What are the exact axioms to uniquely define the Minkowski metric tensor as a bilinear map? I have read that the definition of a metric tensor is a map with the following axioms:

*

*a bilinear form from the tangent vector space (of a smooth manifold) to the real field

*symmetric

*nondegenerate

[Question] Now, from a purely mathematical prospective: given a map X (defined on a 4D tangent space), is it enough to say that:

*

*$X$ is a metric tensor

*$X$ has signature $(-, +, +, +)$ or $(+, -, -, -)$
to deduce that X is the Minkowski metric tensor?
Note: if the answer is yes, it would mean that Minkowski is the only metric tensor that as a bilinear form has the signature $(-, +, +, +)$.
I think that these axioms are not enough, because in GR we work with metric tensors with the same signature (see this question). Therefore:
[Subquestion part a] Which additional axioms should we include to uniquely define the Minkowski metric tensor as a map?
[Subquestion part b] Would the additional axiom simply be explicitly stating that the coefficients of the bilinear form are all 1 (so -1,+1,+1,+1)?
 A: Up to isomorphisms,  the Minkowski spacetime is a real  four-dimensional  affine space $M^4$ equipped with a Lorentzian scalar product $g$ in the vector space $V^4$ of translations of the affine space.
If $V$ is a real four-dimensional vector space,  a Lorentzian scalar product is a symmetric bilinear map $g: V\times V\to \mathbb{R}$ whose Sylvester's canonical form is $\text{diag}(-1,+1,+1,+1)$.
Given a real four-dimensional vector space $V$ and  a vector basis $e_1,e_2,e_3,e_4$, there exists a unique Lorentzian scalar product whose matrix representation on that basis is $\text{diag}(-1,+1,+1,+1)$.
Therefore, to uniquely fix a Lorentzian scalar product it is sufficient to single out a basis and to declare that the scalar product has the canonical form in that basis.
On the other hand if you have a Lorentzian scalar product, there are infinitely many bases as above. These special bases are related to each other through the transformations of the Lorentz group. (That is the definition of the Lorentz group.)
A: Let $p$ be a point in the manifold. By means of coordinate transformations, any Lorentzian metric tensor can be put in the form $\text{diag}(-+++)$ at $p$ by definition. Hence, your axioms are not enough to define the Minkowski metric.
Referring to the components of the tensor won't work, since they change a lot between different coordinate systems. For example, in spherical coordinates, the same Minkowski metric can be written as $\text{diag}(-1, 1, r^2, r^2 \sin^2\theta)$. Instead, we need to provide some definition that is coordinate invariant, so that it holds regardless of the particular coordinate system we choose to work with.
A property that only the Minkowski metric satisfies is that it is the flat metric, i.e., the Riemann tensor associated with its Levi-Civita connection vanishes. This property, if added to the ones you mentioned, characterizes the Minkowski metric uniquely.
In short, the Minkowski metric is the only flat Lorentzian metric. Notice that this is not enough to characterize the whole manifold as Minkowski spacetime: Minkowski spacetime is topologically $\mathbb{R}^4$, but one can have a flat spacetime with a four-torus topology, for example (namely, space looks like Pacman's world, in which you go out on one end and come back through the other side, and the same holds for time).
A: Given a non-degenerate, symmetric, bi-linear form over the tangent space, expressed as $g_{μν} = g(∂_μ, ∂_ν)$, or equivalently as the tensor $g = g_{μν} dx^μ ⊗ dx^ν$ (summation convention used), where $\{x^μ: μ = 0, 1, ..., N\}$ are the coordinates (at least, locally) and $∂_μ = ∂/∂x^μ$ are the partial differential operators comprising the tangent frame, impose the following extra conditions expressed in terms of the Lie derivatives $ℒ_X$ of certain vector fields $X$:
(1) Homogeneity: $ℒ_{∂_ρ} g = 0$, for all $ρ = 0, 1, ..., N$,
(2) N+1-Isotropy: $ℒ_{x_σ ∂_ρ - x_ρ ∂_σ} g = 0$, for all $ρ, σ = 0, 1, ..., N$ (without loss of generality, you can take $ρ ≠ σ$ or even $ρ < σ$); where $x_0 = x^0$ and $x_i = -x^i$, for $i = 1, 2, ..., N$. That gives you bona fide spatial isotropy with respect to the space-like dimensions $1, 2, ..., N$ and non-accelerationosity (for lack of a better term) with respect to the mixed combinations of the time-like dimension $0$ with each of the spatial dimensions.
Then, the metric is a Minkowski metric (up to a non-zero constant multiple), if $N > 1$.
(The Minkowski metric $η = η_{ρσ} dx^ρ ⊗ dx^σ$ is sneaked into the conditions as the constant diagonal matrix $(+1,-1,-1,-1)$ of coefficients in $x_ρ = η_{ρσ} x^σ$. There is no escaping The $η$.)
For $N = 3$ and 3+1 dimensions, the 10 Lie vectors, in 3D vector notation are:
$${∂ \over ∂t},$$
$$∇ = \left({∂ \over ∂x}, {∂ \over ∂y}, {∂ \over ∂z}\right),$$
$$×∇ = \left(y {∂ \over ∂z} - z {∂ \over ∂y}, z {∂ \over ∂x} - x {∂ \over ∂z}, x {∂ \over ∂y} - y {∂ \over ∂x}\right),$$
$$t∇ +  {∂ \over ∂t} = \left(t {∂ \over ∂x} + x {∂ \over ∂t}, t {∂ \over ∂y} + y {∂ \over ∂t}, t {∂ \over ∂z} + z {∂ \over ∂t}\right),$$
where $t = x^0$ and $ = \left(x, y, z\right) = \left(x^1, x^2, x^3\right)$. The four sets of Lie vectors are, respectively, for Stationarity, Spatial Homogeneity, Spatial Isotropy and Non-Accelerationosity. The metric is to be stationary, spatially homongeneous, isotropic and non-accelerating (for lack of a better term).
First, we do (1). Since
$$ℒ_{∂_ρ} dx^μ = ∂_ρ ˩ ddx^μ + d(∂_ρ ˩ dx^μ) = ∂_μ ˩ 0 + d(δ_ρ^μ) = 0,$$
and $ℒ_{∂_ρ} g_{μν} = ∂_ρ g_{μν}$, then using the product rule for $ℒ_{∂_ρ}$, we have:
$$0 = ℒ_{∂_ρ} g_{μν} dx^μ ⊗ dx^ν = \left(∂_ρ g_{μν}\right) dx^μ ⊗ dx^ν + g_{μν} (0) ⊗ dx^ν + dx^μ ⊗ (0) = ∂_ρ g_{μν} dx^μ ⊗ dx^ν,$$
from which it follows that $∂_ρ g_{μν} = 0$ or that the components $g_{μν}$ are all constant.
Second, we do (2). In general
$$ℒ_X dx^μ = X ˩ ddx^μ + d(X ˩ dx^μ) = ∂_μ ˩ 0 + dX^μ = dX^μ,$$
so for $X = x_σ ∂_ρ - x_ρ ∂_σ$, we have $X^μ = x_σ δ_ρ^μ - x_ρ δ_σ^μ$ and, thus:
$$ℒ_{x_σ ∂_ρ - x_ρ ∂_σ} dx^μ = d\left(x_σ δ_ρ^μ - x_ρ δ_σ^μ\right) = δ_ρ^μ dx_σ - δ_σ^μ dx_ρ.$$
Also, since the components $g_{μν}$ are constant, then we have $ℒ_X g_{μν} = X^ρ ∂_ρ g_{μν} = 0$, regardless of what $X$ is. Thus, using the product rule, again, we have:
$$0 = ℒ_{x_σ ∂_ρ - x_ρ ∂_σ} g = (0) dx^μ ⊗ dx^ν + g_{μν} \left(δ_ρ^μ dx_σ - δ_σ^μ dx_ρ\right) ⊗ dx^ν + g_{μν} dx^μ ⊗ \left(δ_ρ^ν dx_σ - δ_σ^ν dx_ρ\right),$$
or
$$0 = g_{ρν} dx_σ ⊗ dx^ν - g_{σν} dx_ρ ⊗ dx^ν + g_{μρ} dx^μ ⊗ dx_σ - g_{μσ} dx^μ ⊗ dx_ρ,$$
or, componentwise, using the symmetry of $η$ and (assumed) symmtry of $g$ to swap indices:
$$0 = g_{νρ} η_{μσ} - g_{νσ} η_{μρ} + g_{μρ} η_{νσ} - g_{μσ} η_{νρ}.$$
This condition is trivial if $ρ = σ$ or $μ = ν$; particularly, if $N = 0$. If $N = 1$, then without loss of generality, we could take $(ρ,σ) = (0,1) = (μ,ν)$ and write
$$0 = g_{10} (0) - g_{11} (-1) + g_{00} (+1) - g_{01} (0) = g_{00} + g_{11}.$$
That's the best you can do. The metric forms a constant symmetric trace-free $2×2$ matrix.
If $N > 1$, choose any $μ$, $ρ ≠ μ$ and $σ = ν ≠ μ, ρ$. Then, we have:
$$0 = g_{νρ} (0) - g_{νσ} (0) + g_{μρ} η_{νσ} - g_{μσ} (0) = ±g_{μρ}.$$
Thus, $g_{μρ} ≠ 0$ for $ρ ≠ μ$ and $g$ forms a diagonal matrix. Next, choose any $ρ = μ$ and $σ = ν ≠ μ, ρ$. Then, we have:
$$0 = g_{νρ} (0) - g_{νσ} η_{μρ} + g_{μρ} η_{νσ} - g_{μσ} (0) = -g_{νσ} η_{μρ} + g_{μρ} η_{νσ}.$$
From this, it follows that $g_{μρ}/η_{μp} = g_{νσ}/η_{νσ}$. Therefore, $g$ is a constant multiple of the Minkowski metric $η$. Since $g$ is assumed to be non-degenerate, the constant multiple must be non-zero. Otherwise, if it's degenerate, the constant multiple is 0, and then $g$ must be 0 and totally degenerate.
A: Yes, that's enough.
To be pedantic, though, a metric always has a positive definite signature, aka (+,+,+, ..,+) whilst a semi-metric can have arbitrary signature. A manifold with a metric is called a Riemannian manifold whilst a manifold with a semi-metric is called a semi-Riemannian manifold. Often the qualifier 'pseudo' is used instead of 'semi', but I prefer to not use that as the conventional understanding of pseudo means false or fake. A Lorentzian manifold is semi-Riemannian manifold with signature (-+++...+) or (+----...-) and this is what you are after. Minkowski space is simply a flat 4d Lorentzian manifold.
