Is the metric tensor relative to a reference coordinate system? I am fairly new to this topic, and I don't know anything about tensors, but I have to know what a metric tensor is and this is how I have been thinking about it:
1)A metric tensor is a mathematical entity that expresses distances between points in generalized coordinates;
This part I got quite well I think, but now here's my question:
I am a being that lives in a flat 3D space, that is, everything I see and experience is according to cartesian coordinates and even if I represent something with other coordinate system (say, with spherical coordinates) that new system arises from my experience on a 3D flat space. Another way of saying this is that the "instinctive" metric tensor I live by is:
$$g_1=\begin{bmatrix}
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    0 & 0 & 1
\end{bmatrix}$$
But if I then, somehow, want to be in a spherical space, that is, a space where the coordinates are $(r,\theta ,\phi )$, then I say (well the books do) that my new metric space is:
$$g_2=\begin{bmatrix}
    1 & 0 & 0 \\
    0 & r^2 & 0 \\
    0 & 0 & r^2sin^2(\theta)
\end{bmatrix}$$
But that results seems to me that is biased,that is, $g_2$ exists as it is because I want that a given distance between 2 points in 3D space to have the same value in the spherical space between those 2 same points when they are parametrized from one space to the other. But what if it all went backwards? Suppose now that I live in a 3D spherical space, I could represent all of space with a 3 orthogonal axis space (like the figure below) and simply replace x,y and z by $(r,\theta ,\phi)$ and I would and my "natural" metric tensor would still be $g_1$ but when I went to the 3D flat space, then $g_2$ would not be the same.
My question/argument is that all metric tensors are relative to our own experience of day to day life, and all other metric spaces exist as they do to agree with our experience, to be in conformity with our reference ($g_1$), thus we cannot talk of metric tensors as absolute entities, we have to mention the reference as well.

 A: Writing a $2$-tensor as a matrix does depend on "reference". Formally, it goes like this:


*

*We already have a smooth manifold $M$ with a $2$-tensor field $g$ on $M$. For no particular reasons, $g$ happens to be a metric tensor.

*We have an open set $U\subseteq M$ and a parameterisation $\varphi:V\to U$ for some open set $V\subseteq\Bbb R^n$.

*There is a unique metric tensor $\varphi^*g$ on $V$ that makes $\varphi$ an isometry, i.e. $\varphi$ is a function that preserves distance.

*As a $2$-tensor field in $\Bbb R^n$ space, $\varphi^*g$ can be written as an $n$ by $n$ matrix (more properly, a matrix-valued function on $V$).
In the case you mentioned, $g_1$ and $g_2$ are two examples of $\varphi^*g$ (with different $\varphi$). The matrix (like $g_1$ and $g_2$) simply tells you how to describe the same metric in different coordinates. This is the logic flow of how they are obtained:


*

*We already have $\Bbb R^3$ as a smooth manifold with the usual metric $g$.

*We have two parameterisations $\varphi_1,\varphi_2$ of (a part of) $\Bbb R^3$, where $\varphi_1$ is the identity function on $\Bbb R^3$ and $\varphi_2$ is spherical coordinates on $\Bbb R^3$.

*$g_1$ is defined as $\varphi_1^*g$, while $g_2$ is defined as $\varphi_2^*g$.
If you want to, you can consider another metric $g_0$ on (a part of) $\Bbb R^3$ such that $$\varphi_2^*g_2=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix},$$ but then you would face the problem that you are not describing the same metric $g$, but instead you are describing another metric $g_0$.

Notes:


*

*Physicists quite often refrain from teaching students about smooth manifolds even when there is one being investigated. When you see the word "generalized coordinates", they are almost always talking about coordinates on a smooth manifold.

*A parameterisation $\varphi:V\to U$ is also called generalized coordinates among physicists, and called smooth coordinate chart among mathematicians.

*$\varphi^*g$ is called the pullback metric on $V$.
