## Hypersurface in Hyperbolic space and its tangent horospheres

Hypersurface in hyperbolic space and its tangent horospheres

## Hypersurface in hyperbolic space and conformal metric on Sphere

Suppose $\Omega$ is a domain in $\mathbb{S}^n$, $g=e^{2\rho}g_0$, where $g_0$ is the standard metric on $\mathbb{S}^n$. One can construct a hypersurface in hyperboloid $\mathbb{H}^{n+1}\subset \mathbb{L}^{n+2}$ by

$\phi(x)=\frac{1}{2}(1+\sigma^2+|\nabla^0\sigma|^2)\psi(x)-\sigma(0,x)-(0,\nabla^0 \sigma)$

where $\sigma=e^{-\rho}$ and $\psi=e^{\rho}(1,x)=\sigma^{-1}(1,x)$. What are the induced metric on $\phi$? One can calculate as the following,

Suppose $u\in T_x\Omega$ is a unit eigenvector associated to the eigenvalue $s$ of $Hess(\sigma)_x$. We need to calculate $d\psi_x(u)$. Firstly it is easy to see

$d\psi_x(u)=-\sigma^{-1}u(\sigma)\psi+\sigma^{-1}(0,u)$

and

$d(\nabla^0 \sigma)_x(u)=\ \nabla_u^0\nabla^0 \sigma+(d \nabla^0\sigma_{x}(u))^{\perp}=Hess(\sigma)_x(u)+\langle d \nabla^0\sigma_{x}(u),x\rangle x$
$= \ su-\langle\nabla^0\sigma_{x},u\rangle x= \ su-u(\sigma)x$

Then

$d\phi_x(u)=u(\sigma)\sigma \psi+\frac{1}{2}u(|\nabla \sigma|^2)\psi+\frac{1}{2}(1+\sigma^2+|\nabla \sigma|^2)d\psi_x(u)-\sigma(0,u)-(0,su)$

So this gives the following formula

$d\phi_x(u)=\ \left(\frac{1}{2}-\lambda\right)d\psi_x(u).$

where $\lambda=s\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2)$. Suppose $u_i$ is the eigenvector corresponding to $s_i$ and $u_i,i=1,\cdots, n$ are orthonormal basis, then

$\ll d\phi_x(u_i),d\phi_x(u_j)\gg=\left(\frac{1}{2}-\lambda_{i}\right)\left(\frac{1}{2}-\lambda_{j}\right)\frac{1}{\sigma^{2}}\delta_{ij}$

where $\lambda_i=s_i\sigma+\frac{1}{2}(\sigma^2-|\nabla^0\sigma|^2)$. Recall that Schouten tensor has the following formula under conformal change,

$Sch_g=Sch_0+d\rho\otimes d\rho-\nabla^{2}_0\rho-\frac{1}{2}|\nabla^0\rho|^2g_0$

$=\frac{1}{2}g_0+\frac{1}{\sigma}\nabla_0^2\sigma-\frac{1}{\sigma^2}|\nabla_0\sigma|^2$

for $g=e^{2\rho}g_0$ and $Sch_0=\frac{1}{2}g_0$. Therefore the eigenvalues of $Sch_g$ are $\lambda_i$.

$\ll d\phi,d\phi\gg=\left(\frac{1}{2}g-Sch_g\right)^2{\sigma^{2}}\delta_{ij}$

$\textbf{Remark:}$ I am using the thesis of Dimas Percy Abanto Silva and some communication with him.

## Hyperbolic translation in Poincare disc model

Hyperboloid model ${\mathbb{H}^{n+1}=(x_0,x_1,\cdots,x_{n+1})}$ is hyperquadric in ${\mathbb{L}^{n+2}}$, with

$\displaystyle -x_0^2+\sum_{i=1}^{n+1} x_i^2=-1$

Half-space model ${\mathbb{R}^{n+1}_+=\{(y_1,y_2,\cdots,y_{n+1}),y_{n+1}>0\}}$. Denote ${(y_1,\cdots,y_{n},y_{n+1})=(y,y_{n+1})}$,

Poincaré ball model ${\mathbb{B}^{n+1}=(u_1,\cdots,u_{n+1})\subset\mathbb{R}^{n+1}}$ There are some transformation formula between these hyperbolic models, say

$\Psi_{12}:\mathbb{H}^{n+1}\longmapsto \mathbb{R}^{n+1}_+$

$(x_0,\cdots,x_{n+1})\longmapsto \frac{1}{x_0+x_{n+1}}\left(x_1,\cdots,x_{n},1\right)$

$\Psi_{21}:\mathbb{R}^{n+1}_+\longmapsto \mathbb{H}^{n+1}$

$(y_1,\cdots,y_{n+1})\longmapsto \left(\frac{1+|y|^2+y_{n+1}^2}{2y_{n+1}},\frac{y}{y_{n+1}},\frac{1-|y|^2-y_{n+1}^2}{2y_{n+1}}\right)$

$\Psi_{23}:\mathbb{R}^{n+1}_+\longmapsto \mathbb{B}^{n+1}$

$(y_1,\cdots,y_{n+1})\longmapsto\frac{1}{|y|^2+(y_{n+1}+1)^2}(2y,|y|^2+y_{n+1}^2-1)$

$\Psi_{32}:\mathbb{B}^{n+1}\longmapsto\mathbb{R}^{n+1}_+$
$(u_1,\cdots, u_{n+1})\longmapsto \frac{1}{|u|^2+(u_{n+1}-1)^2}\left(2u,2(1-u_{n+1})\right)-(0,\cdots,0,1)$

On ${\mathbb{R}^{n+1}_+}$, there are vertical scaling transformations which are isometries of ${\mathbb{R}^n_+}$

$\displaystyle \tau(y_1,\cdots,y_{n+1})= (ty_1,\cdots, ty_{n+1}),\quad t>0$

On ${\mathbb{B}^{n+1}}$, they are called translation in ${e_{n+1}}$ direction through the map

$\displaystyle (u,u_{n+1})\mapsto\Psi_{23}\circ\tau\circ\Psi_{32}(u,u_{n+1})$

In component,

$\displaystyle u_i\mapsto \frac{4tu_i}{(1-t)^2|u|^2+[u_{n+1}(t-1)+t+1]^2}$
$\displaystyle u_{n+1}\mapsto \frac{(t^2-1)|u|^2+4t^2u_{n+1}+(t^2-1)(u_{n+1}-1)^2}{(1-t)^2|u|^2+[u_{n+1}(t-1)+t+1]^2}$

On $\mathbb{H}^{n+1}$,

$(x_0,x,x_{n+1})\longmapsto \Psi_{21}\circ \tau\circ\Psi_{12}(x_0,x,x_{n+1})$

In component,

$\displaystyle x_0=\frac{(1+t^2)x_0^2+(1-t^2)x_{n+1}^2}{2t}$

$x_i=x_i,i=1,\cdots,n$

$\displaystyle x_{n+1}=\frac{(1-t^2)x_0^2+(1+t^2)x_{n+1}^2}{2t}$

## Conformal killing operator and divergence transformation under conformal change

Suppose ${(M,g)}$ is a Riemannian manifold. For each vector field ${V}$, we can define the conformal killing operator ${\mathcal{D}}$ to be the trace free part of Lie derivative ${\mathcal{L}_Vg}$, more precisely

$\displaystyle \mathcal{D}V=\mathcal{L}_Vg-\frac{2}{n}(div_g V)g$

Obviously ${\mathcal{D}}$ maps vector field to trace free symmetric two tensors. Now suppose we have a conformal transformation ${\tilde{g}=e^{2f}g}$, then what happen to the conformal killing operator ${\tilde{\mathcal{D}}}$? Notice that

$\displaystyle \mathcal{L}_V\tilde g=\mathcal{L}_V(e^{2f}g)=2e^{2f}V(f)g+e^{2f}\mathcal{L}_v g$

By using the identity ${\mathcal{L}_V d\mu_g=(div_g V)d\mu_g}$ for any vector field ${V}$, here ${d\mu_g}$ is the volume element, one can get the transfromation of divergence under confromal change

$\displaystyle div_{\tilde g}V=div_g V+nV(f)$

therefore

$\displaystyle \tilde{\mathcal{D}}V=\mathcal{L}_V\tilde g-\frac{2}{n}(div_{\tilde g} V)\tilde g=e^{2f}\mathcal{D}V.$

${\mathcal{D}}$ induces a formal adjoint ${\mathcal{D}^*}$ on trace free 2-tensors. Suppose we have a symmetric 2-tensor ${h=h_{ij}dx^i\otimes dx^j}$, where ${x^i}$ are coordinates on ${M}$. If one has

$\displaystyle \tilde{\mathcal{D}}^*(h-\tilde{\mathcal{D}}V)=0$

for some vector field ${V}$ and trace free 2-tensor ${h}$. What does this coorespond to under metric ${g}$? To see that, we first need a formula about symmtric 2-tensors,

$\displaystyle \langle h,w\rangle_{\tilde g}=\int_M h_{ij}w_{kl}\tilde g^{ik}\tilde g^{jl}d\mu_{\tilde {g}}=\langle e^{(n-4)f}h,w\rangle_g$

Now choose any vector field ${W}$, then

$\displaystyle 0=\langle h-\tilde{\mathcal{D}}V,\tilde{\mathcal{D}}W\rangle_{\tilde g}=\langle h-e^{2f}\mathcal{D}V, e^{2f}\mathcal{D}w \rangle_{\tilde g}=\langle e^{nf}(e^{-2f}h-\mathcal{D} V),\mathcal{D} W\rangle_{g}$

This is equivalent to

$\displaystyle \mathcal D^*(e^{nf}(e^{-2f}h-\mathcal DV))=0$

Next consider the divergence operator ${\delta:\mathscr{S}^{p+1}M\rightarrow \mathscr{S}^p M}$ and its adjoints ${\delta^*}$.

$\displaystyle \delta T=-g^{ik}\nabla_iT_{k....}$

It is well know that on 1-forms

$\displaystyle \delta^*\alpha(X,Y)=\frac{1}{2}{\nabla_X\alpha(Y)+\nabla_Y\alpha(X)}=\frac 12 (L_{\alpha^\sharp}g)(X,Y).$

where ${\sharp}$ operator turns 1-form to a vector field by using metric ${g}$. What is the relation of ${\delta h}$ and ${\tilde \delta h}$? To find that, choose any 1-form ${\alpha}$,

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle \tilde \delta h,e^{-(n-2)f}\alpha\rangle_{\tilde g}=\langle h,\tilde\delta^*(e^{-(n-2)f}\alpha)\rangle_{\tilde g} \ \ \ \ \ (1)$

using the formula about ${\delta^*}$, one gets

$\displaystyle \tilde \delta^*(e^{-(n-2)f}\alpha)=e^{-(n-2)f}\tilde \delta^*\alpha-(n-2)e^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

$\displaystyle =e^{-(n-2)f}\delta^*\alpha+e^{-(n-2)f}\alpha(f)g-ne^{-(n-2)f}\frac{1}{2}(df\otimes\alpha+\alpha\otimes df)$

then using ${h}$ is symmetric, continue from (1)

$\displaystyle \langle\tilde \delta h,\alpha\rangle_{g}=\langle e^{(n-4)f},\tilde \delta^*(e^{-(n-2)f}\alpha)\rangle_{g}=\langle e^{-2f}h,\delta^*\alpha+\alpha(f)g-ndf\otimes\alpha\rangle_{g}$

$\displaystyle =\langle\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f,\alpha\rangle_g$

In other words,

$\displaystyle \tilde \delta h=\delta(e^{-2f}h)-ne^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f$

$\displaystyle =\delta h-(n-2)e^{-2f}h(\nabla f,\cdot)+e^{-2f}(tr_g h)\nabla f \ \ \ \ \ (2)$

In the other way, we can calculate more directly

$\displaystyle \nabla_k h_{ij}=\frac{\partial}{\partial x^k}h_{ij}-\Gamma^p_{ki}h_{pj}-\Gamma^p_{kj}h_{ip}$

$\displaystyle \tilde \Gamma^k_{ij} = \Gamma^k_{ij}+ \delta^k_i\partial_j f + \delta^k_j\partial_i f -g_{ij}\nabla^k f$

We get

$\displaystyle \tilde \delta h=-\tilde g^{ki}\tilde \nabla_kh_{ij}=-e^{-2f}g^{ki}\nabla_kh_{ij}+e^{-2f}g^{ki}h_{pj}(\delta_k^p\partial_if+\delta^p_i\partial_kf-g_{ki}\nabla^pf)$

$\displaystyle +e^{-2f}g^{ki}h_{ip}(\delta^p_k\partial_j f+\delta^p_j\partial_kf-g_{kj}\nabla^pf)$

$\displaystyle \tilde \delta h=e^{-2f}\delta h+e^{-2f}(g^{ki}h_{kj}\partial_if+g^{ki}h_{ij}\partial_kf-nh_{pj}\nabla^p f+g^{ki}h_{ik}\partial_j f+g^{ki}h_{ij}\partial_k f- h_{jp}\nabla^p f)$

therefore

$\displaystyle \tilde \delta h=e^{-2f}\delta h-(n-2)e^{-2f}g^{ki}h_{kj}\partial_if+e^{-2f}(tr_gh)\nabla f$

One can compare this with (2).

## Product metric on product manifold

Suppose we have two Riemannian manifolds $(M^m,g)$ and $(N^n,\tilde g)$, what happened to their product manifold with product metric?
$(M\times N, g\times\tilde g)$.

We will use $a,b,c,..=1..m$ and $A,B,C,..=1..n$. Then by definition $g_{aA}=0$. Therefore

$\Gamma_{aA}^b=\frac 12g^{bc}(\partial_Ag_{ac}+\partial_ag_{Ac}-\partial_cg_{aA})=0$

Similarly for all mixed indices on $\Gamma$. This implies

$\nabla_{\partial_a}\partial_A=\Gamma_{aA}^b\partial_b+\Gamma_{aA}^B\partial_B=0.$

Therefore the Riemannian curvature operator is

$R(\partial_a,\partial_A)\partial_B=R(\partial_a,\partial_A)\partial_b=0.$

and Ricci tensor

$R_{aA}=0.$

When you have the warped product metric like $\mathbb{R}\times N$ with metric $dr^2+\phi^2(r)\tilde g$, the curvature is given like the following

$R(\partial _r,X,\partial_r,Y)=-{\phi''}\phi\tilde g(X,Y)$

$R(\partial_r,X,Y,Z)=0$

$R(X,Y,Z,W)=\phi^2\tilde R(X,Y,Z,W)-(\phi'\phi)^2(\tilde g(X,Z)\tilde g(Y,W)-\tilde g(X,W)\tilde g(Y,Z))$

## BV function and its property involves translation

Theorem 1 Suppose ${u\in L^1(\mathbb{R})}$, then ${u\in \text{BV}}$ if and only if ${\exists\, C}$ such that

$\displaystyle ||\tau_hu-u||_{L^1(\mathbb{R})}\leq C|h|,\quad \forall\, h$

Moreover, one can take ${C=|u|_{BV}}$. Here ${\tau_hu(\cdot)=u(\cdot+h)}$ is the tanslation operator.

Proof: Firstly suppose ${u\in \text{BV}}$. Let us prove

$\displaystyle \left|\int_{\mathbb{R}}(\tau_hu(x)-u(x))\phi(x)dx\right|\leq |u|_{BV}|\phi|_{L^\infty}|h|,\quad \forall \phi\in C^\infty_c(\mathbb{R}) \ \ \ \ \ (1)$

To show that

$\displaystyle LHS=\left|\int_{\mathbb{R}}u(x)(\phi(x-h)-\phi(x))dx\right|$

$\displaystyle =\left|\int_{\mathbb{R}}u(x)\psi(x)hdx\right|$

$\displaystyle \leq |u|_{BV}|\psi|_{L^\infty}|h|$

where

$\displaystyle \psi(x)=\int^x_{-\infty}\frac{\phi(s-h)-\phi(s)}{h}ds\in C_c^\infty(\mathbb{R})$

it is easy to verify ${|\psi|_\infty=|\phi|_\infty}$ therefore (1) is proved. Next one can choose such ${\phi_n\rightarrow sign(\tau_hu-u)\in L^1}$ with ${|\phi_n|\leq 1}$(it is easy to show by mollification). By dominating theorem, one get

$\displaystyle \int_{\mathbb{R}}|\tau_hu(x)-u(x)|dx\leq |u|_{BV}|h|.$

The other direction need more analysis. $\Box$

## Automorphism of annulus

Suppose $\mathbb{R}_\mu=\{z\in \mathbb{C}:1 is the annulus for $\mu>1$. Then

$\bf{Thm:}\mathbb{R}_\mu$ is not biholomorphic to $\mathbb{R}_{\mu'}$ whenever $\mu\neq \mu'$.

The proof goes like following: Suppose there exists a biholomorphic map between them, then this map can be extended to the boundary by the so called Kellog theorem, then one can reflect the annulus the inner part and outside part. Near the origin, one can use the removable singularity theorem to prove the extended map is holomophic. Thus we can get a biholomorphic map between $\mathbb{C}$ and $\mathbb{C}$ which must have the form $az+b$. First $b=0$, because the map keeps the origin. Second $a$ must have the unit length to keep the inner boundary to be $|z|=1$. Thus the map can only be rotation around origin.

This actually tells us the automorphism of an annulus can only be the rotation. Let us verify this fact by another perspective.

As we all know, $\mathbb{R}_\mu$ is biholomorphic to $\mathbb{H}/\Gamma_\lambda$, where $\mathbb{H}$ is the upper half plane and $\Gamma_\lambda(z)=\lambda z$. Here $\ln \lambda\ln \mu=2\pi^2$. $\Gamma_\lambda$ acts on the $\mathbb{H}$ freely discontinuously.

Any automorphism of $\mathbb{H}/\Gamma_\lambda$ can be lift to an automphism of $\mathbb{H}$ which commutes with the action $\Gamma_\lambda$. In the matrix language, these automphisms are the matrix in $PSL(2,\mathbb{R})$ commutes with $\left(\begin{matrix}\sqrt{\lambda}&0\\0&\sqrt{\lambda^{-1}}\end{matrix}\right)$. Some elementary calculation shows that these matrix have the form $\left(\begin{matrix}a&0\\0&a^{-1}\end{matrix}\right)$ where $a\in \mathbb{R}$. Any one of these  is corresponding to a rotation of the annulus.

## The perturbation of and metric and standard bubble

Suppose ${\bar u_\varepsilon=\left(\frac{\varepsilon}{\varepsilon^2+|x|^2}\right)^{\frac{n-2}{2}}}$, then it is well know that it satisfies

$\displaystyle -\Delta \bar u_\varepsilon=n(n-2)\bar u^{\frac{n+2}{n-2}}_\varepsilon$

on ${\mathbb{R}^n}$. Now we want to perturb the Euclidean metric a little bit and an approximate solution of above equation. Suppose ${g(t)=\delta+tS+O(t^2)}$ where ${S=\frac{d}{dt}g(t)|_{t=0}}$ is a symmetric 2-tensor and ${\delta}$ is the Euclidean metric. The reasonable equation of ${u_\varepsilon=u_{\varepsilon}(t)}$ must satisfy should be

$\displaystyle -\Delta_{g}u_{\varepsilon}+\frac{n-2}{4(n-1)}R_gu_\varepsilon=n(n-2)u^{\frac{n+2}{n-2}}_\varepsilon$

Suppose ${u_\varepsilon(t)=\bar u_\varepsilon+tw+O(t^2)}$. Taking derivative on both sides of the above equation and restrict to ${t=0}$, one can get the equation ${w}$ must satisfy

$\displaystyle \Delta w+n(n+2)\bar u_\varepsilon^{\frac{4}{n-2}}w=\frac{n-2}{4(n-1)}\bar u_\varepsilon(\partial_i\partial_j S_{ij}-\Delta tr S)+\partial_{i}(\partial_j\bar u_\varepsilon S_{ij})-\partial_i(tr S)\partial_i\bar u_\varepsilon$

where we have use the fact that

$\displaystyle \frac{d}{dt}\big|_{t=0}\Delta_{g(t)}u_\varepsilon(t)=\Delta w-\partial_i(S_{ij}\partial_j\bar u_\varepsilon)+\frac 12\partial_i(tr S)\partial_i \bar u_\varepsilon$

$\displaystyle \frac{d}{dt}\big|_{t=0}R_{g(t)}u_\varepsilon(t)=(\partial_{i}\partial_jS_{ij}-\Delta tr S)\bar u_\varepsilon$

## Dual spaces

Suppose ${(A,||\cdot||_A)}$ and ${(B,||\cdot||_B)}$ are Banach space, ${A^*}$ and ${B^*}$ are their dual spaces. If ${A\subset B}$ with ${||\cdot||_B\leq C||\cdot||_A}$, then

$\displaystyle i:A\mapsto B$

$\displaystyle \quad x\rightarrow x$

is an embedding. Let us consider the relation of two dual spaces. For any ${f\in B^*}$

$\displaystyle |\langle f,x\rangle|=|f(x)|\leq ||f||_{B^*}||x||_B\leq C||f||_{B^*}||x||_A\quad \forall\, x\in A$

Then ${f|_{A}}$ will be a bounded linear functional on ${A}$

$\displaystyle i^*:B^*\mapsto A^*$

$\displaystyle \qquad f\rightarrow f|_A$

is a bounded linear operator.

In a very special case that ${A}$ is a closed subset of ${B}$ under the norm ${||\cdot||_B}$, one can prove ${i^*}$ is surjective. In fact ${\forall\,g\in A^*}$ can be extended to ${\bar{g}}$ on ${B}$ by Hahn-Banach thm such that ${i^*\bar{g}=g}$. Then

$\displaystyle A^*=B^*/\ker i^*.$

Let us take ${A=H^1_0(\Omega)}$ and $\displaystyle B=H^1(\Omega)$ for example. Define the inner product to be

$\displaystyle \langle u,v\rangle=\int_\Omega uv+D_iuD_iv$

When ${\Omega}$ is a bounded subset of ${\mathbb{R}^n}$, ${H^1_0(\Omega)}$ is a proper closed subset of ${H^1(\Omega)}$. From the above explanation,

$\displaystyle H^{-1}(\Omega)=(H^1_0(\Omega))^*=(H^1(\Omega))^*/\ker i^*$

Define the continuous linear functional ${f(u)= \int_{\partial \Omega}u}$ for ${u\in H^1(\Omega)}$. One can see that ${i^*f=0}$.

$\textbf{Remark:}$ Should attribute this to Lun Zhang. I am always confused that $B^*\subset A^*$.

## Geodesic Normal Coordinates, Gauss lemma and Identity

Suppose ${p}$ is a point on ${n-}$dim manifold ${M}$. Consider the exponential map near ${p}$, at suitable neighborhood any point within it can be expressed uniquely as

$\displaystyle \text{exp}(x^1e_1+x^2e_2+\cdots+x^ne_n)$

where ${\{e_i\}}$ is an orthonormal basis. Then ${\{x^i\}}$ can be coordinate around ${p}$. Let us deduce useful identities using this coordinate. Obviously, the geodesic in this coordinate will be

$\displaystyle \gamma(t)=tv, \quad v\in T_pM$

In particular ${v=e_i}$,

$\displaystyle g_{ij}(p)=\langle \frac{\partial }{\partial x^i}(0),\frac{\partial }{\partial x^j}(0)\rangle=\langle e_i,e_j\rangle=\delta _{ij}$

The geodesic equation will be

$\displaystyle \Gamma^k_{ij}(\gamma(t))v^iv^j=0$

Multiplying ${t^2}$, we get

$\displaystyle \Gamma^k_{ij}(\gamma(t))x^ix^j=0$

Since the tangent vector has constant length along ${\gamma(t)}$, i.e.

$\displaystyle g_{ij}(\gamma(t))v^iv^j=g_{ij}(p)v^iv^j=v^iv^i$

Multiplying ${t^2}$, it is equivalent to say

$\displaystyle g_{ij}x^ix^j=x^ix^i$

Furthermore, recalling the definition of Christoffel symbol, one can get

$\displaystyle \frac{1}{2}(\partial_j g_{ik}+\partial_i g_{jk}-\partial_k g_{ij})x^ix^j=0$

Or equivalently

$\displaystyle \partial_j g_{ik}x^ix^j=\frac{1}{2}\partial_kg_{ij}x^ix^j=\frac{1}{2}\partial_k(g_{ij}x^ix^j)-g_{kj}x^j=x^k-g_{kj}x^j$

While on the left hand side

$\displaystyle \partial_j g_{ik}x^ix^j=\partial_j(g_{ik}x^i)x^j-g_{ik}x^i$

Combining this two facts,

$\displaystyle \partial_j(g_{ik}x^i)x^j=x^k$

$\displaystyle \partial_j(g_{ik}x^i-x^k)x^j=0$

This means along ${\gamma(t)}$,

$\displaystyle \frac{d}{dt}(g_{ik}x^i-x^k)=0$

Since at ${p}$, ${g_{ik}x^i-x^k=0}$, one get

$\displaystyle g_{ik}x^i=x^k$

on any point in the neighborhood. This identity is called Gauss Lemma.

Suppose ${g(x)=\text{exp}(h(x))}$, ${h_{ij}(x)}$ is a symmetric 2-tensor. Then the above identity means ${h_{ij}x^j=0}$.