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I’m fairly new to the subject to the area of differential geometry, but as I understand it, the metric tensor $g$ is a tensor field that acts on the tangent space $T_{p}M$ to each point $p$ on a (Riemannian) manifold $M$. Specifically, it is defined as a mapping $g:T_{p}M\times T_{p}M\rightarrow\mathbb{R}$ such that $(v,w)\mapsto g(v,w)$ where $v,w \in T_{p}M$. The metric tensor intuitively gives the inner product of two vectors in a vector space, and thus can be used to determine magnitudes of vectors as well as the angle between intersecting curves tangent to two tangent vectors at a given point.

What I’m really unsure about is how the metric actually describes geometry on the manifold $M$? I know that one can choose a set of basis vectors adapted to a given set of coordinates on the manifold, a coordinate basis, such that the metric takes the form $$g=g_{\mu\nu}(x)dx^{\mu}\otimes dx^{\nu}\equiv g_{\mu\nu}(x)dx^{\mu}dx^{\nu}$$ where $g_{\mu\nu}(x)=g\left(\frac{\partial}{\partial x^{\mu}}, \frac{\partial}{\partial x^{\nu}}\right)$. But this just gives the metric tensor at a point. How can one use $g_{\mu\nu}(x)$ in the entire coordinate chart? Or is the point that one evaluates $g_{\mu\nu}(x)$ at each point lying within the domain of the coordinate chart?

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  • $\begingroup$ That expression with $\mu,\nu$ actually is given by a local coordinate system, which means that it holds locally around a point, in a coordinate chart, not only at a point. $\endgroup$ – Gibbs Nov 19 '17 at 16:32
  • $\begingroup$ @Gibbs Ah Ok. So does that mean it will hold in the entire coordinate chart? What is the reason for this? Is it because the coordinate basis vectors are tangent to the coordinate curves and so span the coordinate chart? $\endgroup$ – user35305 Nov 19 '17 at 16:47
  • $\begingroup$ It holds in a coordinate chart as I said, yes. You know that $g$ is a metric tensor, so locally it will be a linear combination of $dx^{\mu} \otimes dx^{\nu}$, so $g = a_{\mu \nu}(x)dx^{\mu} \otimes dx^{\nu}$. You can easily compute $a_{\mu \nu}(x) = g_{\mu \nu}(x)$. $\endgroup$ – Gibbs Nov 19 '17 at 16:56
  • $\begingroup$ @Gibbs So is the origin of the coordinate chart at the point $p$ at which the coordinate basis vectors are tangent to? How does one relate the basis vectors at each point covered by the chart, i.e. what is the exact reason why $g_{\mu\nu}(x)$ is able to describe the geometry in the entire coordinate chart? Is it to do with the coordinates being smooth functions and so the metric is just a smooth function of coordinates as one moves from point to point $\endgroup$ – user35305 Nov 19 '17 at 17:10
  • $\begingroup$ As stated by eranreches, the metric is essentially the same thing as the geometry. $\endgroup$ – AccidentalFourierTransform Nov 19 '17 at 19:48
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What I’m really unsure about is how the metric actually describes geometry on the manifold $M$?

Geometry is nothing but measuring distances within the manifold, and the metric tells you how to do that. Once you have a ruler, you can do integrals, measure curvature and etc..

I know that one can choose a set of basis vectors adapted to a given set of coordinates on the manifold, a coordinate basis, such that the metric takes the form $$g=g_{\mu\nu}(x)dx^{\mu}\otimes dx^{\nu}\equiv g_{\mu\nu}(x)dx^{\mu}dx^{\nu}$$ where $g_{\mu\nu}(x)=g\left(\frac{\partial}{\partial x^{\mu}}, \frac{\partial}{\partial x^{\nu}}\right)$. But this just gives the metric tensor at a point. How can one use $g_{\mu\nu}(x)$ in the entire coordinate chart? Or is the point that one evaluates $g_{\mu\nu}(x)$ at each point lying within the domain of the coordinate chart?

You can evaluate the metric tensor at each point in the domain of your coordinate chart. If you have an atlas of charts that covers the entire manifold, then you can evaluate the metric entirely.

EDIT: I'll give an example that I hope will solve your doubts in the comments. Lets look at the manifold $S^{2}$ embedded in $\mathbb{R}^{3}$. It can be described by the chart

$$\Phi:\left[0,\pi\right]\times\left[0,2\pi\right)\rightarrow S^{2}$$

$$\Phi\left(\theta,\varphi\right)=\begin{pmatrix}\sin\theta\cos\varphi\\ \sin\theta\sin\varphi\\ \cos\theta\end{pmatrix}$$

You can check the Jacobian of this chart and see that it is singular at the poles, but this can be neglected since those are just two points. Lets now calculate the metric. You have

$$\frac{\partial\Phi}{\partial\theta}=\begin{pmatrix}\cos\theta\cos\varphi\\ \cos\theta\sin\varphi\\ -\sin\theta\end{pmatrix}$$

$$\frac{\partial\Phi}{\partial\varphi}=\begin{pmatrix}-\sin\theta\sin\varphi\\ \sin\theta\cos\varphi\\ 0\end{pmatrix}$$

so the metric is given by

$$g_{\mu\nu}=\begin{pmatrix}\left<\frac{\partial\Phi}{\partial\theta},\frac{\partial\Phi}{\partial\theta}\right>&\left<\frac{\partial\Phi}{\partial\theta},\frac{\partial\Phi}{\partial\varphi}\right>\\\left<\frac{\partial\Phi}{\partial\varphi},\frac{\partial\Phi}{\partial\theta}\right>&\left<\frac{\partial\Phi}{\partial\varphi},\frac{\partial\Phi}{\partial\varphi}\right>\end{pmatrix}=\begin{pmatrix}1&0\\0&\sin^{2}\theta\end{pmatrix}$$

Does the explicit calculation help?

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    $\begingroup$ Thanks for your answer. So is the point that, because one can simply evaluate the metric tensor at each point in a coordinate chart, it’s components $g_{\mu\nu}(x)$ are smooth functions of the coordinates in that chart? Is one implicitly acting on the coordinate basis vectors at each point within the chart to determine $g_{\mu\nu}(x)$ at each point? Does one measure lengths by using the metric to measure lengths of tangent vectors to curves on the manifold? $\endgroup$ – user35305 Nov 19 '17 at 17:20
  • $\begingroup$ ... Is it also correct to say that one can use the metric to measure angles between intersecting curves at a given point by using the metric to measure the angle between their tangent vectors at that point? $\endgroup$ – user35305 Nov 19 '17 at 17:22
  • $\begingroup$ Yes. The metric's components are smooth functions of the coordinates. As for measuring lengths, suppose $\gamma^{\mu}:[0,1]\rightarrow M$ is a curve on your manifold, then $\gamma^{\prime\mu}$ is in the tangent space and $L(\gamma)=\int_{0}^{1}\sqrt{g_{\mu\nu}\gamma^{\prime\mu}\gamma^{\prime\nu}}dt$ is its length. Regarding measuring angles, that's also correct, as the tangent space is a standard inner product space. $\endgroup$ – eranreches Nov 19 '17 at 17:24
  • $\begingroup$ In a given coordinate chart, to construct $g_{\mu\nu}(x)$ for the entire chart is one implicitly evaluating the metric on the coordinate basis at each point within the chart? When integrating the norm of the tangent vector to a curve to find out the curves length, is one using that the curve has a unique vector tangent to it at each point such that one “transitions” from one tangent space to another at consecutive points determining the norm of the associated tangent vector at each point? $\endgroup$ – user35305 Nov 19 '17 at 17:48
  • $\begingroup$ You indeed need to evaluate the metric at each point on the manifold using the chart. As for the integration, there is a unique tangent to the curve. This tangent exists at each tangent space, and is not related to other tangent spaces. You don't need a connection in order to evaluate distances. $\endgroup$ – eranreches Nov 19 '17 at 17:57

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