Example. Rationally isometric forms with different trace ratios [fdip-0043]
Example. Rationally isometric forms with different trace ratios [fdip-0043]
Take the rational alternating four-cycle, with counting pairing \(g_0=I\), fixed \(m^2=1\), and opposite vertices \((0,2)\) retained and \((1,3)\) eliminated. For rational \(t>0\), the eigenvalues of its precision are \[1,\quad 3,\quad 2t+1,\quad 2t+3.\] Thus \(q_t(x)=x^TQ_tx\) is positive, and \(L_{\mathrm {rel}}=Q_t-I\) is the earlier \(L_t\), with nonnegative eigenvalues \(0,2,2t,2t+2\). Its trace is \(4(t+1)>0\). These data lie in the full enriched observable domain.
The arithmetic domain calculation gives \[\begin {aligned} \det Q_t&=3(2t+1)(2t+3),\\ R(t)&=\frac {t^2+t+1}{2(t+1)^2}. \end {aligned}\] Here the determinant is also that of the half-polar pairing, because \(g_0=I\). At the two positive rational parameters, \[\begin {aligned} \det Q_1&=45=5\cdot 3^2,\\ R(1)&=\frac 38,\\ \det Q_{29/2}&=2880=5\cdot 24^2,\\ R(29/2)&=\frac {903}{1922}. \end {aligned}\] For example, at \(t=29/2\) the eigenvalues of \(L_{\mathrm {rel}}\) are \(0,2,29,31\); their sum is \(62\) and the sum of squares is \(1806\). Hence the second ratio is \(1806/62^2=903/1922\). It differs from \(3/8\), since \(8\cdot 903=7224\) and \(3\cdot 1922=5766\).
In fact these forms are rationally isometric. The rational matrix \(H/2\) from Example [fdip-000U] diagonalizes \(Q_t\) by congruence as \(\operatorname {diag}(1,3,2t+1,2t+3)\). The change of basis \[\begin {aligned} P&=\begin {pmatrix}1/2&5/2\\1/2&-3/2\end {pmatrix}, &\det P&=-2,\\ P^T\begin {pmatrix}3&0\\0&5\end {pmatrix}P &=\begin {pmatrix}2&0\\0&30\end {pmatrix} \end {aligned}\] therefore gives \[q_1\simeq \langle 1,3,3,5\rangle \simeq \langle 1,3,2,30\rangle .\] At \(t=29/2\), rescaling the last coordinate by \(1/4\) and then permuting the last two coordinates gives \[\begin {aligned} q_{29/2}&\simeq \langle 1,3,30,32\rangle \\ &\simeq \langle 1,3,30,2\rangle \simeq \langle 1,3,2,30\rangle . \end {aligned}\] All these changes of basis are invertible over \(\mathbb Q\). Thus the fiber-constancy criterion already rules out recovery of \(R\) from the rational isometry class of \(q\).
Both signed discriminant classes are \([5]\). Consequently both even Clifford centers are \(\mathbb Q(\sqrt 5)\), and both have the same quadratic Galois character and the same primitive Dirichlet representative \(\chi _5\), of conductor 5. In the fixed uncompleted convention, the values \(L(2,\chi _5)\) are identical. The fiber-constancy criterion therefore rules out a fixed function of this value recovering \(R\) on this family. The same conclusion holds for any fixed character-only collection of L-values. It uses equality of the character, not a numerical approximation to an L-value.
The loss already occurs when the supplied reference pairing is forgotten. The exhibited quadratic isometry cannot preserve that pairing: if it did, the relative precisions would be conjugate and the trace ratios would agree. Hence no invariant determined solely by the quadratic-isometry class, including an invariant of its Clifford algebra or center, recovers \(R\) on this family. Arithmetic data retaining the reference pairing and mass are not excluded. No physical QFT parameter has been identified with \(R\).