Zero sampled constraint residual with unresolved smooth data

Let $N\ge 4$ be even, let $N_2,N_3\ge 3$, let $L_{\ast }>0$ be a length, and choose $0<\epsilon <1$. On ${\mathbb{T}}_{2\pi }^3$ with dimensionless angular coordinates $x^i$, set

$$\tilde{h}=L_{\ast }^2((d{x}^1{)}^2+(d{x}^2{)}^2+(d{x}^3{)}^2),\tilde{\sigma }=0,\tau =0,{\varphi }_N(x)=1+\epsilon \sin (N{x}^1\mathrm{/}2)\mathrm{.}$$

Take the vacuum branch with $\mathrm{\Lambda }=0$ and $\rho =j=0$. Fix any $\kappa >0$ with the SI units specified by the discrete diagnostic; its value drops out because $j=0$. The conformal data are $h={\varphi }_N^4\tilde{h}$ and $K=0$. Use a periodic mesh with $N_1=N$ and counts $N_2,N_3$ in the other directions. Put ${\mathrm{\Delta }}_i=2\pi \mathrm{/}{N}_i$ and use nodes $x_{j_i}^i=2\pi j_i\mathrm{/}{N}_i$, $0\le j_i<N_i$, with indices wrapped modulo each $N_i$. For $i=1,2,3$, let ${\mathbf{e}}_i$ be the $i$th standard basis vector of ${\mathbb{Z}}^3$. For a scalar grid array $f_{\mathbf{j}}$, use the centered periodic maps

$$({\delta }_{i}f{)}_{\mathbf{j}}:=\frac{f_{\mathbf{j}+{\mathbf{e}}_i}-f_{\mathbf{j}-{\mathbf{e}}_i}}{2{\mathrm{\Delta }}_i},i=1,2,3,$$

with indices reduced modulo $(N_1,N_2,N_3)$, and apply them componentwise. For reconstruction, partition each periodic coordinate circle into the half-open cells

$$C_{\mathbf{j}}:=\prod _{i=1}^3[\frac{2\pi (j_i-1\mathrm{/}2)}{N_i},\frac{2\pi (j_i+1\mathrm{/}2)}{N_i})(\mathrm{m}\mathrm{o}\mathrm{d}2\pi ),$$

and assign every component on $C_{\mathbf{j}}$ its value at node $\mathbf{j}$. This specifies the piecewise-constant map ${\mathcal{I}}_{\mathbf{\Delta }}$ in the constraint-residual definition. For the reported residual arrays use uniform dimensionless weights $w_{\mathbf{j}}:=(N{N}_2{N}_3{)}^{-1}$ and the norm

$$\mathrm{\parallel }({\mathcal{H}}_{\mathbf{\Delta }},{\mathcal{M}}_{\mathbf{\Delta }}){\mathrm{\parallel }}_{\mathbf{\Delta },2}:={\left(\sum _{\mathbf{j}}{w}_{\mathbf{j}}(\mathrm{\mid }{\mathcal{H}}_{\mathbf{\Delta }}(\mathbf{j}){\mathrm{\mid }}^2+h^{ik}(\mathbf{j}){\mathcal{M}}_{\mathbf{\Delta },i}(\mathbf{j}){\mathcal{M}}_{\mathbf{\Delta },k}(\mathbf{j}))\right)}^{1\mathrm{/}2}\mathrm{.}$$

The bars around the scalar residual denote the real absolute value. For the sampled scalar and covector arrays, define the bilinear form

$$⟨(H,M),(\widehat{H},\widehat{M}){⟩}_{\mathbf{\Delta }}:=\sum _{\mathbf{j}}{w}_{\mathbf{j}}(H(\mathbf{j})\widehat{H}(\mathbf{j})+h^{ik}(\mathbf{j})M_i(\mathbf{j}){\widehat{M}}_k(\mathbf{j}))\mathrm{.}$$

At the nodes of this example $h^{ik}=L_{\ast }^{-2}{\delta }^{ik}$ and every $w_{\mathbf{j}}=(N{N}_2{N}_3{)}^{-1}$ is positive. Hence this bilinear form is positive definite on the finite-dimensional space of scalar and covector grid arrays, and the displayed quantity is its induced norm. In the fixed SI component convention, each summand has units ${\mathrm{m}}^{-4}$, so the norm has units ${\mathrm{m}}^{-2}$. The recorded mesh hierarchy for this diagnostic is the singleton containing this grid; no refinement or convergence comparison is made.

The discrete residuals ${\mathcal{H}}_{\mathbf{\Delta }}$ and ${\mathcal{M}}_{\mathbf{\Delta }}$ vanish at every grid node, so their stated weighted norm is zero. However, the continuum Hamiltonian residual at $x^1=\pi \mathrm{/}N$ is

$$\mathcal{H}(h,K)=\frac{2\epsilon N^2}{L_{\ast }^2(1+\epsilon {)}^5}>0.$$

Thus a smooth high-frequency field can have the same samples and zero discrete residual as flat data while violating the continuum constraint between grid points. A small or vanishing grid residual alone does not certify continuum constraint satisfaction or solution accuracy.

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Christoffel formula for the levi civita connection

Curvature of an affine connection

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The zero sets of sine and cosine and the least positive common period 2 pi

Quarter-turn values and shifts by pi/2 and pi

The derivatives of sine and cosine are cosine and minus sine

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