]

\displaystyle =\left(\frac{u'}{r}\right)^{n}[1+\frac{r}{u'}(u''-\frac{u'}{r})]=\left(\frac{u'}{r}\right)^{n-1}u''

If we use the polar coordinates {(r,\theta_1,\cdots, \theta_{n-1})}, and {g=dr^2+r^2\sum_{i=1}^{n-1}d\theta_i^2} and the following fact

\displaystyle \nabla_X\partial_r=\begin{cases}\frac{1}{r}X&\text{if } X\text{ is tangent to }\mathbb{S}^{n-1}\\0 \quad &\text{if } X=\partial_r\end{cases}

then one can calculate the Hessian of {u} under this coordinates

\displaystyle Hess (u)(\partial_r,\partial_r)=u''

\displaystyle Hess (u)(\partial_{\theta_i},\partial_r)=0

\displaystyle Hess (u)(\partial_{\theta_i},\partial_{\theta_j})=ru'\delta_{ij}.

Then

\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'}{r}\right)^{n-1}u''

If the metric is g=dr^2+\phi^2ds_{n-1}^2, then we will have

\displaystyle \frac{\det Hess (u)}{{\det g}}=\left(\frac{u'\phi'}{\phi}\right)^{n-1}u''

Unit normal to a radial graph over sphere

Consider \Omega\subset \mathbb{S}^n is a domain in the sphere. S is a radial graph over \Omega.

\boldsymbol{F}(x)=\{v(x)x:x\in \Omega\}

What is the unit normal to this radial graph?

Suppose \{e_1,\cdots,e_n\} is a smooth local frame on \Omega. Let \nabla be the covariant derivative on \mathbb{S}^n. Tangent space of S consists of \{\nabla_{e_i}\boldsymbol{F}\}_{i=1}^n which are

\nabla_{e_i}\boldsymbol{F}=v(x)e_i+e_i(v)\cdot x

In order to get the unit normal, we need some simplification. Let us assume \{e_i\} are orthonormal basis of the tangent space of \Omega and \nabla v=e_1(v)e_1. Then

\nabla_{e_1}\boldsymbol{F}=v(x)e_1+e_1(v)x, \quad \nabla_{e_i}\boldsymbol{F}=v(x)e_i, \quad i\geq 2

Then we obtain an orthonormal basis of the tangent of S

\{\frac{1}{\sqrt{v^2+|\nabla v|^2}}\nabla_{e_1}\boldsymbol{F},\nabla_{e_2}\boldsymbol{F},\cdots,\nabla_{e_n}\boldsymbol{F}\}

We are able to get the normal by projecting x to this subspace

\nu=x-\frac{1}{v^2+|\nabla v|^2}\langle x,\nabla_{e_1}\boldsymbol{F}\rangle\nabla_{e_1}\boldsymbol{F}=\frac{v^2x-v\nabla v}{v^2+|\nabla v|^2}.

After normalization, the (outer)unit normal can be written

    \frac{vx-\nabla v}{\sqrt{v^2+|\nabla v|^2}}

Remark: Guan, Bo and  Spruck, Joel. Boundary-value Problems on \mathbb{S}^n for Surfaces of Constant Gauss Curvature.

Laplacian on graph

Suppose {\Omega\subset \mathbb{R}^n} is a domain and {\Sigma=graph(F)\subset \mathbb{R}^{n+1}} is a hypersurface, where {F=F(u_1,\cdots,u_n)} is a function on {\Omega}. Define {f=f(u_1,\cdots,u_n)} on {\Omega}. Then {f} also can be considered as a function on {\Sigma}. How do we understand {\Delta_\Sigma f}?

Denote {\partial_i=\frac{\partial}{\partial{u_i}}} for short. If we pull the metric of {\mathbb{R}^{n+1}} back to {\Omega}, denote as {g}, then

\displaystyle g_{ij}=g(\partial_i,\partial_j)=\delta_{ij}+F_{u_i}F_{u_j},\quad g^{ij}=\delta_{ij}-\frac{F_{u_i}F_{u_j}}{W^2},\quad \det g=W^2

where {W=\sqrt{1+|\nabla F|^2}} and {F_{u_i}=\frac{\partial F}{\partial u_i}}. Then one can use the local coordinate to calculate {\Delta_\Sigma f}

\displaystyle \Delta_\Sigma f=\frac{1}{\sqrt{\det g}}\partial_{u_i}\left(\sqrt{\det g}\,g^{ij}f_{u_j}\right)

\displaystyle =g^{ij}f_{u_iu_j}+f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}

also one can see from another definition of Laplacian

\displaystyle \Delta_\Sigma f=g^{ij}Hess(f)(\partial_i,\partial_j)=g^{ij}[\partial_j(\partial_i(f))-(\nabla_{\partial_j}\partial_i)f]

\displaystyle =g^{ij}f_{u_iu_j}-g^{ij}(\nabla_{\partial_j}\partial_i)f

By using the expression of {g^{ij}} stated above, we can calculate

\displaystyle \frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=\frac{-F_{u_j}F_{u_{i}u_{k}}g^{ik}}{W^2}

It follows from the definition of tangential derivative on {\Sigma}, see, that

\displaystyle \langle\nabla^\Sigma f, e_{n+1}\rangle=g^{ij}f_{u_{i}}F_{u_{j}}=f_{u_{i}}F_{u_{i}}-\frac{f_{u_i}F_{u_i}|\nabla F|^2}{W^2}=\frac{f_{u_{i}}F_{u_{i}}}{W^2}

then

\displaystyle f_{u_j}\frac{\partial_{u_i}(\sqrt{\det g}\,g^{ij})}{\sqrt{\det g}}=-\langle \nabla^\Sigma f,e_{n+1}\rangle F_{u_iu_k}g^{ik}=-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

where {H} is the mean curvature of the {\Sigma}. Combining all the above calculations,

\displaystyle \Delta_\Sigma f=g^{ij}f_{u_iu_j}-\langle \nabla^\Sigma f,e_{n+1}\rangle HW

Stereographic projection from center

Suppose we are doing stereographic projection at the center. Namely consider the following map

\displaystyle \phi:\mathbb{S}^2\rightarrow \mathbb{R}^2

\displaystyle (x,y,z)\mapsto\frac{(u_1,u_2,-1)}{\lambda}

where {\lambda=\sqrt{u_1^2+u_2^2+1}}. Then one can see {u_1=\frac{x}{z}}, {u_2=\frac{y}{z}}. Under this stereographic projection, a strip will be equivalent to some lens domain on the sphere.

new_lens

Let us pull the standard metric of {\mathbb{S}^2} to the {\mathbb{R}^2}. For the following statement, we will always omit {\phi^*} and {\phi_*}. Calculation shows,

\displaystyle dx=\left(\frac{1}{\lambda}-\frac{u_1^2}{\lambda^3}\right)du_1-\frac{u_1u_2}{\lambda^3}du_2=z(-1+x^2)du_1+xyzdu_2

\displaystyle dy=-\frac{u_1u_2}{\lambda^3}du_1+\left(\frac{1}{\lambda}-\frac{u_2^2}{\lambda^3}\right)du_2=xyzdu_1+z(-1+y^2)du_2

\displaystyle dz=\frac{1}{\lambda^3}(u_1du_1+u_2du_2)=z^2(xdu_1+ydu_2)

therefore

\displaystyle dx^2+dy^2+dz^2=z^2(1-x^2)du_1^2-2xyz^2du_1du_2+z^2(1-y^2)du^2_2

\displaystyle =\frac{1}{\lambda^2}(\delta_{ij}-\frac{u_iu_j}{\lambda^2})du_idu_j

here we used the fact that {(x,y,z)\in \mathbb{S}^2}. Suppose {\bar\nabla} is the connection on {\mathbb{S}^2} equipped with the standard metric. We want to calculate {\bar \nabla_{\partial_{u_i}}\partial_{u_j}}. To that end, it is better to use the {x,y,z} coordinates in {\mathbb{R}^3}

\displaystyle \partial_{u_1}=z(x^2-1)\partial_x+xyz\partial_y+xz^2\partial_z

\displaystyle \partial_{u_2}=xyz\partial_x+z(y^2-1)\partial_y+yz^2\partial_z

Since we know

\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=\left(\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}\right)^T=\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}-\langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle\partial_r

where {T} means the tangential part to {\mathbb{S}^2} and {\partial_r=x\partial_x+y\partial_y+z\partial_z} is the unit normal to {\mathbb{S}^2}. Using the connection in {\mathbb{R}^3}, we get

\displaystyle \nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}=z^2\left[3x(x^2-1)\partial_x+y(3x^2-1)\partial_y+z(3x^2-1)\partial_z\right]

\displaystyle \langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle=z^2(x^2-1)

One can verify from the above equalities that

\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=2xz\partial_{u_1}

Similarly

\displaystyle \bar \nabla_{\partial_{u_1}}\partial_{u_2}=yz\partial_{u_1}+xz\partial_{u_2}

\displaystyle \bar\nabla_{\partial u_2}\partial_{u_2}=2yz\partial_{u_2}

Scalar curvature of cylinder

Consider the scalar curvature {R_g} of a cylinder {\mathbb{R}\times \mathbb{S}^{n-1}} with the standard product metric {g_{prod}}, where {n\geq 2}.

First approach, the cylinder is diffeomorphic to the puncture plane,

\displaystyle \phi:\mathbb{R}^n\backslash\{0\}\longmapsto \mathbb{R}\times \mathbb{S}^{n-1}

\displaystyle x\rightarrow (\log|x|,\frac{x}{|x|})

It is easy to verify that {(\phi^{-1})^*g_{prod}=|x|^{-2}|dx|^2}. One can calculate {R_g} from the conformal change formula

\displaystyle R_g=-\frac{4(n-1)}{n-2}|x|^{-\frac{n+2}{2}}\Delta |x|^{-\frac{n-2}{2}}=(n-2)(n-1)

Second approach, remember the Ricci curvature splits on product manifold with product metric. Therefore

\displaystyle R_g=R_{g_{S^n}}=(n-2)(n-1)

From the above, we know that if {n=2}, then {R_g=0} while {R_g>0} if {n\geq 3}. I always get confused with this two cases. There is a nice way to see this result for {n=2}. {\mathbb{S}^1\times \mathbb{R}} can be covered locally isometrically by {\mathbb{R}^2} with Euclidean metric. Therefore {R_g=0} in this case. However, {\mathbb{R}^n} should cover {\mathbb{S}^1\times \mathbb{R}^{n-1}} instead of {\mathbb{R}\times \mathbb{S}^{n-1}}.

Cylinder, Puncture plane and Cone

Suppose \phi: \mathbb{R}^n-\{0\}\longmapsto \mathbb{R}\times S^{n-1} defined by \phi(x)=(\log|x|,\frac{x}{|x|}). Use (t,\xi) denote the points on \mathbb{R}\times S^{n-1}, then

\displaystyle dt=\frac{x_i}{|x|^2}dx^i,\quad d\xi^i=\frac{dx^i}{|x|}-\frac{x_ix_j}{|x|^3}dx^j

This means

\displaystyle dt^2+d\xi^2=\frac{1}{|x|^2}|dx|^2

then \phi is a conformal diffeomorphism from (\mathbb{R}^n-\{0\},|dx|^2) to (\mathbb{R}\times S^{n-1}, dt^2+d\xi^2). Suppose g is a conic metric such that g(x)=|x|^\beta|dx|^2 on \mathbb{R}^n-\{0\}. Thus (\mathbb{R}^n-\{0\},g) is isometric to (\mathbb{R}\times S^{n-1},e^{t(2+\beta)}(dt^2+d\xi^2)) through \phi. Therefore Y(\mathbb{R}^n-\{0\},[g])=Y(\mathbb{R}\times S^{n-1},[dt^2+d\xi^2]). It is easy to see that Y(\mathbb{R}\times S^{n-1},[dt^2+d\xi^2])=Y(S^{n}) because

\displaystyle g_{S^n}=\frac{4}{(1+|x|^2)}|dx|^2=(\cosh t)^{-2}(dt^2+d\xi^2)

Bach flat four dimensional manifold and sigma2 functional

We want to find the necessary condition of being the critical points of {\int\sigma_2} on four dimensional manifold.

1. Preliminary

Suppose {(M^n,g)} is a Riemannian manifold with {n=4}. {P_g} is the Schouten tensor

\displaystyle P_g=\frac{1}{n-2}\left(Ric-\frac{R}{2(n-1)}g\right)

and denote {J=\text{\,Tr\,} P_g}. Define

\displaystyle \sigma_2(g)=\frac{1}{2}[(\text{\,Tr\,} P_g)^2-|P_g|_g^2]

\displaystyle I_2(g)=\int_M \sigma_2(g)d\mu_g

where {|P|_g^2=\langle P,P\rangle_g}. It is well known that {I_2(g)} is conformally invariant.

Suppose {g(t)=g+th} where {h} is a symmetric 2-tensor. We want to calculate the first derivative of {I_2(g(t))} at {t=0}. To that end, let us list some basic facts (see the book of Toppings). Firstly denote {(\delta h)_j=-\nabla^i{h_{ij}}} the divergence operator and

\displaystyle G(h)=h-\frac{1}{2}(\text{\,Tr\,} h) g

\displaystyle (\Delta_L h)_{ij}=(\Delta h)_{ij}-h_{ik}Ric_{jl}g^{kl}-h_{jk}Ric_{il}g^{kl}+2R_{ikjl}h^{kl}

where {\Delta_L} is the Lichnerowicz Laplacian. Then the first variation of Ricci curvature and scalar curvature are

\displaystyle \dot{R}=\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle \ \ \ \ \ (1)

\displaystyle \dot{Ric}=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta G(h))^\sharp}g=-\frac{1}{2}\Delta_Lh-\frac{1}{2}L_{(\delta h)^\sharp}g-\frac{1}{2}Hess(\text{\,Tr\,} h)

\displaystyle =-\frac{1}{2}\Delta_Lh-d(\delta h)-\frac{1}{2}Hess(\text{\,Tr\,} h)

where we were using upper dot to denote the derivative with respect to {t}.

2. First variation of the sigma2 functional

Lemma 1 {(M^4,g)} is a critical point of {I_2(g)} if and only if

\displaystyle \Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g=0 \ \ \ \ \ (2)

where {(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}.

Proof:

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\int_M J\dot J-\langle\dot P,P\rangle+\langle h,P\wedge P\rangle+\frac{1}{2}\sigma_2\text{\,Tr\,} h \,d\mu_g

where {(P\wedge P)_{ij}=P_{ik}P_{jl}g^{kl}}. Since we have

\displaystyle \int_M\langle P,\dot P\rangle\\ =\frac{1}{n-2}\int_M\langle P, \dot Ric-\dot Jg-Jh\rangle =\frac{1}{n-2}[\langle P, \dot Ric\rangle-\dot J J-J\langle h,P\rangle]

\displaystyle =\frac{1}{n-2}[-\frac{1}{2}\langle h,\Delta_L P\rangle+\langle h,Hess(J)\rangle-\frac{1}{2}\Delta J \text{\,Tr\,} h-\dot J J-J\langle h,P\rangle]

Plugging this into the derivative of {I_2} to get

\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t))\\ =\int_M\frac{1}{2}\langle h,\Delta_L P\rangle-\langle h,Hess(J)\rangle+\frac{1}{2}\Delta J \text{\,Tr\,} h\\

\displaystyle \quad +(n-1)\dot J J+J\langle h,P\rangle+(n-2)\langle h,P\wedge P\rangle+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g

In order to simplify the above equation, we recall the definition of Lichnerowicz Laplacian {\Delta_L}

\displaystyle (\Delta_LP)_{ij}=(\Delta P)_{ij}-2P_{ik}Ric_{jl}g^{kl}+2R_{ikjl}P^{kl}

\displaystyle =(\Delta P)_{ij}-2(n-2)P_{ik}P_{jl}g^{kl}-2JP_{ij}+2R_{ikjl}P^{kl}

Apply (1) to get

(n-1)\dot J J=\frac12J\dot R=\frac12J[\delta^2h-\Delta(\text{\,Tr\,} h)-\langle h,Ric\rangle]
=\frac{1}{2}[\langle h, Hess(J)\rangle-\text{\,Tr\,} h\Delta J-(n-2)J\langle h, P\rangle-J^2\text{\,Tr\,} h]

Therefore we can simplify it to be

\displaystyle (n-2)\frac{d}{dt}\big|_{t=0} I_2(g(t)) =\int_M\frac{1}{2}\langle h,\Delta P\rangle-\frac{1}{2}\langle h, Hess(J)\rangle+h^{ij}R_{ikjl}P^{kl}

\displaystyle -\frac{n-2}{2}J\langle h,P\rangle-\frac{1}{2} J^2 \text{\,Tr\,} h+\frac{n-2}{2}\sigma_2\text{\,Tr\,} h d\mu_g

Let us denote {(\mathring{Rm}(P))_{ij}=R_{ikjl}P^{kl}}. Using the fact {n=4} and the definition of {\sigma_2},

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g

where

\displaystyle Q=\Delta P-Hess(J)+2\mathring{Rm}(P)-2JP-|P|_g^2g

\Box

Remark 1 It is easy to verify {\text{\,Tr\,} Q=0}, this is equivalent to say {I_2} is invariant under conformal change. More precisely, letting {h=2ug}, then

\displaystyle \frac{d}{dt}\big|_{t=0} I_2(g(t))=\frac{1}{4}\int_M\langle h,Q\rangle d\mu_g=\frac{1}{2}\int_M u\text{\,Tr\,} Q d\mu_g=0.

Remark 2 If {g} is an Einstein metric with {Ric=2(n-1)\lambda g}, then {P=\lambda g}, {J=n\lambda} and

\displaystyle \mathring{Rm}(P)=\lambda Ric=2(n-1)\lambda^2 g

It is easy to verify that {Q=0}. In other words, Einstein metrics are critical points of {I_2}.

Are there any non Einstein metric which are critical points of {I_2}?

Here is one example. Suppose {M=\mathbb{S}^2\times N}, where {\mathbb{S}^2} is the sphere with standard round metric and {(N,g_N)} is a two dimensional compact manifolds with sectional curvature {-1}. {M} is endowed with the product metric. We can prove {Ric=g_{S^2}-g_N}, {P=\frac{1}{2}g_{S^2}-\frac{1}{2}g_N}, {J=0}, {\mathring{Rm}(P)=g_{prod}} and consequently {Q=0}.

Note that the above example is a locally conformally flat manifold. For this type of manifold, we have the following lemma which can say

Lemma 2 Suppose {g} is locally conformally flat and {Q=0}, then

Proof: When {g} is locally conformally flat,

\displaystyle \mathring{Rm}(g)=JP+|P|_g^2g-2P\wedge P

{Q=0} is equivalent to

\displaystyle \Delta P-Hess(J)+|P|_g^2g-4P\wedge P=0

Actually this is equivalent to the Bach tensor {B} is zero. \Box

3. Another point of view

We have the Euler Characteristic formula for four dimensional manifolds

\displaystyle 8\pi^2\chi(M)=\int_M (|W|_g^2+\sigma_2) d\mu_g

therefore the critical points for {\int_M \sigma_2d\mu_g} will be the same as the critical points of {\int_M |W|_g^2d\mu_g}. However, the functional

\displaystyle g\rightarrow \int_M |W|_g^2d\mu_g

is well studied by Bach. The critical points of this functional satisfy Bach tensor equal to 0.

\displaystyle B_{ij}=\nabla^k\nabla^l W_{likj}+\frac{1}{2}Ric^{kl}W_{likj}

Obviously, {B=0} for Einstein metric, but not all Bach flat metrics are Einstein. For example {B=0} for any locally conformally flat manifolds.

Bubble functions under different setting

Bubble function can be defined either on \mathbb{R}^n, \mathbb{S}^n or \mathbb{B}^n. For the following notations, c_n will denote suitable constants which may be different from line to line.

  • For every \epsilon>0 and  \xi\in\mathbb{R}^n, define

\displaystyle u_{\epsilon,\xi}=c_n\left(\frac{\epsilon}{\epsilon^2+|x-\xi|^2}\right)^{\frac{n-2}{2}}

It is well know that -\Delta u= c_nu^{\frac{n+2}{n-2}}. Moreover (\mathbb{R}^n,u^{\frac{4}{n-2}}_{\epsilon,\xi}g_E) is isometric to the standard sphere minus one point.

  • For any a\in \mathbb{S}^n and \lambda>0 define

\displaystyle\delta(a,\lambda)=c_n\left(\frac{\lambda}{\lambda^2+1+(\lambda^2-1)\cos d(a,x)}\right)^{\frac{n-2}{2}}

where d(a,x) is the geodesic distance of a and x on \mathbb{S}^n. Actually \cos d(a,x)=a\cdot x

  • For each p\in \mathbb{B}^{n+1}, define \delta_p(x):\mathbb{S}^n\to \mathbb{R} by

\displaystyle\delta_p(x)=c_n\left(\frac{1-|p|^2}{|x+p|^2}\right)^{\frac{n-2}{2}}

Both the second and third one satisfy

\displaystyle \frac{4(n-1)}{n-2}\Delta_{\mathbb{S}^n}\delta-n(n-1)\delta+c_n\delta^{\frac{n-2}{n+2}}=0

If we make p=\frac{\lambda-1}{\lambda+1}a, the third one will be changed to the second one.

To get the second one from the first one, let us deonte \Phi_a:\mathbb{S}^n\to \mathbb{R}^n be the stereographic projection from point a. Then

\displaystyle \delta(a,\lambda)\circ \Phi^{-1}_a=c_n\left(\frac{\lambda(1+|y|^2)}{\lambda^2|y|^2+1}\right)^{\frac{n-2}{2}}=c_nu_{\lambda,0}u_{1,0}^{-1}

It should be able to see the third one from hyperbolic translation directly.

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