Compactly Supported Metric Directions Stay Lorentzian for Small Amplitude

By [F1], any Lorentzian form on a ten-dimensional vector space has a basis with matrix $D=\mathrm{diag}(-1,1,\dots ,1)$. Let a symmetric form in this basis have matrix $D+E$, where ${\max }_{a,b}\mathrm{\mid }{E}_{ab}\mathrm{\mid }<1\mathrm{/}18$. For $v={\sum }_{a=1}^9{v}_a{e}_a$ in the positive coordinate subspace, $(D+E)(v,v)\ge \mathrm{\mid }v{\mathrm{\mid }}^2-\max \mathrm{\mid }{E}_{ab}\mathrm{\mid }({\sum }_{a=1}^9\mathrm{\mid }{v}_a\mathrm{\mid }{)}^2\ge (1-9\max \mathrm{\mid }{E}_{ab}\mathrm{\mid })\mathrm{\mid }v{\mathrm{\mid }}^2>0$. Also $(D+E)(e_0,e_0)=-1+E_{00}<0$. Thus the form is positive on a 9-dimensional subspace $W$ and negative on the line spanned by $e_0$. By [F1], write its inertia normal form as $\mathrm{diag}(I_p,-I_q,0_r)$, where $p+q+r=10$. Projection from $W$ to the $p$ positive coordinates is injective: a nonzero vector in its kernel would have quadratic value at most zero, contrary to positivity on $W$; hence $p\ge 9$. Projection from the negative line to the $q$ negative coordinates is also injective, since a vector supported only in positive and zero coordinates has nonnegative quadratic value; hence $q\ge 1$. Therefore $p=9,q=1,r=0$. This proves that the Lorentzian forms form an open subset of the symmetric matrices near $D$. An invertible change of basis acts by an invertible linear congruence map on the space of symmetric matrices, so this open-set conclusion holds near every Lorentzian matrix in every basis.

We first prove the required algebra on an arbitrary open $U\subseteq {\mathbb{R}}^m$, with continuity understood jointly in all coordinates. If $u,v:U\to \mathbb{R}$ are continuous at $z_0$, write $u_0=u(z_0)$ and $v_0=v(z_0)$. For any $\eta >0$, their joint continuity gives one neighborhood where $\mathrm{\mid }u-u_0\mathrm{\mid }<\eta$ and $\mathrm{\mid }v-v_0\mathrm{\mid }<\eta$. On this neighborhood, $\mathrm{\mid }(u+v)-(u_0+v_0)\mathrm{\mid }<2\eta$; if $\eta \le 1$, then $\mathrm{\mid }uv-u_0{v}_0\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }\mathrm{\mid }v-v_0\mathrm{\mid }+\mathrm{\mid }{v}_0\mathrm{\mid }\mathrm{\mid }u-u_0\mathrm{\mid }<(\mathrm{\mid }{u}_0\mathrm{\mid }+1+\mathrm{\mid }{v}_0\mathrm{\mid })\eta$. Choosing $\eta$ small proves joint continuity of sums and products. If $v_0\mathrm{\ne }0$, additionally require $\mathrm{\mid }v-v_0\mathrm{\mid }<\mathrm{\mid }{v}_0\mathrm{\mid }\mathrm{/}2$, so $\mathrm{\mid }v\mathrm{\mid }>\mathrm{\mid }{v}_0\mathrm{\mid }\mathrm{/}2$ and $\mathrm{\mid }{v}^{-1}-v_0^{-1}\mathrm{\mid }\le 2\mathrm{\mid }v-v_0\mathrm{\mid }\mathrm{/}\mathrm{\mid }{v}_0{\mathrm{\mid }}^2$; this proves joint continuity of the reciprocal on its nonzero domain, and products give quotients. These arguments use neighborhoods in $U$, not separate continuity along lines. For smooth inputs, restrict to an open interval in one coordinate direction and apply [F6]. Induction on the length of an ordered derivative word writes every derivative of a finite sum or product as a finite sum of products of input derivatives. For a nowhere-zero smooth $v$, the formula ${\mathrm{\partial }}_i(v^{-1})=-v^{-2}{\mathrm{\partial }}_{i}v$ and the same induction write every derivative of $v^{-1}$ as a finite sum of products of derivatives of $v$ divided by positive integer powers of $v$. Each such expression is jointly continuous by [F5] and the joint-continuity estimates just proved. Thus finite sums, products, and quotients with nonvanishing smooth denominators are smooth on $U$ in the exact ordered-partial sense of [F5].

Let $K=\mathrm{supp}h$. If $K=\mathrm{\varnothing }$, then $G+sh=G$ for every $s$. Otherwise define $\mathcal{C}$ to be the set of all tuples $(p,\phi ,V,\delta ,B)$ such that $p\in K$, $(V,\phi )$ is a smooth chart with $p\in V$, $\delta ,B>0$, the maximum-entry ball of radius $\delta$ around the coordinate matrix $G(p)$ consists of Lorentzian forms, and throughout $V$ one has $\mathrm{\parallel }G(x)-G(p){\mathrm{\parallel }}_{\max }<\delta \mathrm{/}2$ and $\mathrm{\parallel }h(x){\mathrm{\parallel }}_{\max }<B$. Step 1.1 and continuity imply that for every $p\in K$ at least one such tuple exists. Thus the chart neighborhoods occurring in $\mathcal{C}$ form an open cover of $K$; this is the family of all admissible tuples, not a pointwise choice function. By [F3] there is a finite subcover with tuples $(p_j,{\phi }_j,V_j,{\delta }_j,B_j)$, $1\le j\le N$. Set $ϵ:={\min }_{1\le j\le N}{\delta }_j\mathrm{/}(2(1+B_j))>0$. If $x\in V_j$ and $\mathrm{\mid }s\mathrm{\mid }<ϵ$, then $\mathrm{\parallel }G(x)+sh(x)-G(p_j){\mathrm{\parallel }}_{\max }<{\delta }_j$, so $G(x)+sh(x)$ is Lorentzian in that chart. These neighborhoods cover $K$, and off $K$ the metric is unchanged.

Given compactly supported smooth symmetric $q^{MN}$, set $h_{MN}:=-G_{MA}{G}_{NB}{q}^{AB}$; step 1.2 in a manifold chart makes this tensor smooth, and it is symmetric and compactly supported. By step 2.1, $G(s):=G+sh$ is Lorentzian for small $\mathrm{\mid }s\mathrm{\mid }$. On an open parameter-chart domain $U\subseteq \mathbb{R}\times {\mathbb{R}}^{10}$, the functions $G_{MN}(x)$ and $h_{MN}(x)$ lift smoothly: their ordered $x$ derivatives are pulled back by the continuous coordinate projection, and any derivative in $s$ is zero. The coordinate function $s$ is smooth, so step 1.2 makes every entry of $A(s,x):=(G(s{)}_{MN}(x))$ jointly smooth. By [F4], its determinant and adjugate entries are finite polynomials in these entries and hence smooth by step 1.2. Lorentzian nondegeneracy gives $\det A\mathrm{\ne }0$ throughout $U$, so that same result makes $(\det A{)}^{-1}\mathrm{adj}(A)=A^{-1}$ jointly smooth. Differentiate $G(s{)}^{MA}G(s{)}_{AN}={\delta }^M{}_N$ in $s$ at zero using [F6] to obtain ${\mathrm{\partial }}_{s}G(s{)}^{MN}{\mathrm{\mid }}_0=-G^{MA}{G}^{NB}{h}_{AB}=q^{MN}$. This is the required smooth Lorentzian family.