Timeline for Can Minkowski spacetime be redefined as a non-flat riemannian manifold?
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| Apr 10, 2020 at 16:15 | comment | added | Qmechanic♦ | Related: math.stackexchange.com/q/1658393/11127 math.stackexchange.com/q/2026110/11127 mathoverflow.net/q/125960/13917 Autonne-Takagi factorization. Horn & Johnson, Matrix Analysis, Corollary 2.6.6. | |
| Apr 10, 2020 at 15:08 | comment | added | Qmechanic♦ | 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! $\Box$ | |
| Apr 10, 2020 at 13:30 | comment | added | Qmechanic♦ | 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 $|F|=1$. | |
| Apr 10, 2020 at 12:44 | comment | added | Qmechanic♦ | 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. | |
| Apr 9, 2020 at 15:20 | comment | added | Qmechanic♦ | 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 $M_|^{t^g}=\pm M_|$. | |
| Apr 9, 2020 at 13:56 | comment | added | Qmechanic♦ | 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 $f(v) \perp v$. | |
| Apr 2, 2020 at 13:57 | comment | added | Qmechanic♦ | 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, $m_i\geq 0$. | |
| Mar 29, 2020 at 11:15 | comment | added | Qmechanic♦ | 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. $n_{\pm}=n^{\prime}_{\pm}$. $\Box$ | |
| Mar 29, 2020 at 11:01 | comment | added | Qmechanic♦ | 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. | |
| Aug 9, 2014 at 15:00 | comment | added | Qmechanic♦ | 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. | |
| Aug 9, 2014 at 13:42 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added explanation
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| Aug 9, 2014 at 13:40 | comment | added | massimo | 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 | |
| Aug 9, 2014 at 13:31 | comment | added | massimo | 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. .... | |
| Aug 9, 2014 at 13:27 | comment | added | Qmechanic♦ | Dear @massimo: A Riemannian manifold has by definition Euclidean signature. A pseudo-Riemannian manifold can have other signatures. | |
| Aug 9, 2014 at 13:25 | comment | added | massimo | 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 | |
| Aug 8, 2014 at 23:12 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 5 characters in body
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| Aug 8, 2014 at 23:06 | history | answered | Qmechanic♦ | CC BY-SA 3.0 |