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Consider Riemann Normal coordinates on a manifold. Consider a point other than the origin. Given that the metric has vanishing derivatives at this point, is it correct to deduce that the metric is Euclidean at this point? If the deduction is correct, how to prove/ argue this?

Kong
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  • If this were true, wouldn't that imply that all Riemannian manifolds are locally Euclidean? – Giuseppe Negro Feb 06 '18 at 23:41
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    Manifolds are locally isomorphic to $R^n$ (I mean real n-manifolds) by definition, by virtue of the coordinate charts. So yes they are Euclidean, if they were not, we couldn't do calculus on manifolds. – Kong Feb 08 '18 at 10:41
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    Yes, I agree, but it depends on what we mean by "isomorphic". In my previous comment I should have written: "If this were true, wouldn't that imply that all Riemannian manifolds are locally isometric to Euclidean space?" Because this last claim is false. – Giuseppe Negro Feb 08 '18 at 12:07

2 Answers2

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If you consider a general point let it be $ x$ in the domain of the normal coordinates, we can't say much. One important relation is given by the Gauss Lemma that can be stated as $$\sum_i x^iu^i = \sum _ix^ig_{i j}(x) u^j \quad \forall u \in \mathbb{R}^n$$

From the Gauss Lemma we have that $$1)\quad \quad x^j = x^ig_{ij}(x).$$ Deriving 1) along $\partial_k$ we obtain $$\partial_k x^j = \delta_k^j = \partial_k x^i g_{i j}(x) + x^i g_{i j,k}(x) = g_{k j}(x) + x^i g_{i j,k}(x)$$ If the derivatives of the metric vanish at $x$, then $g_{ij,k}(x)= 0$ so we obtain that $$\delta_k^j = g_{k j}(x).$$ Notice that also the Christoffel symbols for the Levi-Civita connection vanish at $x$ in this case so you have that for a vector field $X$, $\nabla_{\partial_k} X(x) = \frac {\partial X^i(x)} {\partial x^k} \frac \partial {\partial x^i}|_x$. This can help in computations, also here you can find a proof of an interesting relation The metric tensor at $p$ satisfies $g_{a b, c d} + g_{a d,b c} + g_{a c, d b} = 0$ in a normal coordinate system centered at $p$ .

Overflowian
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  • Thanks for the reply ... this sorts it out :) – Kong Feb 01 '18 at 12:16
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    You are welcome, do not forget to rate the answers that you found useful and if you are satisfied with an answer it is a good practice to accept it (by clicking on the checkmark sign). – Overflowian Feb 01 '18 at 12:40
  • I am looking at it but now find this a little confusing. Let’s refer to 1) above: $x^j=x^i g_{ij}⇒x^j=x_j$. As in would this hold for all points on the manifold? I mean isn’t this the same as saying that the metric is the Kronecker Delta at all points? Excuse the syntax, am still learning to write math at this – Kong Feb 06 '18 at 19:43
  • It was a notational confusion, what you are saying is this: $x_k=x^ig_{ik}\Rightarrow\partial_j x_k=(\partial_j x^i)g_{ik}+x^i\partial_j g_{ik}=\delta_j^ig_{ik}=g_{jk}$. So that: $\partial_j x_k=g_{jk}\Rightarrow\delta_{jk}=g_{jk}$ – Kong Feb 06 '18 at 22:53
  • Yes but again it is more clear IMHO to write $g_{ij}(x)$ since the derivatives vanish there not in general. By the way, the example I gave above of a metric is not so good because the metric should be symmetric, so you can consider for example a diagonal matrix that is not the identity . – Overflowian Feb 07 '18 at 07:33
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    Agreed dear Warlock. – Kong Feb 08 '18 at 10:35
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Clearly: $x_k=x^ig_{ik}$. Consider differentiating on both sides: $\partial_j x_k=(\partial_j x^i)g_{ik}+x^i\partial_j g_{ik}$. Now if the metric derivatives vanish then the last term is zero and we have: $\partial_j x_k=(\partial_j x^i)g_{ik}\Rightarrow\delta_{jk}=\delta_j^ig_{ik}\Rightarrow\delta_{jk}=g_{jk}$.

Kong
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