Constraint preservation does not imply bounded leapfrog evolution

The matrices in [A1] are the signed boundary-incidence maps of one oriented tetrahedron, with edges ordered $01,02,03,12,13,23$, faces ordered $012,013,023,123$, and cell orientation $0123$. Matrix multiplication gives $D_1{D}_0=0_{4\times 4}$ and $D_2{D}_1=0_{1\times 6}$. Also $D_{2}x=-1+1=0$ and $D_0^{\mathsf{T}}{d}_{1\mathrm{/}2}=0_{4\times 1}$, so the initial state satisfies both constraints.

Put $c:=(-1,1,-1,1{)}^{\mathsf{T}}$, so $D_2=c^{\mathsf{T}}$. Direct multiplication gives $D_1{D}_1^{\mathsf{T}}=\left(\begin{array}{cccc}3& 1& -1& 1\\ 1& 3& 1& -1\\ -1& 1& 3& 1\\ 1& -1& 1& 3\end{array}\right)=4I_4-c{c}^{\mathsf{T}}$. For every $z\in {\mathbb{R}}^4$, this gives $\parallel D_1^{\mathsf{T}}z{\parallel }_2^2=4\parallel z{\parallel }_2^2-(c^{\mathsf{T}}z{)}^2\le 4\parallel z{\parallel }_2^2$. For real coordinate vectors $u,v\in {\mathbb{R}}^m$, direct expansion gives $\parallel u{\parallel }_2^2\parallel v{\parallel }_2^2-(u^{\mathsf{T}}v{)}^2={\sum }_{1\le i<j\le m}(u_i{v}_j-u_j{v}_i{)}^2\ge 0$, hence $\mathrm{\mid }{u}^{\mathsf{T}}v\mathrm{\mid }\le \parallel u{\parallel }_2\parallel v{\parallel }_2$. Applying this with $u=D_1^{\mathsf{T}}z$ and $v\in {\mathbb{R}}^6$ gives $\mathrm{\mid }{z}^{\mathsf{T}}{D}_{1}v\mathrm{\mid }\le 2\parallel z{\parallel }_2\parallel v{\parallel }_2$. For $w=D_{1}v$, applying the same identity with $u=z$ and $v=w$ gives $\mathrm{\mid }{z}^{\mathsf{T}}w\mathrm{\mid }\le \parallel w{\parallel }_2$ for unit $z$; if $w\mathrm{\ne }0$ equality is attained at $z=w\mathrm{/}\parallel w{\parallel }_2$, while if $w=0$ both sides are zero. Thus $\parallel w{\parallel }_2={\max }_{\parallel z{\parallel }_2=1}\mathrm{\mid }{z}^{\mathsf{T}}w\mathrm{\mid }$ and $\parallel D_{1}v{\parallel }_2\le 2\parallel v{\parallel }_2$ for every $v\in {\mathbb{R}}^6$. Since $c^{\mathsf{T}}x=0$, $e:=x\mathrm{/}\sqrt{2}$ is unit and $D_1{D}_1^{\mathsf{T}}e=4e$; consequently $f:=D_1^{\mathsf{T}}e\mathrm{/}2$ is unit and $D_{1}f=2e$. Thus the operator norm of $D_1$ is exactly $2$, attained at $f$. Here $M_1=M_2=I$, so the $\sigma$ defined in [F3] is $2$. For a state $b=ue$, $d=vf$, the update in [F2] with $\tau =2$ gives $u^{+}=u-4v$ and $v^{+}=v+4u^{+}=4u-15v$, hence the invariant two-dimensional subspace has matrix $\left(\begin{array}{cc}1& -4\\ 4& -15\end{array}\right)$. The initial state has modal coordinates $(u,v)=(\sqrt{2},0)$.

The matrix in step 2.1 has trace $-14$, determinant $1$, and eigenvalues ${\lambda }_{-}=-7-4\sqrt{3}$ and ${\lambda }_{+}=-7+4\sqrt{3}$, found by solving ${\lambda }^2+14\lambda +1=0$. Since $6<4\sqrt{3}<7$, ${\lambda }_{-}<-1$ and $-1<{\lambda }_{+}<0$. Direct substitution gives eigenvectors $v_{-}=(1,2+\sqrt{3}{)}^{\mathsf{T}}$ and $v_{+}=(1,2-\sqrt{3}{)}^{\mathsf{T}}$; their determinant is $-2\sqrt{3}\mathrm{\ne }0$, so write the initial modal vector as $(\sqrt{2},0{)}^{\mathsf{T}}=a{v}_{-}+b{v}_{+}$. The coefficient $a$ is nonzero, since otherwise $(\sqrt{2},0{)}^{\mathsf{T}}$ would be a multiple of $v_{+}$, whose second coordinate $2-\sqrt{3}$ is nonzero. The first modal coordinate after $n$ steps is therefore $u_n=a{\lambda }_{-}^n+b{\lambda }_{+}^n$. If $u_n$ were bounded, then $b{\lambda }_{+}^n$ would be bounded because $\mathrm{\mid }{\lambda }_{+}\mathrm{\mid }<1$, so their difference $a{\lambda }_{-}^n$ would be bounded, contradicting $a\mathrm{\ne }0$ and $\mathrm{\mid }{\lambda }_{-}\mathrm{\mid }>1$. Since $\parallel b_n{\parallel }_2\ge \mathrm{\mid }{u}_n\mathrm{\mid }$, the state orbit is unbounded.

By [F3], each update preserves both constraints, and step 1.1 sets them to zero initially. Explicitly, $D_2{b}^{+}=D_{2}b-\tau D_2{D}_1{M}_1^{-1}d=D_{2}b$ and $D_0^{\mathsf{T}}{d}^{+}=D_0^{\mathsf{T}}d+\tau (D_1{D}_0{)}^{\mathsf{T}}{M}_2{b}^{+}=D_0^{\mathsf{T}}d$, so induction gives zero constraints at every iterate. Thus exact constraint preservation holds at every step although this rational finite-complex algorithm has an unbounded orbit for $\tau \sigma =4>2$. This refutes the stated bounded-orbit claim.