Gaussian elimination at every prime [fdip-008M]

Which parts of Gaussian elimination depend on continuous quantities? The arithmetic families place the Clifford data over every ring in which \(2\) is a unit, and the bad fibers separate the domains of the form, the elimination, and the trace ratio. The Gaussian law asks a further question. A moment can lie in \(\mathbb Q\) while the constant that normalizes the real density does not.

Take the unit-weight four-cycle with mass squared equal to \(1\). In vertex order \((0,1,2,3)\) its precision is \[Q=\begin {pmatrix}3&-1&0&-1\\-1&3&-1&0\\0&-1&3&-1\\-1&0&-1&3\end {pmatrix},\] so \(\det Q=45\). For these 4 coordinates the real normalizer is \[(2\pi )^{2}(\det Q)^{-1/2},\] as in the finite positive Gaussian normalization. That quantity involves both \(\pi \) and \(\sqrt {45}\). Retain the adjacent vertices \(R=\{0,1\}\) and eliminate \(U=\{2,3\}\). This split is not the opposite-vertex split of the four-cycle elimination. The Schur complement and one moment computed in the next section are \[S=\begin {pmatrix}21/8&-9/8\\-9/8&21/8\end {pmatrix},\qquad \omega _Q(x_2^2x_3x_0+x_2^3x_1)=32/75.\] Both lie in \(\mathbb Q\). In this example the continuous quantity is the real normalizer.

The next section defines the moment rule over a ring and proves that elimination of those moments is the Schur complement, with no real integral. The following section constructs the additive character sum on an odd finite field and its elimination identity. The section after that identifies the Euler product of every nondegenerate integral form with an \(L\)-function of its discriminant character. The last section sums the Epstein zetas of the classes of a negative fundamental discriminant. The Weil index is outside that identity. This chapter does not identify the character sum with the real normalizer.