Difference between vielbein and the Jacobian matrix In math books, I saw the metric tensor is defined with the use of the Jacobian matrix as
$$g_{\mu \nu}=J_{\mu}^a \: J_{\nu}^b \: \eta_{ab}\tag{1}$$
where $J_{\mu}^a=\frac{\partial \bar{x}^a}{\partial x^{\mu}}$ (Added: where barred symbols denote Minkowskian coordinates and unbarred ones stand for curvilinear coordinates). And with the matrix notation $\mathrm{g}= \mathrm{J^T} \cdot \eta \cdot \mathrm{J} .$
In 1928 Einstein introduced the $n$-Bein which was further developed and it is known as "tetrad formalism" of GR. The metric tensor in terms of the vierbein (tetrad) field is
$$g_{\mu \nu}={e_{\mu}}^a \: {e_{\nu}}^b \: \eta_{ab}.\tag{2}$$
They both satisfy the orthonormality condition
$${e^{\mu}}_a \: {e_{\nu}}^a=\delta^{\mu}_{\nu} \quad {e_{\mu}}^a \: {e^{\mu}}_b=\delta^{b}_{a}.$$
As the Jacobian matrix is bijective iff $\mathrm{J} \neq 0$ so
$\bar{\mathrm{J}}=\mathrm{J}^{-1}$ and we also have the same as above.
As (1) and (2) look identical, the question is: what is the difference between the Jacobian matrix and the vielbein matrix? Do they represent the same math objects in the application to 4-dimensional space?
My guess is that it is just a matter of terminology and that the Jacobian matrix is used for a broader range of coordinate transformations, though the "vierbein" is the term from the GR that applies to the 4-dimensional case.
References:

*

*Taha Sochi, "Tensor Calculus", https://arxiv.org/abs/1610.04347.

 A: In a nutshell, vielbeins $e^a_{\mu}$ work more generally for abstract manifolds (up to topological obstructions), and generalize the Jacobian $J^a_{\mu}=\partial y^a/\partial x^{\mu}$, which only works for affine spaces. Unlike the vielbeins, the Jacobian always satisfies an integrability condition $\partial J^a_{\mu}/\partial x^{\nu}=(\mu\leftrightarrow \nu)$.
A: A choice of coordinates $x^\mu$ for some patch of spacetime automatically defines a corresponding basis for the tangent space at each point, with basis vectors $\frac{\partial}{\partial x^\mu}$. This is referred to as a coordinate basis, or sometimes as a holonomic basis.
Of course, a choice of basis is in principle independent of a choice of coordinates. The fact that there is a natural coordinate-induced basis available doesn’t mean we have to use it.
This might lead one to wonder if there are choices of basis which cannot be induced by a coordinate chart, and the answer is a resounding yes. As an example, one can show that the familiar orthonormal polar unit vectors $\hat r$ and $\hat \theta $ are such a choice.
When we go from one coordinates chart to another, the Jacobian matrix provides the corresponding transformation between coordinate-induced bases. However, if a non-holonomic basis is involved then there’s obviously no corresponding Jacobian because the non-holonomic basis doesn’t correspond to a choice of coordinates. Therefore, the change of basis needs to be described by a more general object. This is the vielbein matrix $e_\mu^{\ \ \nu}$.

Consider the following example for the standard Euclidean plane with Cartesian coordinates $(x,y)$.  This choice of coordinates corresponds to the (holonomic) basis $\left\{\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right\}$.
If we shift to polar coordinates $(r,\theta)$, we can find a corresponding polar basis $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$. Since we have
$$x = r\cos(\theta) \qquad y = r\sin(\theta)$$
it follows that
$$\frac{\partial}{\partial r} = \frac{\partial x}{\partial r} \frac{\partial}{\partial x} + \frac{\partial y}{\partial r} \frac{\partial}{\partial y} = \cos(\theta)\frac{\partial}{\partial x}+\sin(\theta)\frac{\partial}{\partial y}$$
and similarly for $\frac{\partial}{\partial \theta}$.  Letting $y\equiv (r,\theta)$, this can be written compactly as
$$\frac{\partial}{\partial y^\mu} = \frac{\partial x^\nu}{\partial y^\mu} \frac{\partial}{\partial x^\nu} \equiv J^\nu_{\ \mu} \frac{\partial}{\partial x^\nu}$$
with $J$ the Jacobian.  In this basis, the metric takes the form
$$g = \pmatrix{1& 0 \\ 0 & r^2}$$
which means that this polar basis is orthogonal but not orthonormal.  In contrast, consider the basis
$$\hat r \equiv \cos(\theta)\frac{\partial}{\partial x} + \sin(\theta)\frac{\partial}{\partial y}$$
$$\hat \theta \equiv -\sin(\theta)\frac{\partial}{\partial x} + \cos(\theta)\frac{\partial}{\partial y}$$
One can show without much effort that these basis vectors are orthonormal.  They are not holonomic, however; one can see this by noting that for a smooth function $f$, $\hat r(\hat \theta f) \neq \hat \theta(\hat r f)$, which means that they cannot be expressed as
$$\hat r = \frac{\partial}{\partial u} \qquad \hat \theta = \frac{\partial}{\partial v}$$
for some coordinates $(u,v)$.  Therefore, we cannot write a Jacobian for this coordinate transformation.  Instead, writing $(\hat r,\hat\theta) \equiv (\hat e_r, \hat e_\theta)$, the change of basis is provided by
$$e_\mu^{\ \ \nu} = \pmatrix{\cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta)}$$
$$\hat e_\mu = e_{\mu}^{\ \ \nu} \frac{\partial}{\partial x^\nu}$$
