I'm reading Escobar's The Yamabe Problem On Manifolds With Boundary.


He says



> 
> Let $(y\_{1},\cdots,y\_{n})$ be normal coordinates around $0\in \partial M$, such that $\eta(0)=-\frac{\partial}{\partial y\_{n}}$, and second fundamental form of $\partial M$ at 0 has a diagonal form.
> 
> 
> 


Here $M$ is a Riemannian manifold with boundary and $0$ is a nonumbilic point on $\partial M$.$\eta$ represents the outward normal.


I wonder how can we take such a normal coordinates. As I know,normal coordinates can be taken at boundary point only when the boundary is locally totally geodesic. And in this case the second fundamental form must be 0.


In fact if we take a normal coordinates at $0$, which satisfies $g\_{ij}(0)=\delta\_{ij},\Gamma\_{ij}^{k}(0)=0.$


Then we compute
$$h\_{ij}(0)=g(\nabla\_{\partial y^{i}}\partial y^{n},\partial\_{y^{j}})=\Gamma\_{in}^{k}g\_{kj}=0.$$


So how can we take such a normal coordinate? Any help will be thanked.



Local geometry of nonumbilic points
You can—at least locally—find a normal coordinate system adapted to any submanifold $N \subset M$. (You can extend $M$ past its boundary to have $N = \partial M$ lie in the interior, but really that is not necessary here.)


Now suppose you have normal coordinates $(y^1,\dots,y^n)$ near a point $0 \in \partial M$, with $\partial y^n = -\eta$ in a neighbourhood of said point.


The formula you give for the second fundamental form does not look right; it should be in terms of the induced connection $\nabla'$ on $\partial M$:
\begin{equation}
h\_{ij}
= g(\nabla'\_{\partial y^i}\partial y^j,\eta)
= g(\partial y^j,\nabla'\_{\partial y^i} \partial y^n),
\end{equation}
where the sign is switched around because $\partial y^n = -\eta$.


As $h\_{ij}(0)$ is a symmetric $(n-1) \times (n-1)$ matrix, we can diagonalise it via an orthonormal change of basis. You leave $y^n$ unchanged, and modify the rest of the coordinates $(y^1,\dots,y^{n-1})$ through this same change of basis to obtain the desired coordinate system.


