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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)?

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  • $\begingroup$ Are you allowing similarity transformations in your definition of what the Minkowski metric "is"? Any bilinear form whose components (in a given basis) are $$\begin{bmatrix}-a^2 & 0 & 0 & 0 \\ 0 & b^2 & 0 & 0 \\ 0 & 0 & c^2 & 0 \\ 0 & 0 & 0 & d^2 \end{bmatrix}$$has signature $(-,+,+,+)$. But it's only equal to the Minkowski metric (in this basis) if $a = b = c = d = 1$. $\endgroup$ Feb 13, 2022 at 14:41
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    $\begingroup$ Or, for another example, consider the bilinear form with components$$\begin{bmatrix} -a^2 + c^2 & -ab + cd & 0 & 0 \\ -ab+cd & -b^2 + d^2 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ which is what you get when you apply an arbitrary similarity transformation on the $t$ and $x$ coordinates of the Minkowski metric. Is this still the Minkowski metric by your definition? $\endgroup$ Feb 13, 2022 at 14:45
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    $\begingroup$ The Minkowski spacetime in is a four-dimensional affine space whose vector space of translations is equipped with a non-singular bilinear form with signature -,-,-,+. $\endgroup$ Feb 13, 2022 at 14:52
  • $\begingroup$ @MichaelSeifert Hi! In my question, I am not defining Minkowski metric because that would be circular reasoning from my side: i am asking readers to define what it is, via telling me what are its axioms. The idea is to pretend that i don't even know what Minkowski metric is, and that readers tell me what it is, by defining the axioms that uniquely characterise the map. (Are the axioms that i wrote enough to do so, or are more needed? If so, which ones?) $\endgroup$
    – TrentKent6
    Feb 13, 2022 at 15:00
  • $\begingroup$ @ValterMoretti Hi! I am asking the axioms of the bilinear form, not of the spacetime. I can however deduce from your definition, that for you the answer to my question is yes, and therefore according to you, the Minkowski metric tensor is the only metric tensor that as a bilinear form has the signature (-, +, +, +). $\endgroup$
    – TrentKent6
    Feb 13, 2022 at 15:02

4 Answers 4

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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.)

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Feb 14, 2022 at 9:02
  • $\begingroup$ what is the difference between saying that the scalar product has diag(−1,+1,+1,+1) as Sylvester's canonical form (as you do in your answer) and saying that the coefficients of the bilinear form are (-1,+1,+1,+1) (as I do at the end of my question)? $\endgroup$
    – TrentKent6
    Feb 17, 2022 at 16:50
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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).

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  • $\begingroup$ i've a question. In the case of the real 4D affine space of special relativity, we know that Lorentz transformations (+ translations) are the unique transforms that leave invariant the line element of the manifold. If my coordinate system is always spherical (so starts&ends spherical) the inner product of the manifold (i.e. minkowski metric tensor, which can give the line element) should be written accordingly (as displayed in this answer). My question is then: how would the Lorentz transformation look like, in order to leave this inner product (and consequently, the line element) invariant? $\endgroup$
    – TrentKent6
    Feb 16, 2022 at 15:20
  • $\begingroup$ @TrentKent6 I think that is complicated enough to deserve a new question hahaha. Not all of the Lorentz transformations would keep that line element invariant: if you make a boost in some direction, you're choosing a direction in which you're boosting, so the previously isotropic coordinate system will have to adapt to deal with that. My >guess< is that only rotations of the coordinate system would satisfy this, but I'm not entirely sure $\endgroup$ Feb 16, 2022 at 16:44
  • $\begingroup$ I was thinking back at your answer, and there's something unclear in your definition of minkowski ≡ metric tensor with (R=0). Consider a completely flat spacetime (so R=0 everywhere) and a non-inertial (aka accelerating) observer. GLOBALLY (for example, at a point located 1 km away; so i stress: not locally!) this observer will see the line element there be described not by the minkowski metric, but by a different metric (since he's not inertial). And yet, it's still R=0 there. If they could describe the global point with minkowski, they would be inertial, contradicting the starting point... $\endgroup$
    – TrentKent6
    Feb 17, 2022 at 11:35
  • $\begingroup$ [...] it follows that, since the metric tensor used for the far away point is not minkowski and since it has R=0, minkowski can't be the only metric tensor with (R=0). $\endgroup$
    – TrentKent6
    Feb 17, 2022 at 11:37
  • $\begingroup$ @TrentKent6 The tensor is coordinate and observer independent. An accelerated observer does not see a Minkowski metric in the same sense that spherical coordinates do not write $(-,+,+,+)$. The tensor is a geometric object defined without the need of coordinates or observers, only its components depend on the coordinates $\endgroup$ Feb 17, 2022 at 17:01
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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: $${∂ \over ∂t},$$ $$∇ = \left({∂ \over ∂x}, {∂ \over ∂y}, {∂ \over ∂z}\right),$$ $$𝐫×∇ = \left(y {∂ \over ∂z} - z {∂ \over ∂y}, z {∂ \over ∂x} - x {∂ \over ∂z}, x {∂ \over ∂y} - y {∂ \over ∂x}\right),$$ $$t∇ + 𝐫 {∂ \over ∂t} = \left(t {∂ \over ∂x} + x {∂ \over ∂t}, t {∂ \over ∂y} + y {∂ \over ∂t}, t {∂ \over ∂z} + z {∂ \over ∂t}\right),$$ where $t = x^0$ and $𝐫 = \left(x, y, z\right) = \left(x^1, x^2, x^3\right)$. 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.

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  • $\begingroup$ 1) do you have sources? i would TRULY like to read more about this exact answer. $\endgroup$
    – TrentKent6
    Oct 27, 2022 at 23:16
  • $\begingroup$ 2) do these axioms also work for the "minkowski metric tensor" which appears ONLY LOCALLY in any possible general relativity geometries? $\endgroup$
    – TrentKent6
    Oct 27, 2022 at 23:17
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    $\begingroup$ That's actually the standard way to get solutions possessing a range of symmetries. Start with a general metric, require that its Lie bracket with the relevant symmetries vanish. That defines the isometries of the metric and the vector fields are called its Killing Fields. For the black hole (or Schwarzschild) solution, for instance, you impose only the isotropy and stationary conditions, not spatial homogeneity (since it has a center) nor non-accelerationess (since the center is fixed). The construction is not local; particularly, the homogeneity (i.e. "translation") symmetries are not. $\endgroup$
    – NinjaDarth
    Nov 2, 2022 at 18:58
  • $\begingroup$ Adding to my previous reply, in the "Exact Solutions" book/catalogue, they actually do the analyses, in a similar way, with Killing Fields to lay out the general solutions families, before going into the catalogue entries. cambridge.org/core/books/… $\endgroup$
    – NinjaDarth
    Sep 25, 2023 at 3:05
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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.

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  • $\begingroup$ hi, to which part of my question is your answer "yes" referring to? is it the "question" or the "subquestion part b" (where i introduce the coefficients -1,1,1,1)? $\endgroup$
    – TrentKent6
    Feb 17, 2022 at 18:32
  • $\begingroup$ @TrentKent6: To the main question. $\endgroup$ Feb 20, 2022 at 22:40
  • $\begingroup$ but the Schwarzschild metric tensor also satisfies those 2 axioms, and yet it is not the Minkowski metric tensor... $\endgroup$
    – TrentKent6
    Feb 20, 2022 at 22:43
  • $\begingroup$ @TrentKent6: Ah, ok. I see what you are driving at. I'd say that the additional axiom you need is that the space is flat, that is the Riemann curvature vanishes. $\endgroup$ Feb 20, 2022 at 22:47
  • $\begingroup$ but in curved spacetime, locally (at any point) you still have minkowski metric tensor, and yet at at that point the Riemann curvature is not required to vanish... $\endgroup$
    – TrentKent6
    Feb 20, 2022 at 22:53

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