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Let $M$ be a manifold, $(x^1,...,x^n)$ a coordinate system on $U \subset M$, and $g$ a metric tensor, then the components of metric tensor on $U$ are:

$g_{ij} = <\partial^i,\partial^j>$

Now the book I am reading from does not show how to calculate $g_{ij}$ for an explicit example, so I want to learn how to do that. Given the simplest case of $\mathbb{R}^n$ with the coordinate chart $x^i(x_1,...,x_n) = x_i$, I want to obtain the dot product. Given 2 vectors $(u_1,...,u_n)$

$\partial^i(x_1,...,x_n) = (0,..,1,...,0)$ at $i$-position.

So $g_{ij} = <\partial^i,\partial^j> = 1$ if $i=j$ and $0$ otherwise, so $g_{ij}= \delta_{ij}$, the kronecker-delta function.

My questions is what I just did correct? Can I apply it to any manifold with any charts?

CuriousAlpaca
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  • well, the fact that for $\Bbb{R}^n$ in the chart you've chosen $g_{ij}=\delta_{ij}$ is almost true by definition. If you really want to practice, calculate the components of $g$ relative to the spherical coordinate chart in $\Bbb{R}^3$. Then, repeat the same thing for $\Bbb{R}^2$ with parabolic coordinates (this is a not so commonly taught example, so if you want to practice calculating, you should try this). – peek-a-boo Feb 27 '21 at 07:14
  • I guess my question is whether the method I employed correct? For any manifold, can I take the derivative of each component of the chart, and then take their dot product to obtain these coefficients? – CuriousAlpaca Feb 27 '21 at 07:20
  • Well, you have to figure out what $\frac{\partial}{\partial x^i}$ is, and then by definition, $g_{ij}=g\left(\frac{\partial}{\partial x^i}, \frac{\partial}{\partial x^j}\right)$. Honestly, I can't comment on whether what you did is right, because this example is so trivial that it is very possible that you may have gotten the right answer for not entirely correct reasoning (I mean it looks like you're on the right track, but then again I can't really be sure whether you actually understood it). This is why I suggest you try a slightly more non-trivial problem. – peek-a-boo Feb 27 '21 at 07:23
  • I couldn't obtain the correct result using the above method, so I must be wrong. – CuriousAlpaca Feb 27 '21 at 07:31
  • take a look at this answer where I carry out (most of) the calculations in the two situations I described above. Hopefully that helps you with learning how to calculate these things. – peek-a-boo Feb 27 '21 at 07:32
  • looking at your examples, and then looking at this answer https://math.stackexchange.com/questions/2527765/how-does-a-metric-tensor-describe-geometry-on-a-manifold I think I understand how to write these coefficients. Given a manifold $M$ and chart $(U,\phi)$, $\phi:U\rightarrow \mathbb{R^n}$, we need to obtain $\phi^{-1}$, compute the partial with respect to each $x^i$, and then take the dot product of these partials according to the above definition. Assuming that our manifold lies in $\mathbb{R^m}$ of course. – CuriousAlpaca Feb 27 '21 at 14:13
  • But then I wonder what if our manifold does not lie in $\mathbb{R^m}$, at least not explicitly so? How would we do these calculations? – CuriousAlpaca Feb 27 '21 at 14:18
  • You need to have the metric defined somehow; what is $\langle \partial_i,\partial_j\rangle$ without a Riemannian metric? BTW, you should get your indices as subscripts, not superscripts. Unless the metric is induced from a the metric on an ambient space, you need to provide it yourself. – Ted Shifrin Feb 27 '21 at 18:54
  • So let's say our manifold lies in some $\mathbb{R}^m$, and we use the standard metric on it. I am still unclear why finding $\partial_i$ would be equivalent (in this case) to finding the inverse of the coordinate chart and take derivative with its argument. Can you elaborate why this is justified? – CuriousAlpaca Feb 28 '21 at 03:54
  • The issue is not calculating the metric. You need to learn what $\oartial_i$ actually means. – Ted Shifrin Feb 28 '21 at 05:29
  • My (possibly incorrect) understanding of $\partial_i$ of a coordinate chart $(\phi,U)$ at a point $x$ is that it is a tangent vector, which takes any $f \in C^\infty(M)$ to $\frac{\partial (f \circ \phi^-1)}{dx^i }(x)$. I am not sure how it helps. If you can elaborate, that would be great. – CuriousAlpaca Feb 28 '21 at 06:04

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