# Metric induced by euclidean vs. normal coordinates (Riemannian Geometry) [closed]

Hey,

I've just started my way in Differential Geometry and been having alot of questions but my first has to do with the difference in metrics.

Say I have a diff. Manifold $S^2$. Embedded in $R^3$ the euclidean Metric induces this metric on $S^2$.

$g_{ij} = \begin{pmatrix} 1 & 0 \newline 0 & \sin^2\theta \end{pmatrix}$

with $\theta$ being part of the polar coordinates.

If I introduce riemannian normal coordinates to $S^2$ the metric should look like this:

$g_{ij} = \begin{pmatrix} 1 & 0 \newline 0 & 1/r^2 \end{pmatrix}$

(with $r,\phi$ being riemannian normal coordinates) Or at least that is what I think it should.

Is my assumption correct? Those Metrics should still give the same result for 2 vectors, correct?

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## closed as not a real question by Anton Petrunin, Deane Yang, Igor Rivin, Will Jagy, Bill ThurstonApr 17 '11 at 20:01

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You are confusing things. Given any metric $g$ on a manifold $M$, there are always local coordinated (called normal coordinates) centered at any point $p$ in which the metric $g_{ij}(p) = \delta_{ij}$ and also $\frac{\partial}{\partial x^k}g_{ij} (p) = 0$. Now on $S^2$, one can place the round metric (which you call the "Euclidean" metric) or one can place some other metric. Any metric will admit local normal coordinates centered at each point. – Spiro Karigiannis Apr 17 '11 at 18:58
Your second matrix $g_{ij}$ is not a metric in normal coordinates. If you have a metric $g$ and look at it in any two local coordinate systems (whether normal or not) then, yes, they will give the same result for the "same" two tangent vectors at a point $p$, but of course the expressions for these "same" two tangent vectors as $2$-tuples of real numbers will be different because you are using two different bases (of coordinate vector fields) for the tangent space $T_p M$. – Spiro Karigiannis Apr 17 '11 at 19:01