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It is well-known that we can construct an orthonormal basis at any point $p\in M$, where (M,g) is a Riemannian manifold using exponential mapping. This coordinate is called normal coordinates.

My question is: Apart from this construction, can we get a coordinate $\{x^i\}$ such that $$ g_{ij}(p)=\delta_{ij} ? $$ For example using some linear transformations. Here we don’t ask the Christoffel symbol vanishes. In other words we want to construct an orthonormal basis at a fixed point.

It’s worthy pointing out that even using Gram-Schmidt Algorithm, we can get a orthonormal basis but we cannot get the corresponding coordinates.

HONGHAOJIE YOU
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  • If you choose a coordinate system around $p$, then using a linear transformation you can assume that only at $p$ that $g_{ij}(p)=\delta_{ij}$, but generally you cannot achieve more, although I don’t have an example where you cannot have orthonormal coordinate system in an open set. – Laci Sep 16 '22 at 14:56
  • @Laci I’m sorry but could you give an example of such linear transformations or explain it more specific? – HONGHAOJIE YOU Sep 16 '22 at 15:26
  • I am stuck in eliminating the cross item $g_{ij},i\neq j$. – HONGHAOJIE YOU Sep 16 '22 at 15:28
  • If you have a coordinate system $x_1,\dots,x_n$, then you can take linear combination of these functions. Let $y_j=\sum a_{ij}x_i$ then $\partial_{y_j}=\sum a_{ij} \partial_{x_i}$. Now if the matrix $a_{ij}$ is nonsingular, then $y_1,\dots,y_n$ is a coordinate system. Now in $p$ you can use Gramm-Schmidt and use that matrix. – Laci Sep 16 '22 at 15:42
  • If $y_j=a_{ij}x_i, \partial_{yj}\neq a_{ij}\partial_{xi}$. For example, we can take $y=x^1+x^2$, then $\partial_y\neq \partial_{x1}+\partial_{x2}$ by taking $y=x_1+x_2$. – HONGHAOJIE YOU Sep 16 '22 at 16:19
  • You are right, but if my memory is right the partial derivative $\partial_{y_j}$ will be a linear combination of $\partial_{x_i}$. I think $A^{-1}$ will be right. – Laci Sep 16 '22 at 16:56
  • I still can find a counter example. I think the your method is right, but the matrix may not be so easy to find. – HONGHAOJIE YOU Sep 17 '22 at 03:34
  • After Gram-Schmidt Algorithm， you have a orthonormal basis ${e_1,...,e_n}\subset T_pM$. Choose smooth curves $\alpha_1(t),...,\alpha_n(t)$ such that $\alpha_i(0) =p$, and $\alpha'i(0) =e_i$. Then, when $|t|$ is small enough, $(\alpha_1(t),...,\alpha_n(t))$ will be a local coordinate of $p$ such that $g{ij}(p)=\delta_{ij}$. (I have not proof, just feel.) – Enhao Lan Sep 17 '22 at 06:59
  • Besides, in your above example about transformation of coordinates, if $y=x_1+x_2$, then $\partial_y = \partial_{x1}+\partial_{x2}$. Maybe, you need the section $2$ of chapter $0$ of do Carmo's Riemannian Geometry. – Enhao Lan Sep 17 '22 at 07:05
  • If you have a coordinate system $x_1,\dots,x_n$ then $\partial_{x_i}(x_j)=\delta_{ij}$, but if $\partial_y=\partial_{x_1}+\partial_{x_2}$, then $\partial_y(y)=2$. – Laci Sep 17 '22 at 09:37
  • I think the problem with your construction is first of all you only define $n$ different curves, so $(\alpha_1(t_1),\dots,\alpha_n(t_n))$ doesn’t really makes sense. It makes sense if you are mapping to $\mathbb{R}^n$. We can say that use local chart $\varphi$ and then consider $\varphi^{-1}(\alpha_1(t_1),\dots,\alpha_n(t_n))$ but then for an element $q\neq p$ you cant guarantee that $g_{ij}(q)=\delta_{ij}$. – Laci Sep 17 '22 at 09:46
  • Also if my memory is correct, on an open set $U$ the existence of an orthonormal frame coming from a coordinate system would imply that the Riemann-curvature of $g$ is zero on $U$. – Laci Sep 17 '22 at 09:49
  • Thanks for all your help, this problem maybe need more efforts to handle. – HONGHAOJIE YOU Sep 17 '22 at 09:56

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