I was wondering how do we get the components of the metric tensor? Why, in euclidian 3D space, the metric tensor is represented like this :
$$ g_{\mu\nu}=\left[\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right] $$
And how to find the value for a specific curved space-time?
The metric tensor is a bilinear map that takes in vectors of the tangent space to the manifold. We can expand the metric tensor as $$g(X_i, X_j) = g_{ij}dx^idx^j$$ Now, say the metric is a function of two coordinates, then there are four input combinations for $i,j$. Let's expand the case where $i = 1, j = 1$, then we have $$g(X_1, X_1) = g_{11}(dx^1)^2$$ $$= g_{11}(dx^1(X^1))$$ Recall the dual basis vectors were constructed according to $dx_iX^j = \delta^j_i$. Thus, we have $$g(X_1, X_1) = g_{11}$$
We can repeat this for the other three pairs of $i,j$ to get the remaining components. This is how we can express the metric explicitly in coordinates from an abstract point of view.
If we know information about the embedding of the manifold, we can also use pullbacks to find the induced metric. This method will give the explicit values of the metric, but again you need extra information to do this. Namely, this information would be about the manifold being embedded in and the chart of the submanifold.
