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Minkowski space time is defined in terms of a flat pseudo-Riemannian manifold. I have wondered if it can be redefined as Riamannian manifold and in the case what type of curvature would there appear.

Formally:

Let M be a semi-Riemannian manifold of dimension 4, corresponding to the Minkowski space, and let g be the metric tensor (non positive definite), T be the Riemann curvature tensor and P a generic point of M.

Question 1

Which (if any) of the following is true?

a. at every P there exists one system of coordinates for which the metric g becomes Riemannian (positive definite) in a ball of radius R non infinitesimal centred in P

b. there exists one system of coordinates for which g is Riemannian (positive definite) at all P of M

Comment: in words, can we, with a change of coordinates, get rid of semi-Riemannianity – either in finite region or globally?

If this is the case, how do we pay it in terms of curvature? This the next question:

Question 2

c. if previous a) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

d. if previous b) is true, is it true that T cannot be null in the entire ball? And what type of curvature T "displays"?

Thanks a lot

massimo
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    Um...homeomorphic does not mesh well with Euclidean, since homeomorphy does not care for metrics or inner product, so all your questions about homeomorphy make no sense, since the very definition of a manifodl is that it is locally homeomorphic to $\mathbb{R}^n$. For the others, Minkowski space is flat but it is not Euclidean. You should start at the basics of differential geometry and SR/GR before asking such huge questions which are borderline unanswerable. – ACuriousMind Aug 02 '14 at 15:33
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    Comment to the question (v2): Echoing @ACuriousMind's comment, it seems that OP might be using non-conventional terminology. For example, Minkowski spacetime usually refers to $\mathbb{R}^4$ endowed with the Minkowski metric, not general Lorentzian manifolds. See also e.g. this Phys.SE answer. – Qmechanic Aug 02 '14 at 15:35
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    Dear ACuriousMind, homeomorphic was the only transformation I could think of for: isometric would not make sense, because it is "the change of metric" which would enable you to pass from a flat to a curved representation. When you represent the geometry of a curved emisphere (inheriting tridimensional euclidean metric) onto a flat euclidean plan, by orthogonal projection of the first on the second, your metric is no longer Euclidean. – massimo Aug 02 '14 at 15:48
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    Dear Qmechanic, you said right: my use of non-euclidean may not correspond to common use. Not only: it is exactly the analysis of how it changed the non euclidean concept from Gauss/ Lobacevski/ Bolyai to Riemann that made me rethink of Minkowski's model interpretation. – massimo Aug 02 '14 at 16:09

2 Answers2

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We interpret OP's question (v3) as essentially asking

Can a Lorentzian manifold (with Minkowski signature) by coordinate transformations be redefined as a Riemannian manifold (with Euclidean signature)?

The answer is No since the metric signature of a pseudo-Riemannian manifold is invariant under general coordinate transformations. This follows e.g. from Sylvester's law of inertia. Recall that the metric tensor in any coordinate system is a real symmetric matrix, and therefore diagonalizable.

Qmechanic
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  • Dear Qmechanic, thks for your answer BUT my question was significantly different in one point: I did not ask that the Riemannian manifold be "with Euclidean signature"! otherwise I would not have even asked about curvature tensor. Thks anyhow, M – massimo Aug 09 '14 at 13:25
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    Dear @massimo: A Riemannian manifold has by definition Euclidean signature. A pseudo-Riemannian manifold can have other signatures. – Qmechanic Aug 09 '14 at 13:27
  • sorry Qmechanic, may be we use different terminologies here: Sylvester applies to diagonalisable matrices; for me only euclidean riemannian are diagonalisable; "curved" riemannian are not and are those I am after. .... – massimo Aug 09 '14 at 13:31
  • Dear Qmechanic, may be I have understood you now  you would say: every Riemannian is in an infinitesimal ball of its points Euclidean; if Minkowski manifold were changeable into a Riemannian manifold in the infinitesimal ball of one of its points it would therefore be also Euclidean - that it its metric signature would be Euclidean; but this cannot be because of Sylvester ... Have I got you right? thanks – massimo Aug 09 '14 at 13:40
  • Dear @massimo: You are right that small enough neighbourhood of an $n$-dimensional manifold $M$ are diffeomorphic to a neighbourhood of $\mathbb{R}^n$. But focusing on neighbourhoods seems like a distraction here. Consider for simplicity just a single point $p\in M$. Study the metric tensor $g_p: T_p M \times T_p M \to \mathbb{R}$. In any coordinate system the metric $g_{\mu\nu}(p)=g_p(\partial_{\mu},\partial_{\nu})$ is a real symmetric matrix, which is diagonalizable. – Qmechanic Aug 09 '14 at 15:00
  • Sketched proof of Sylvester's law of inertia (in lin. alg. setting): Define kernel ${\rm ker}(g):={v\in V\mid g(v,V)\subseteq{0}}$. Define range ${\rm ran}(g):={0}\cup{v\in V \mid \exists w\in V:~g(v,w)\neq 0}={\rm span}{v\in V \mid \exists w\in V:~g(v,w)\neq 0}$ subspace. ${\rm ran}(g)+ {\rm ker}(g)=V$. ${\rm ran}(g)\cap {\rm ker}(g)={0}$. $n_0:={\rm dim}({\rm ker}(g))$ and ${\rm rank}(g):={\rm dim}({\rm ran}(g))$. Matrix congruence. – Qmechanic Mar 29 '20 at 11:01
  • We can diagonalize $g$ in some basis. Assume that there exist 2 diagonalizing bases. Change of diagonalizing basis: $e_i=e^{\prime}j A^j{}_i$. Then the 2 diagonal metrics are connected as $g{i\ell}=(A^T)i{}^j g^{\prime}{jk}A^k{}{\ell}$. Then $V++V_-={\rm ran}(g)=V^{\prime}++V^{\prime}-$ and $V_+\cap V_- = {0}=V^{\prime}+\cap V^{\prime}-$. But the cross-relations $V_{\pm}\cap V^{\prime}{\mp} = {0}$ must also hold, so that $n{\pm}+n^{\prime}{\mp}\leq {\rm rank}(g)$, which leads to $0\leq n{\pm}-n^{\prime}{\pm}\leq 0$, i.e. $n{\pm}=n^{\prime}_{\pm}$. $\Box$ – Qmechanic Mar 29 '20 at 11:15
  • Notes for later: Canonical form of complex (anti)symmetric $n\times n$ matrix $M^i{}j$. Given positive definite sesqui-linear form $\langle\cdot|\cdot\rangle$. We assume that its matrix $g{ij}$ is real symmetric. (In fact we could have assumed it is $\delta_{ij}$.) Define transposed matrix $M^{t^g}:=g^{-1}M^tg=\pm M$ and Hermitian conjugate matrix $M^{\dagger^g}:=g^{-1}M^{\dagger}g$. The matrix $P:=M^{\dagger^g}M=\pm \overline{M}M$ is semipositive definite and orthonormally diagonalizable. Let $E_i:={\rm Ker}(P-m_i^2\mathbb{1})$ be the mutually orthogonal eigenspaces, $m_i\geq 0$. – Qmechanic Apr 02 '20 at 13:57
  • Define $f(v):=\overline{M}\bar{v}\Rightarrow \overline{f(v)}=Mv$. For $v\in E_i$ note that $\pm f(f(v))=\pm\overline{M}\overline{f(v)}=Pv=m_i^2 v$ $\Rightarrow \pm Mf(v)=m_i^2\bar{v}$ $\Rightarrow Pf(v)=m_i^2\overline{M}\bar{v}=m_i^2f(v)\Rightarrow f(v)\in E_i$. The norm is $||f(v)||^2=||\overline{f(v)}||^2=||Mv||^2=v^{\dagger}M^{\dagger} g M v=\langle v |Pv\rangle = m_i^2||v||^2$. Next $\langle f(v)| v\rangle =f(v)^{\dagger}g v = v^tM^t gv$ which both is equal to $v^tgM v$ and $v^t g M^{t^g}v=\pm v^tgMv$. In antisymmetric case $f(v) \perp v$. – Qmechanic Apr 09 '20 at 13:56
  • Consider the subspace $W:={\rm span}{\mathbb{C}}{v, f(v)}$. Then $M|:W\to \overline{W}$. (In the symmetric case we may assume $f(v)$ is not proportional to $v$. Otherwise $M_|$ already diagonal.) Choose an orthonormal basis $(e_1,e_2)$ for $W$ and an orthonormal basis $(\bar{e}1,\bar{e}_2)$ for $\overline{W}$. (In the antisymmetric case we can choose the orthonormal basis $(e, f(e)/m_i)$ for $W$. The $M|$-matrix takes the form $\begin{pmatrix} 0& m_i\cr -m_i & 0\end{pmatrix}$. ) We have $M_|^{t^g}=\pm M_|$. – Qmechanic Apr 09 '20 at 15:20
  • Construct $g$-unitary matrix $U\in U(2)$ with columns given by an orthonormal basis. $gM_|=U^{t}gD U$, or $M_|=U^{t^g}D U$ where $D=\pm D^{t^g}$ is the canonical block form. $P=U^{\dagger^g}D^{\dagger^g}DU$. Enough to consider $U=\begin{pmatrix} a & -\bar{b} \cr b & \bar{a} \end{pmatrix}\in SU(2)$ as a $U(1)$ phase factor is irrelevant for the format of $D$. The antisymmetric case is already proven. – Qmechanic Apr 10 '20 at 12:44
  • In the remainder we consider the symmetric case. Let $M_|=\begin{pmatrix} A & B \cr B & C \end{pmatrix}.$ (We can assume $B\neq 0$. Else $M_|$ already diagonal.) Then $m_i^2 \mathbb{1}{2\times 2}=P|=\overline{M}|M|=\begin{pmatrix} |A|^2+|B|^2 & \bar{A}B+\bar{B}C \cr A \bar{B}+B\bar{C} & |B|^2+|C|^2 \end{pmatrix}$. $\Rightarrow |A|=|C|$, $A \bar{B}=-B\bar{C}$. Let $\beta:=B/|B|$. Let $\alpha:=A/|A|$ and $\gamma:=C/|C|$. Then $\alpha\gamma=-\beta^2$. So $\alpha=F\beta$ and $\gamma=-\bar{F}\beta$ with $|F|=1$. – Qmechanic Apr 10 '20 at 13:30
  • The off-diagonal element of $D_|$ is $bC\bar{a}-aA\bar{b} +(|a|^2-|b|^2)B\stackrel{?}{=}0$. Choose $a=f\cos\theta$ and $b=\sin\theta$, where $|f|=1$. Then $|A|\beta(F f +\bar{F}\bar{f})=|A|(\alpha f -\gamma\bar{f})=Af-C\bar{f}\stackrel{?}{=} 2B \cot 2\theta$. The LHS is a degenerate ellipse since $|A|= |C|$. Can we chose $D$ to be diagonal? Yes! $\Box$ – Qmechanic Apr 10 '20 at 15:08
  • Related: https://math.stackexchange.com/q/1658393/11127 https://math.stackexchange.com/q/2026110/11127 https://mathoverflow.net/q/125960/13917 Autonne-Takagi factorization. Horn & Johnson, Matrix Analysis, Corollary 2.6.6. – Qmechanic Apr 10 '20 at 16:15
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As mentioned by Qmechanic, the answer to your questions is no.

However, assuming space-time is oriented, we have the following:

For any pseudo-Riemannian metric $g$, there exists a normalized time-like one-form $h^0$ and a Riemannian metric $g^R$ so that $$ g = 2h^0\otimes h^0 - g^R $$

This yields a locally Euclidean topology compatible with the manifold topology.

Christoph
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