# Metric compatibility from existence of normal coordinate

It is well known that for a Riemannian manifold $$(M, g)$$, one can define a torsionless and metric-compatible connection $$\nabla$$, and then use $$\nabla$$ to construct geodesics and normal coordinates. In these normal coordinates, $$g = \delta$$ (or in Lorentzian signature, $$g = \eta$$).

I wonder if one can turn the logic around. Suppose I am given a torsionless connection $$\nabla$$, and I use it to define geodesics and normal coordinates. Then I grab a metric $$g$$ and declare that at any point, $$g = \delta$$ in the normal coordinates (of course I should only focus on those coordinates with orthonormal basis vector w.r.t. $$g$$). Am I guaranteed that $$\nabla g = 0$$?

Naively, I would think that: in the normal coordinate at point $$p$$, $$\Gamma|_p = 0$$. Then $$\nabla g = \partial g$$ at $$p$$. So I should study $$\partial g|_p$$ in the normal coordinate, which I think naively is zero (from the intuition that in the usual discussion on normal coordinates, $$g = \delta + O(x^2)$$).

• I do not understand well the question. You have a $\nabla$-geodesic coordinate system centered at some point $p$ and defined in an open neighborhood $U\ni p$. So $\Gamma|_p = 0$. Are you assuming that $g|_p=\delta$ simply or, more strongly as I think, $g|_q = \delta$ for every $q\in U$? Similarly, are you interested in $\nabla_q g=0$ everywhere in $U$, right? Mar 1 at 8:43
• However, if I have understood well your question, $\nabla_q g \neq 0$ in general except for $q=p$. This is because $\partial g=0$ in $U$, but $\nabla_q g = \partial_q g$ only for $q=p$, in general, from the standard action of affine connections on $(0,2)$ tensors and from the fact that in geodesic coordinates of a torsionless connection, the $\Gamma$s vanish at the center of the coordinates only. Mar 1 at 8:53
• @ValterMoretti Yes indeed, fix any $p$ and find a normal coordinate around $p$ means that only at $p$, $\Gamma_p = 0$. So in my question, I want to assume the following: for any $p$ and then pick any $\nabla$-geodesic coordinate centered at $p$, I want $g_p = \delta$ (just at the center $p$). Note that I also want this to hold when we exhaust all $p$ (and the normal coordinates as well). When $g$ is compatible with $\nabla$, I think this is just the standard result: $g_p = δ$ for any $p$ and any associated normal coordinates around $p$. But what happens when $g$ is NOT compatible? Mar 1 at 9:27
• @ValterMoretti Maybe there is some issue in normalization of the basis vector which I am not careful about, when I say "$g_p = \delta$". Mar 1 at 9:32
• OK, I think I have understood... Mar 1 at 9:45