Gaussian elimination at every prime [fdip-008M]
✍️sourceAGENTDRAFTED
- November 2, 2026
Gaussian elimination at every prime [fdip-008M]
✍️sourceAGENTDRAFTED
- November 2, 2026
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.
Theorem 1. Wick states over a ring [fdip-008N]AGENTDRAFTED
- November 2, 2026
Theorem 1. Wick states over a ring [fdip-008N]AGENTDRAFTED
- November 2, 2026
Let \(A\) be a commutative ring in which \(2\) is a unit, and let \(Q\) be a symmetric \(n\times n\) matrix over \(A\) with an inverse \(C=Q^{-1}\) in \(M_n(A)\). Write \(A[x]=A[x_1,\ldots ,x_n]\). The Wick functional \(\omega _Q\) is the \(A\)-linear map \(A[x]\to A\) with \(\omega _Q(1)=1\) determined by Isserlis' pairing rule. On a product of an odd number of factors the value is \(0\). On a product of \(2m\) factors, not necessarily distinct, sum over the pairings of those \(2m\) factor positions the product of the \(m\) entries of \(C\) on the paired positions. There are \((2m-1)!!\) such pairings, so \(\omega _Q(x_i^4)=3C_{ii}^2\). The value depends only on those entries. It uses no order on \(A\) and no measure on \(A^n\). The same pairing rule is the one applied to centered real Gaussians in the quartic elimination, which cites [etingof2024mathematical, v2, Theorem 3.1 and Example 3.2, p. 29] for these contractions. Here the rule is the definition of \(\omega _Q\).
Order the coordinates as a retained set \(R\) and an eliminated set \(U\), and assume that the block \(Q_{UU}\) is invertible over \(A\). Define \[\begin {aligned} L&=Q_{UU}^{-1}Q_{UR},\\ B&=-L,\\ S&=Q_{RR}-Q_{RU}L. \end {aligned}\] Symmetry of \(Q\) makes \(S\) symmetric. For \(f\in A[x]\), substitute \(x_U=Bx_R+z\) and apply the Wick functional of covariance \(Q_{UU}^{-1}\) in the indeterminates \(z\), treating the retained variables as coefficients. Call the result \(E[f\mid x_R]\in A[x_R]\).
The block \(S\) is invertible, and \(C_{RR}=S^{-1}\). Restriction of \(\omega _Q\) to \(A[x_R]\) is the Wick functional \(\omega _S\). Furthermore \(E[f\mid x_R]\) is a polynomial and \[\omega _Q(f)=\omega _S(E[f\mid x_R]).\] The identity \(2^{-1}x^TQx=2^{-1}z^TQ_{UU}z+2^{-1}x_R^TSx_R\), with \(z=x_U-Bx_R\), is the quadratic completion that names these blocks. The moment proof does not integrate that quadratic.
Proof.
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Proof.
- November 2, 2026
Factor the symmetric matrix in retained and eliminated blocks as \[\begin {aligned} Q&=\begin {pmatrix}I&Q_{RU}Q_{UU}^{-1}\\0&I\end {pmatrix} \begin {pmatrix}S&0\\0&Q_{UU}\end {pmatrix} \begin {pmatrix}I&0\\L&I\end {pmatrix}. \end {aligned}\] The two outer factors are invertible over \(A\): each is undone by changing the sign of its off-diagonal block. The hypothesis already makes \(Q_{UU}\) invertible. If \(S\) is invertible, the block-diagonal matrix of \(S^{-1}\) and \(Q_{UU}^{-1}\) inverts the middle factor, so that factor and \(Q\) are invertible. Conversely, if \(Q\) is invertible then the middle factor \(D\) is invertible. If \(DE=I=ED\), the retained block of \(DE\) is \(S\) times the retained block of \(E\), and the retained block of \(ED\) is that block times \(S\). Both products are the identity, so \(S\) is invertible. The same reading applies to \(Q_{UU}\). Thus, once \(Q_{UU}\) is invertible, \(Q\) is invertible if and only if \(S\) is. The running hypotheses give both, so \(S\) is invertible. The inverse is \[\begin {aligned} C&=\begin {pmatrix}I&0\\-L&I\end {pmatrix} \begin {pmatrix}S^{-1}&0\\0&Q_{UU}^{-1}\end {pmatrix} \begin {pmatrix}I&-Q_{RU}Q_{UU}^{-1}\\0&I\end {pmatrix}. \end {aligned}\] Reading blocks gives \(C_{RR}=S^{-1}\) and \(C_{UR}=-LS^{-1}\). A polynomial in the retained variables alone is paired only on retained indices, so its value under \(\omega _Q\) uses only \(C_{RR}\). That is \(\omega _S\).
The same inverse gives \[C_{UU}=Q_{UU}^{-1}+LS^{-1}Q_{RU}Q_{UU}^{-1}.\] Symmetry makes \(Q_{RU}Q_{UU}^{-1}=L^T\). For the linear forms \(z=x_U+Lx_R\), the cross covariance is \[C(z,x_R)=C_{UR}+LC_{RR}=-LS^{-1}+LS^{-1}=0.\] The unshifted cross blocks are \[C_{RU}=-S^{-1}L^T,\qquad C_{UR}=-LS^{-1}.\] Expanding the residual in those blocks gives \[\begin {aligned} C(z,z)&=C_{UU}+LC_{RR}L^T+LC_{RU}+C_{UR}L^T\\ &=Q_{UU}^{-1}+LS^{-1}L^T+LS^{-1}L^T\\ &\qquad +L(-S^{-1}L^T)+(-LS^{-1})L^T\\ &=Q_{UU}^{-1}. \end {aligned}\] Thus, under \(\omega _Q\), the linear forms \((x_R,z)\) have covariance \(\operatorname {diag}(S^{-1},Q_{UU}^{-1})\).
If \(\ell =a^Tx\) and \(\ell '=b^Tx\), the pairing rule gives \(\omega _Q(\ell \ell ')=a^TCb\). By multilinearity the same holds for a product of linear forms: its moment is the Isserlis value of the matrix of those pairwise covariances. Since \(x_U=z-Lx_R\), one has \(f(x)=f(x_R,Bx_R+z)\) as polynomials. The moment \(\omega _Q(f)\) is therefore the Isserlis moment of \((x_R,z)\) at the block diagonal covariance above. A matching that joins a retained factor to a residual factor contributes \(0\). Every surviving matching is a retained matching times a residual matching, and the sum factors. Averaging the residual variables first, with coefficients in \(A[x_R]\), is \(E[f\mid x_R]\). The retained Wick functional of that polynomial is \(\omega _Q(f)\).
For the quadratic completion expand \(x_U=z+Bx_R=z-Lx_R\). The cross term produced by \(Q_{UR}\) cancels against \(Q_{UU}L\), and the retained square that remains is \(S\). Each half is legitimate because \(2\) is a unit. This is the same block arithmetic as the real elimination theorem, without positivity and without the factor \((2\pi )^{|U|/2}(\det Q_{UU})^{-1/2}\).
The correspondence with formal libraries is partial. Mathlib's
Matrix.fromBlocks_eq_of_invertible₂₂ is the block
factorization over a commutative ring, and
Matrix.isUnit_fromBlocks_iff_of_invertible₂₂ is the criterion
that, once the eliminated block is invertible, the matrix is a unit if
and only if the Schur complement is. Matrix.det_fromBlocks₂₂
is the determinant identity. No Mathlib or TauCeti declaration supplies
the Isserlis pairing rule or the tower identity over a commutative ring.
TauCeti's Matrix.PosDef.gaussianCondCov is the real
positive-definite case only, as already recorded for the
real elimination theorem.
On the four-cycle matrix of the chapter, \(Q_{UU}=\begin {pmatrix}3&-1\\-1&3\end {pmatrix}\) has determinant \(8\), and \[B=\frac 18\begin {pmatrix}1&3\\3&1\end {pmatrix},\qquad \det S=45/8.\] The product of the determinants is \(\det Q_{UU}\cdot \det S=45=\det Q\). Take \(f=x_2^2x_3x_0+x_2^3x_1\). Exact rational evaluation of the two Wick functionals gives \(\omega _Q(f)=\omega _S(E[f\mid x_R])=32/75\). The conditional expectation lies in \(\mathbb Q[x_0,x_1]\).
The rule is not the odd Wick rule, whose matrix is alternating and whose value is a Pfaffian. It defines no measure on \(A^n\), no finite-field character sum, and no Euler product. Characteristic \(2\) is outside the hypothesis; the half-turn example is the place where an algebra in that characteristic is treated. Nothing here evaluates the real normalizer or compares these moments with the logarithm of a determinant.
Theorem 2. Finite-field partition functions [fdip-008O]AGENTDRAFTED
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Theorem 2. Finite-field partition functions [fdip-008O]AGENTDRAFTED
- November 2, 2026
What takes the place of a Gaussian partition function when field values lie in a finite field? Let \(p\) be an odd prime and let \(e(t)=\exp (2\pi i t/p)\) for the representative of \(t\in \mathbb F_p\) in \(\{0,\ldots ,p-1\}\). For a symmetric \(n\times n\) matrix \(Q\) over \(\mathbb F_p\), define \[Z_p(Q)=\sum _{x\in \mathbb F_p^n}e(x^TQx).\] The exponent is the full quadratic form \(x^TQx\), not half of it. The sum is a complex number. It is not a probability.
Write \(\bigl (\frac {a}{p}\bigr )\) for the Legendre symbol, and \[g=\sum _{t\in \mathbb F_p}e(t^2).\] If \(\det Q\neq 0\) in \(\mathbb F_p\), then \[Z_p(Q)=\Bigl (\frac {\det Q}{p}\Bigr )g^n,\qquad g^2=\Bigl (\frac {-1}{p}\Bigr )p.\] In particular, when \(n\) is divisible by \(4\), \[Z_p(Q)=\Bigl (\frac {\det Q}{p}\Bigr )p^{n/2},\] an ordinary integer. If a coordinate split has \(Q_{UU}\) invertible, with Schur complement \(S\) and \(B=-Q_{UU}^{-1}Q_{UR}\) as in the Wick state, then for every retained vector \[\sum _{x_U}e(x^TQx)=Z_p(Q_{UU})\,e(x_R^TSx_R).\] The same block factorization gives \(\det Q=\det Q_{UU}\det S\), so the Legendre symbol of the determinant factors as a product.
Proof.
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Proof.
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For \(a\in \mathbb F_p\), put \(g(a)=\sum _te(at^2)\). Then \(g(0)=p\). The value \(g(a)\) depends only on the square class of \(a\): if \(a=c^2b\) with \(c\neq 0\), the substitution \(t=c^{-1}s\) gives \(g(a)=g(b)\). There are \((p-1)/2\) nonzero squares and the same number of nonsquares. Summing the nontrivial additive character, \[\sum _{a\neq 0}g(a)=\sum _t\sum _{a\neq 0}e(at^2)=(p-1)+(p-1)(-1)=0,\] because the inner sum is \(p-1\) at \(t=0\) and \(-1\) for \(t\neq 0\). Hence \(g(\nu )=-g\) for any nonsquare \(\nu \), which is \(g(a)=\bigl (\frac {a}{p}\bigr )g\) for \(a\neq 0\).
The change of variables \(u=x-y\), \(v=x+y\) is bijective on \(\mathbb F_p^2\) because \(2\) is invertible: \(x=2^{-1}(u+v)\) and \(y=2^{-1}(v-u)\). It turns the difference of squares into a product, so \[\sum _{x,y}e(x^2-y^2)=\sum _{u,v}e(uv)=p.\] The left side is \(g\cdot g(-1)=\bigl (\frac {-1}{p}\bigr )g^2\). Therefore \(g^2=\bigl (\frac {-1}{p}\bigr )p\).
Over a field of characteristic not \(2\), a symmetric matrix is congruent to a diagonal matrix. If \(Q=0\) it is already diagonal. Otherwise some quadratic value is nonzero: if every \(v^TQv\) vanished, the polarization identity \[(x+y)^TQ(x+y)-x^TQx-y^TQy=2x^TQy\] would force every bilinear value to vanish, hence \(Q=0\). Take such a vector first in a basis, and replace each later basis vector \(w\) by \(w-(v^TQw)(v^TQv)^{-1}v\). The change is invertible, those vectors are orthogonal to \(v\), and the form splits off the line through \(v\). Induction diagonalizes the complement.
If \(A\) is invertible and \(R=A^TQA\), then \(y=A^{-1}x\) runs through \(\mathbb F_p^n\) as \(x\) does, and \(x^TQx=y^TRy\). Thus \(Z_p(Q)=Z_p(R)\). Also \(\det R=(\det A)^2\det Q\), so the two matrices have the same Legendre symbol. On a nondegenerate diagonal matrix \(\operatorname {diag}(a_1,\ldots ,a_n)\) the sum factors as \(\prod _ig(a_i)=\bigl (\frac {\det R}{p}\bigr )g^n\). This is the stated formula. When \(4\) divides \(n\), the exponent \(n/2\) is even, so \[g^n=(g^2)^{n/2}=\Bigl (\frac {-1}{p}\Bigr )^{n/2}p^{n/2}=p^{n/2}.\]
For the elimination, set \(z=x_U-Bx_R\). As \(x_U\) runs through the eliminated space, so does \(z\). Expanding the symmetric form gives the polynomial identity \[x^TQx=z^TQ_{UU}z+x_R^TSx_R,\] the same cancellation used for the quadratic completion in the Wick state, applied to the full form rather than to half of it. Additivity of \(e\) factors the sum.
Write \(Q\) in retained and eliminated blocks and factor it as in the Wick state, with unitriangular factors of determinant \(1\) and a block-diagonal factor \(\operatorname {diag}(S,Q_{UU})\). The determinant multiplies, so \(\det Q=\det Q_{UU}\det S\).
The identity \(\det Q=\det Q_{UU}\det S\) is the finite-field form of the block factorization used for the quadratic center characters. It does not prove that rational character identity.
For a symmetric \(n\times n\) matrix over \(\mathbb F_p\), define the signed discriminant \(d_{\pm }(Q)=(-1)^{\frac {n(n-1)}{2}}\det Q\) in \(\mathbb F_p\). If \(\det Q\neq 0\), then \[Z_p(Q)=\Bigl (\frac {d_{\pm }(Q)}{p}\Bigr )g^{\varepsilon }p^{m},\] where \(n=2m+\varepsilon \) and \(\varepsilon \) is \(0\) or \(1\). When \(4\) divides \(n\), one has \(\varepsilon =0\) and \(d_{\pm }(Q)=\det Q\), so this is the power stated above.
Proof.
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Proof.
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The nondegenerate evaluation gives \(Z_p(Q)=\bigl (\frac {\det Q}{p}\bigr )g^n\), and \(g^2=\bigl (\frac {-1}{p}\bigr )p\). Hence \(g^n=g^{\varepsilon }(g^2)^m =g^{\varepsilon }\bigl (\frac {-1}{p}\bigr )^m p^m\), and \[Z_p(Q)=\Bigl (\frac {(-1)^m\det Q}{p}\Bigr )g^{\varepsilon }p^{m}.\] The exponent \(m\) has the same parity as \(\frac {n(n-1)}{2}\). If \(\varepsilon =0\), then \(n=2m\) and \(\frac {n(n-1)}{2}=m(2m-1)\); the factor \(2m-1\) is odd. If \(\varepsilon =1\), then \(n=2m+1\) and \(\frac {n(n-1)}{2}=m(2m+1)\); the factor \(2m+1\) is odd. In both cases \((-1)^m=(-1)^{\frac {n(n-1)}{2}}\), so the symbol is that of \(d_{\pm }(Q)\). When \(4\) divides \(n\), the integer \(m\) is even, and the sign is \(+1\).
Now suppose \(\det Q=0\), and let \(r\) be the rank of \(Q\). The same diagonalization, which does not require a nonzero determinant, produces an invertible change of variables to \(\operatorname {diag}(a_1,\ldots ,a_r,0,\ldots ,0)\) with every \(a_i\neq 0\). In the new coordinates the last \(n-r\) basis vectors span the kernel. Then \[Z_p(Q)=p^{n-r}Z_p(Q'),\] where \(Q'=\operatorname {diag}(a_1,\ldots ,a_r)\). If \(r=0\), the original form is zero and \(Z_p(Q)=p^n\). If \(r\ge 1\), the signed evaluation on \(Q'\) gives \[Z_p(Q)=p^{n-r}\Bigl (\frac {d_{\pm }(Q')}{p}\Bigr )g^{\varepsilon '}p^{m'},\] with \(r=2m'+\varepsilon '\) and \(\varepsilon '\) equal to \(0\) or \(1\).
Proof.
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Proof.
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The diagonalization already proved for a form that is not identically zero splits off a line whenever some quadratic value is nonzero, and it stops when every remaining value is zero. On that diagonal matrix the equation \(Qv=0\) reads \(a_iv_i=0\) for each \(i\le r\), so those coefficients vanish and the zero block is the kernel. An invertible change of variables preserves the partition function, by the same substitution used in the nondegenerate case. On the diagonal matrix the sum factors as a product of one-variable sums. Each nonzero entry \(a_i\) contributes \(g(a_i)\), and each zero entry contributes \(\sum _te(0)=p\). There are \(n-r\) zero entries, so the power of \(p\) is \(p^{n-r}\), and the product of the \(g(a_i)\) is \(Z_p(Q')\). The nondegenerate evaluation and the signed formula apply to \(Q'\).
The symbol \(\bigl (\frac {d_{\pm }(Q')}{p}\bigr )\) does not depend on the choice of splitting. Another invertible substitution multiplies \(\prod _ia_i\) by a square, so it preserves the Legendre symbol of that product. The sign \(\,(-1)^{\frac {r(r-1)}{2}}\) depends only on the rank, and the rank is unchanged by an invertible substitution. If \(r=0\), there is no diagonal product: every one of the \(p^n\) vectors contributes \(e(0)=1\).
Suppose a coordinate split has \(Q_{UU}\) invertible, with Schur complement \(S\), and write \(n_U\) and \(n_R\) for the two block sizes. Then \(Z_p(Q)=Z_p(Q_{UU})Z_p(S)\), and \(d_{\pm }(Q)=(-1)^{n_Un_R}d_{\pm }(Q_{UU})d_{\pm }(S)\). The sign is \(+1\) when \(n_U\) or \(n_R\) is even.
Proof.
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Proof.
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Summing the elimination identity over the retained vectors multiplies the two partition functions. For the signed discriminants, the exponent identity \[\frac {n(n-1)}{2} =\frac {n_U(n_U-1)}{2}+\frac {n_R(n_R-1)}{2}+n_Un_R\] holds for \(n=n_U+n_R\), and the block factorization gives \(\det Q=\det Q_{UU}\det S\). Multiplying the signs produces the factor \((-1)^{n_Un_R}\).
On the massive four-cycle, \(n=4\) and \(\det Q=45\), so \(d_{\pm }(Q)=\det Q\), because the exponent \(\frac {4\cdot 3}{2}\) equals \(6\). For an odd prime that does not divide \(45\), \(Z_p(Q)=\bigl (\frac {45}{p}\bigr )p^2\). Since \(45=9\cdot 5\) and \(9\) is a square, \(\bigl (\frac {9}{p}\bigr )=1\), and the value equals \(\bigl (\frac {5}{p}\bigr )p^2\).
Characteristic \(2\) is outside the argument, because both the polarization step and the change of variables divide by \(2\). The kernel dimension \(n-r\) records the rank drop of \(Q\) itself. Different domains separate the primes at which a chosen elimination block stays invertible from the primes at which a trace ratio stays regular; those loci are not this kernel. This section gives no Euler product and no comparison with the real normalizer.
Theorem 3. The Euler product [fdip-008P]AGENTDRAFTED
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Theorem 3. The Euler product [fdip-008P]AGENTDRAFTED
- November 2, 2026
The finite-field partition functions attach one complex number \(Z_p(Q)\) to a symmetric integral matrix at each odd prime. For a nondegenerate form that number is a signed Legendre symbol times an explicit normaliser. The product of those normalised factors is an \(L\)-function of the discriminant character.
Let \(Q\) be a symmetric \(n\times n\) integral matrix with \(\det Q\neq 0\). Write \(d_{\pm }(Q)\) for its signed discriminant, the integer \((-1)^{n(n-1)/2}\det Q\) named by the finite-field evaluation, and let \(\chi _D\) be the primitive character of that integer in the discriminant note. When \(d_{\pm }(Q)\) is a square, this character is the trivial character of conductor one; write \(D=1\) in that case. For an odd prime \(p\) that does not divide \(\det Q\), the signed evaluation gives \[Z_p(Q)=\Bigl (\frac {d_{\pm }(Q)}{p}\Bigr )g^{\varepsilon }p^{m},\] where \(n=2m+\varepsilon \) and \(\varepsilon \) is \(0\) or \(1\). The Gauss sum satisfies \(g^2=\bigl (\frac {-1}{p}\bigr )p\neq 0\), so the quotient \(\hat Z_p(Q)=\frac {Z_p(Q)}{g^{\varepsilon }p^{m}}\) equals \(\bigl (\frac {d_{\pm }(Q)}{p}\bigr )\). For \(\operatorname {Re}s>1\), \[\prod _{p\nmid 2\det Q}\bigl (1-\hat Z_p(Q)\,p^{-s}\bigr )^{-1} =L(s,\chi _D) \prod _{p\mid 2\det Q,\ p\nmid D} \bigl (1-\chi _D(p)\,p^{-s}\bigr ).\] No prime divides \(D=1\), so in the square case the second product runs over every prime dividing \(2\det Q\). The comparison point with the quadratic center character is \(s=2\).
Proof.
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Proof.
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The signed evaluation is the identity quoted above, and the quotient by \(g^{\varepsilon }p^{m}\) is the displayed Legendre symbol. That symbol equals \(\pm 1\). A prime in the left-hand product does not divide \(2\det Q\), so it is odd and does not divide \(d_{\pm }(Q)\). The discriminant note therefore gives \(\bigl (\frac {d_{\pm }(Q)}{p}\bigr )=\chi _D(p)\), and \(\hat Z_p(Q)=\chi _D(p)\) at every such prime.
The uncompleted product \(L(s,\chi _D)=\prod _p\bigl (1-\chi _D(p)p^{-s}\bigr )^{-1}\) is the one fixed in the center note, absolutely convergent for \(\operatorname {Re}s>1\). For \(\operatorname {Re}s>1\) every factor \(1-\chi _D(p)p^{-s}\) is nonzero. The left-hand product is that product with the finitely many primes dividing \(2\det Q\) removed, so it equals \(L(s,\chi _D)\) divided by \(\prod _{p\mid 2\det Q}\bigl (1-\chi _D(p)p^{-s}\bigr )^{-1}\). Division by an inverted factor is multiplication by the factor itself. At a prime dividing \(D\), \(\chi _D(p)=0\), so \(1-\chi _D(p)p^{-s}=1\) and that prime drops out. The quotient is therefore \(L(s,\chi _D)\) times \(\prod _{p\mid 2\det Q,\ p\nmid D}\bigl (1-\chi _D(p)\,p^{-s}\bigr )\), which is the right-hand side. Both sides are equal as convergent products, not as a truncation.
When \(n\) is odd, \(\varepsilon =1\), and the normaliser retains one factor \(g\). The sign \(\,(-1)^m\) already sits inside the Legendre symbol, so \(\hat Z_p(Q)\) is that real symbol. This section does not choose a square root of \(p\), and it does not replace \(g\) by a complex unit.
The product omits \(2\) and every prime divisor of \(\det Q\). At an odd prime where the rank of \(Q\) drops to \(r\), the same finite-field note gives \(Z_p(Q)=p^{n-r}Z_p(Q')\) for a nondegenerate block \(Q'\) of rank \(r\), and \(p^n\) when \(r=0\). That value is not the quotient by the nondegenerate normaliser of rank \(n\). The prime \(2\) lies outside because the diagonalization divides by \(2\).
On the massive four-cycle, \(n=4\) and \(d_{\pm }(Q)=45\). Since \(45=9\cdot 5\) and \(5\equiv 1\pmod 4\), the discriminant of that integer is \(D=5\). The omitted primes that do not divide \(5\) are \(2\) and \(3\). Under the Kronecker convention named in the center note, \(\chi _5(2)=\bigl (\frac {5}{2}\bigr )=(-1)^{(5^2-1)/8}=-1\). Since \(5\equiv 2\pmod 3\) and \(2^{(3-1)/2}\equiv -1\pmod 3\), \(\bigl (\frac {5}{3}\bigr )=-1\). The character note gives \(\chi _5(3)=\bigl (\frac {5}{3}\bigr )\). The prime \(5\) divides \(D\), so \(\chi _5(5)=0\) and that factor is not restored. The identity specialises to \[L(s,\chi _5)\,(1+2^{-s})\,(1+3^{-s}).\] At \(s=2\) the two restored factors are \(5/4\) and \(10/9\).
The correspondence with formal libraries is partial. Mathlib's
legendreSym.quadratic_reciprocity is the law that exchanges
the two arguments of a Legendre symbol. This section does not use that
law. Mathlib's DirichletCharacter.LFunction agrees, for real
part greater than \(1\), with the Dirichlet series
\(\sum _n\chi (n)n^{-s}\). Neither declaration identifies
\(\hat Z_p(Q)\) with \(\chi _D(p)\) or assembles the product above.
The identity does not evaluate \(L(s,\chi _D)\) in closed form. In particular it does not derive \(4\pi ^2/(25\sqrt 5)\).
A function of \(L(s,\chi _D)\) does not recover a trace ratio. That obstruction is the fiber-constancy criterion, which concerns invariants of the quadratic form alone. The product above does not evaluate a Weil index, and it is not identified with the real normalizer.
Theorem 4. The discriminant character of an integer [fdip-00A5]AGENTDRAFTED
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Theorem 4. The discriminant character of an integer [fdip-00A5]AGENTDRAFTED
- November 2, 2026
The quadratic center character attaches to a square class its primitive Dirichlet character. A partition function produces Legendre symbols of one integer. This note identifies the two on odd primes.
Let \(d\) be a nonzero integer. If \(d\) is a square, the center note uses the trivial character of conductor one. If \(d\) is not a square, let \(d_0\) be its squarefree integer representative, with the sign of \(d\), and set \(D=d_0\) when \(d_0\equiv 1\pmod 4\) and \(D=4d_0\) otherwise. Write \(\chi _D\) for the primitive character of that discriminant in the center note. For an odd prime \(p\) that does not divide \(d\), \(\bigl (\frac {d}{p}\bigr )=\chi _D(p)\).
Proof.
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Proof.
- November 2, 2026
For an odd prime \(p\) and an integer \(a\) not divisible by \(p\), the symbol \(\bigl (\frac {a}{p}\bigr )\) is \(1\) when \(a\) is a nonzero square modulo \(p\), and \(-1\) otherwise. The group \(\mathbb F_p^\times \) has \((p-1)/2\) squares. The equation \(x^2=1\) factors as \((x-1)(x+1)=0\), so the kernel of the squaring map is the two units \(1\) and \(-1\). In that field, every square \(a=b^2\) satisfies \(a^{(p-1)/2}=b^{p-1}=1\). The polynomial \(X^{(p-1)/2}-1\) has at most \((p-1)/2\) roots, so the squares are all of its roots. A nonsquare is not among them. Fermat's identity \(a^{p-1}-1=0\) factors as \((a^{(p-1)/2}-1)(a^{(p-1)/2}+1)=0\), so a nonsquare satisfies \(a^{(p-1)/2}=-1\) in the field. Thus there is a unique residue \(\varepsilon \) among \(1\) and \(-1\) such that \(a^{(p-1)/2}\equiv \varepsilon \pmod p\), and \(\varepsilon \) is the symbol just defined. The integer power itself need not lie in that pair.
The symbol is multiplicative. If \(p\) divides neither \(a\) nor \(b\), the field powers satisfy \[(ab)^{(p-1)/2}\equiv a^{(p-1)/2}b^{(p-1)/2}\pmod p.\] Each of those three powers is congruent to its symbol, so \[\bigl (\frac {ab}{p}\bigr )\equiv \bigl (\frac {a}{p}\bigr )\bigl (\frac {b}{p}\bigr )\pmod p.\] Both sides lie among \(1\) and \(-1\). Those two residues are distinct modulo an odd prime, so the two sides are equal as integers. A square \(c^2\) not divisible by \(p\) therefore has symbol \(1\).
Every nonzero integer has a unique squarefree kernel. Factor \(d=\varepsilon \prod _pp^{a_p}\), with \(\varepsilon \) the sign of \(d\) and each \(a_p\ge 0\). Let \(d_0\) be \(\varepsilon \) times the product of the primes whose exponent is odd, and let \(f^2\) absorb every even part of an exponent. Then \(d=f^2d_0\). No square \(p^2\) divides \(d_0\). The sign is unique, and so is the set of primes of odd exponent, so \(d_0\) is unique.
The odd prime \(p\) does not divide \(d\), so it divides neither \(f\) nor \(d_0\). Multiplicativity gives \(\bigl (\frac {d}{p}\bigr )=\bigl (\frac {d_0}{p}\bigr )\). The center note writes \(\chi _D(n)=\bigl (\frac {D}{n}\bigr )\) under its Kronecker convention, and that value is \(0\) when \(p\) divides \(D\). On an odd prime that does not divide \(D\), this note reads \(\bigl (\frac {D}{p}\bigr )\) as the Legendre symbol just defined. The prime \(2\) is not given that reading. If \(d\) is a square, that note uses the trivial character, and the symbol just proved is \(1\). If \(d\) is not a square and \(d_0\equiv 1\pmod 4\), then \(D=d_0\). The prime \(p\) does not divide \(D\), so \(\chi _D(p)=\bigl (\frac {d_0}{p}\bigr )\). If instead \(D=4d_0\), the same prime is odd and does not divide \(d_0\), so it does not divide \(D\). Then \(\bigl (\frac {D}{p}\bigr )=\bigl (\frac {4}{p}\bigr )\bigl (\frac {d_0}{p}\bigr )\), and \(4\) is a square, so the first factor is \(1\). In every case \(\bigl (\frac {d}{p}\bigr )=\chi _D(p)\).
The center note records that \(\chi _D\) is primitive of conductor \(|D|\). This section uses that conductor and does not re-prove it. The identification is the equality of symbols on odd primes. It is not the law that exchanges the two arguments of a Legendre symbol.
Theorem 5. Class Epstein zetas sum to a Dirichlet value [fdip-008Q]AGENTDRAFTED
- November 2, 2026
Theorem 5. Class Epstein zetas sum to a Dirichlet value [fdip-008Q]AGENTDRAFTED
- November 2, 2026
The preceding section identifies a local factor with the discriminant character. The same character has a global sum over classes of binary forms. Let \(D<0\) be a fundamental discriminant: either \(D\equiv 1\pmod 4\) and \(D\) is squarefree, or \(D=4m\) with \(m\) squarefree and \(m\equiv 2\pmod 4\) or \(m\equiv 3\pmod 4\). A positive definite binary form of discriminant \(D\) is \(Q(x,y)=ax^2+bxy+cy^2\) with \(a>0\) and \(b^2-4ac=D\). Two such forms are equivalent when one is the other after an \(\mathrm {SL}_2(\mathbb Z)\) change of variables. Equivalent forms represent the same integers equally often, because the change of variables is a bijection of \(\mathbb Z^2\).
For \(\operatorname {Re}s>1\) the Epstein zeta of \(Q\) is \[E_Q(s)=\sum _{(x,y)\neq (0,0)}Q(x,y)^{-s}.\] Let \(\chi _D\) be the quadratic center character of \(D\). It is completely multiplicative, it vanishes when \(\gcd (n,|D|)>1\), and on an odd prime it is the Legendre symbol of Theorem 4. The prime \(2\) is not given that reading. When \(D\) is even, \(\chi _D(2)=0\) because \(\gcd (2,|D|)>1\), which is the vanishing clause of the center note. When \(D\) is an odd fundamental discriminant, the value already stated in Proposition [fdip-0044] is \(\chi _D(2)=(-1)^{(D^2-1)/8}\), from [conradQuadraticCharacters, Corollary 3.1, p. 4]. That sign is \(1\) when \(D\equiv 1\pmod 8\) and \(-1\) when \(D\equiv 5\pmod 8\). This is the value of the symbol in the Euler factor \((1-\chi _D(2)\,2^{-s})^{-1}\) of the center note. Let \(w_D\) be the number of integer solutions of \(t^2-Du^2=4\). Write \(\zeta (s)=\sum _{n\ge 1}n^{-s}\) and write \(L(s,\chi _D)\) for the Euler product in the center note. Let \(\mathcal F\) contain one form from each equivalence class.
Then \[\sum _{Q\in \mathcal F}E_Q(s)=w_D\,\zeta (s)\,L(s,\chi _D).\] The weight is \(w_D=6\) when \(D=-3\), \(w_D=4\) when \(D=-4\), and \(w_D=2\) otherwise. For \(D=-4\) the only class is \(x^2+y^2\), so that one zeta equals \(4\zeta (s)L(s,\chi _{-4})\).
Proof.
- November 2, 2026
Proof.
- November 2, 2026
Since \(D<0\), the equation \(t^2+|D|u^2=4\) forces \(|u|\le 1\). The solutions with \(u=0\) are \((t,u)=(\pm 2,0)\). When \(|u|=1\), \(t^2=4-|D|\), so \(t=\pm 1\) when \(|D|=3\) and \(t=0\) when \(|D|=4\). Thus \(|D|=3\) gives 6 solutions, \(|D|=4\) gives 4, and every larger value gives two. These are the three values of \(w_D\).
The matrix of \(Q\) has determinant \(-D/4\) up to the usual factor \(4\). Explicitly, \[\det \begin {pmatrix}2a&b\\b&2c\end {pmatrix}=-D.\] An element \(\gamma \in \mathrm {SL}_2(\mathbb Z)\) replaces that matrix by a congruent matrix, so it preserves \(D\). Let \(\tau =(-b+\sqrt {D})/(2a)\) be the root in the upper half plane, taking the square root of positive imaginary part. The factorization \(Q(x,y)=a(x-\tau y)(x-\overline {\tau } y)\) shows that an automorph permutes the two roots. Determinant \(1\) preserves the upper half plane, so the automorph fixes \(\tau \). If \(\gamma =\begin {pmatrix}p&q\\r&s\end {pmatrix}\), the fixed-point equation is \(r\tau ^2+(s-p)\tau -q=0\). The minimal equation \(a\tau ^2+b\tau +c=0\) is irreducible over \(\mathbb Q\), so there is a rational scalar \(u\) with \(r=au\), \(s-p=bu\), and \(q=-cu\). First, \(\gcd (a,b,c)=1\). A prime dividing all three coefficients square-divides \(D\). The odd part of a fundamental discriminant is squarefree, so that prime is not odd. If the prime is \(2\), then \(D=4m\). Writing the coefficients as twice integers \(a'\), \(b'\), and \(c'\) gives \(m=b'^2-4a'c'\equiv b'^2\pmod 4\), and a square modulo \(4\) is \(0\) or \(1\), contrary to the fundamental condition on \(m\). Bézout's identity therefore writes \(1\) as an integer combination of \(a\), \(b\), and \(c\), and the rational scalar \(u\) is an integer. Set \(t=p+s\). The determinant \(ps-qr=1\) becomes \((t^2-Du^2)/4=1\). Conversely each solution \((t,u)\) determines \[p=\frac {t-bu}{2},\qquad s=\frac {t+bu}{2},\qquad q=-cu,\qquad r=au.\] The same parity of \(t\) and \(bu\) makes \(p\) and \(s\) integers, since \(t^2\equiv Du^2\pmod 2\) and \(D\equiv b^2\pmod 2\). The form produced by \(\gamma \) has the same root and the same discriminant, so it is a positive multiple of \(Q\). The discriminant scales by the square of that multiple, and the multiple is positive, so the multiple is \(1\). Thus these \(w_D\) matrices are exactly the automorphs of every form of discriminant \(D\).
Let \(R(n)\) be the number of pairs \(\bigl (Q,(x,y)\bigr )\) with \(Q\in \mathcal F\) and \(Q(x,y)=n\). Let \(R^\ast (n)\) count only the pairs with \(\gcd (x,y)=1\). A nonzero vector has content \(d=\gcd (x,y)\), and \(Q(dv)=d^2Q(v)\), so \[R(n)=\sum _{d^2\mid n}R^\ast (n/d^2).\] The stabilizer in the automorph group of any primitive vector is trivial: an automorph fixing a vector has eigenvalue \(1\), hence trace \(2\) because the determinant is \(1\), hence \(t=2\) and \(u=0\). The group therefore acts freely on the proper representations of each form. It remains to show that the orbits are indexed by the residues \(b\) modulo \(2n\) with \(b^2\equiv D\pmod {4n}\).
Given a proper representation of \(n\) by a vector \(v\), complete \(v\) to a basis \((v,w)\) of determinant \(1\). In that basis the form is \(nx^2+Bxy+Cy^2\) with \(B^2-4nC=D\). Replacing \(w\) by \(w+tv\) sends \(xv+yw\) to \((x+ty)v+yw\). The term \(n(x+ty)^2\) contributes the extra middle term \(2nt\,xy\), so the middle coefficient changes by \(2nt\) and \(B\) modulo \(2n\) is invariant. An automorph changes the vector inside one orbit and does not change this residue. In the other direction, a residue \(b\) produces the positive definite form \(nx^2+bxy+cy^2\) with \(c=(b^2-D)/(4n)\). That form has discriminant \(D\), so it is equivalent to exactly one representative in \(\mathcal F\). Transporting \((1,0)\) by such an equivalence yields a proper representation with that residue. If two proper representations have the same residue \(b\), replace each second basis vector by a shear so that the middle coefficient equals \(b\). The discriminant identity then forces both coordinate forms to equal \(nx^2+bxy+cy^2\) with the same \(c=(b^2-D)/(4n)\). Those two determinant-\(1\) bases therefore belong to the same class representative. If \(M_1\) and \(M_2\) are the basis matrices, \(M_1M_2^{-1}\) carries one represented vector to the other and preserves the form, since both bases pull it back to that coordinate polynomial. The number of such residues is therefore the number of orbits, and \(R^\ast (n)=w_D\rho (n)\), where \(\rho (n)\) denotes that number of residues.
The count \(\rho \) is multiplicative on coprime arguments. If \(n_2\) is odd and coprime to \(n_1\), a residue modulo \(2n_1n_2\) is a pair of a residue modulo \(2n_1\) and a residue modulo \(n_2\). The square condition splits, and for odd \(n\) the count \(\rho (n)\) equals the number of solutions of \(v^2\equiv D\pmod n\). Indeed \(\gcd (4,n)=1\), so the condition modulo \(4n\) splits into conditions modulo \(4\) and modulo \(n\), and exactly one of the two lifts modulo \(2n\) has the parity required by \(D\) modulo \(4\). Thus \(\rho (n_1n_2)=\rho (n_1)\rho (n_2)\).
For an odd prime \(p\) the same count is the number of roots of \(v^2-D\) modulo \(p^k\). If \(p\) does not divide \(D\) and \(\chi _D(p)=1\), there are two simple roots modulo \(p\), and the derivative \(2v\) is invertible at each, so each lifts uniquely and \(\rho (p^k)=2\). If \(\chi _D(p)=-1\), there is no root and \(\rho (p^k)=0\). If \(p\) divides \(D\), then \(p\) divides \(D\) exactly once. Modulo \(p\) the only root is \(0\), so \(\rho (p)=1\). Modulo \(p^k\) with \(k\ge 2\) the valuation \(v_p(D)=1\) is odd, while a square has even valuation, so \(\rho (p^k)=0\).
At the prime \(2\), first take \(n=1\): exactly one of \(b=0,1\) satisfies \(b^2\equiv D\pmod 4\), so \(\rho (1)=1\). If \(D\) is odd and \(D\equiv 5\pmod 8\), then for \(a\ge 1\) the modulus \(2^{a+2}\) is a multiple of \(8\), and no square is congruent to \(5\) modulo \(8\), so \(\rho (2^a)=0\). If \(D\equiv 1\pmod 8\), the two odd residues modulo \(4\) satisfy the congruence modulo \(8\). A solution modulo \(2^{a+1}\) lifts to a solution modulo \(2^{a+2}\) by \(b=b_0+t2^{a+1}\). The cross term determines \(t\) uniquely because \(b_0\) is odd, and the quadratic term in \(t\) is divisible by \(2^{a+3}\). Thus \(\rho (2^a)=2\) for every \(a\ge 1\). If \(D\) is even, inspect \(b\) modulo \(4\) against the condition modulo \(8\). When \(D\equiv 4\pmod 8\) the only solution is \(b\equiv 2\pmod 4\). When \(D\equiv 0\pmod 8\) the only solution is \(b\equiv 0\pmod 4\). Thus \(\rho (2)=1\). For \(a\ge 2\) write \(D=4m\). If \(m\equiv 2\pmod 4\), then \(v_2(D)=3\). A square cannot have that valuation, and \(3\) is strictly smaller than the modulus \(2^{a+2}\), so there is no solution. If \(m\) is odd, a solution would have valuation exactly \(2\), hence the shape \(b=2c\) with \(c\) odd, and then \(c^2\equiv m\pmod 4\). An odd square is \(1\) modulo \(4\), while \(m\equiv 3\pmod 4\). There is again no solution, so \(\rho (2^a)=0\) for \(a\ge 2\).
Let \(\psi (n)=\sum _{d\mid n}\chi _D(d)\) and \(g(n)=\sum _{d^2\mid n}\rho (n/d^2)\). Both functions are multiplicative, since a square divisor of a product of coprime arguments splits, so it is enough to compare them on prime powers. If \(\chi _D(p)=1\), then \(\rho (p^j)=2\) for \(j\ge 1\). For even \(k=2m\) the sum \(g(p^k)\) is \(m\) terms equal to \(2\) and one term \(\rho (1)\), hence \(k+1\). For odd \(k=2m+1\) it is \(m+1\) terms equal to \(2\), hence again \(k+1\). The character sum is the same geometric sum. If \(\chi _D(p)=-1\), then \(\rho (p^j)=0\) for \(j\ge 1\), so \(g(p^k)\) keeps only \(\rho (1)\) when \(k\) is even and is \(0\) when \(k\) is odd. The alternating character sum has those same two values. If \(\chi _D(p)=0\), then \(\rho (p)=1\) and \(\rho (p^j)=0\) for \(j\ge 2\), so every \(g(p^k)\) equals \(1\), as does \(\psi (p^k)\). Therefore \(g=\psi \), and \[R(n)=w_D\sum _{d\mid n}\chi _D(d).\]
Absolute convergence for \(\operatorname {Re}s>1\) lets the zeta sums be rearranged. The form has no real root and \(a>0\), so its minimum \(\mu \) on the unit circle is positive and \(Q(x,y)\ge \mu (x^2+y^2)\). The comparison series converges for \(\operatorname {Re}s>1\) because, in polar coordinates, the tail is a constant multiple of \(\int _1^\infty r^{1-2s}\,dr\). Hence \[\sum _{Q\in \mathcal F}E_Q(s) =\sum _{n\ge 1}R(n)\,n^{-s}.\] The representation count and the substitution \(n=dm\) give \[w_D\sum _{d,m\ge 1}\chi _D(d)\,(dm)^{-s} =w_D\,\zeta (s)\sum _{d\ge 1}\chi _D(d)\,d^{-s}.\] Complete multiplicativity expands each Euler factor \((1-\chi _D(p)p^{-s})^{-1}\) as a geometric series. The product in the center note is therefore the Dirichlet series, which is the identity claimed above.
For the illustration \(D=-4\), repeat the shear \(\begin {pmatrix}1&k\\0&1\end {pmatrix}\), which replaces \(b\) by \(b+2ak\) and does not change \(a\), until \(|b|\le a\). Whenever \(a>c\), apply \(\begin {pmatrix}0&-1\\1&0\end {pmatrix}\). That matrix exchanges the outer coefficients and changes the sign of the middle coefficient, so it replaces \(a\) by the smaller positive integer \(c\). Each such swap decreases \(a\), and the process stops at a representative with \(|b|\le a\le c\). Then \(b\) is even, \(ac=(b/2)^2+1\), and \(a^2\le ac\le a^2/4+1\), so \(a=1\), \(b=0\), and \(c=1\). The only class is \(x^2+y^2\), and \(w_{-4}=4\).
The identity concerns binary forms. The four-cycle precision has rank \(4\). Schiemann's papers of 1990 and 1997, which have not been read here, are not a step in the argument: this section does not claim that one theta series separates a quaternary form from the other forms in its genus.
The Weil index remains speculative. No product of local indices is proved here, and Weil's paper of 1964 has not been read for this note. The class sum does not identify an Epstein zeta with the real normalizer.