gpt 6 astra pro attempt on Math-174

Tate's rational leading-term Stark conjecture

Definitions and mathematical statement

For a Galois extension $L/K$ with group $G$, let $S_L$ be the places above $S$ and set $X=\ker({\mathbb{Q}}[S_L]\xrightarrow{\sum}{\mathbb{Q}})$ and $U={\mathbb{Q}}\otimes_{{\mathbb{Z}}}\mathcal O_{L,S}^{\times}$. For a ${\mathbb{Q}}[G]$-isomorphism $f:X\to U$, the logarithmic map $\lambda(u)=-\sum_{w\in S_L}\log|u|_w\,w$ uses product-formula normalized absolute values. For an irreducible character $\chi$, take the determinant of $\lambda f$ on $\operatorname{Hom}_{{\mathbb{C}}[G]}(V_{\chi^\vee},{\mathbb{C}}\otimes X)$ to obtain $R_S(\chi,f)$. Let $L_S^*(0,\chi)$ be the first nonzero Laurent coefficient at zero of the Artin $L$-function with the $S$ factors omitted. The target is \[ A_S(\chi,f):=\frac{R_S(\chi,f)}{L_S^*(0,\chi)}\in{\mathbb{Q}}(\chi), \qquad\sigma A_S(\chi,f)=A_S(\chi^\sigma,f). \] The empty determinant is one; the character dual and the sign in $\lambda$ are part of the chosen normalization.

Formulation and concept references: [S1].

Short English statement

After dividing the appropriate logarithmic regulator by an Artin L-function's leading term, is the result algebraic in the character field and compatible with every field automorphism?

Research attempt

Current frontier.

The rationality and equivariance statement here is Conjecture 1.1 of [S1]. Its Remark 1.2 records independence of the choices of $S$ and $f$; Theorem 1.3 treats totally odd characters using Siegel's theorem, and Remark 1.4 records other established cases, including rational-valued characters. These statements must be distinguished from the paper's stronger, integral and prime-specific Stark results. The more recent Brumer--Stark theorem over ${\mathbb{Z}}$ concerns the CM/abelian setting, not a proof of the arbitrary-character target [E1], Theorem 1.1. The calculations below reconstruct normalization reductions explicitly and test whether class-number information could supply the missing character-by-character algebraicity. They are not offered as new versions of the known choice-independence theorem.

Attack route.

One route is descent of the determinant from a complex representation to its character field; this succeeds for rational changes of $f$, but not automatically for the transcendental regulator itself. A second route removes the extra places one at a time; the required local factors can be computed exactly. A third tries to recover individual Stark ratios from products furnished by class-number formulas. The third route has a genuine loss of information, which is exposed below.

Attempt: changing $f$ without assuming Schur index one.

For $a\in\operatorname{Aut}_{{\mathbb{Q}}[G]}X$, let \begin{equation} d_\chi(a)=\det\left(a\mid \operatorname{Hom}_{{\mathbb{C}}[G]}(V_{\chi^\vee},{\mathbb{C}}\otimes X)\right). \end{equation} For every positive integer $r$, character orthogonality gives \begin{equation}\label{wsref-0-33-30-30-30-36-37-36-34-3a-74-72-61-63-65} \operatorname{Tr}(a^r\mid M_\chi) =\frac1{|G|}\sum_{g\in G}\chi(g) \operatorname{Tr}_{{\mathbb{Q}}}(a^r g\mid X),\qquad M_\chi=\operatorname{Hom}_{{\mathbb{C}}[G]}(V_{\chi^\vee},{\mathbb{C}}\otimes X). \end{equation} Indeed, decompose ${\mathbb{C}}\otimes X$ into irreducible representations tensored with multiplicity spaces. The trace of $a^rg$ is the sum of the irreducible characters times the traces on those spaces. Pairing with $\chi(g)$ selects the $\chi^\vee$ summand. All traces on the right other than $\chi(g)$ are rational. Newton's identities applied to the first $\dim M_\chi$ power traces therefore show \begin{equation} d_\chi(a)\in{\mathbb{Q}}(\chi)^\times,\qquad \sigma d_\chi(a)=d_{\chi^\sigma}(a). \end{equation} This argument does not require $V_\chi$ to have a model over ${\mathbb{Q}}(\chi)$; that would be an unjustified Schur-index shortcut.

If $f_2=f_1a$, determinant multiplicativity gives \begin{equation} A_S(\chi,f_2)=d_\chi(a)A_S(\chi,f_1). \end{equation} Thus proving the target for one ${\mathbb{Q}}[G]$-isomorphism proves it for every one. The empty multiplicity space has trace zero and determinant one, so that case is covered. This supplies a direct proof of the $f$ part of the choice-independence statement recorded in [S1].

Attempt: an explicit unramified-place transition.

Let $v\notin S$ be finite. It is unramified, since $S$ already contains all ramified places. Write $D=\langle F_v\rangle$ for its cyclic decomposition group, $f_v=|D|$, and $Y_v={\mathbb{Q}}[G/D]$. Forgetting the new place coordinates and taking their valuations give exact sequences \begin{equation} 0\to X_S\to X_{S\cup\{v\}}\to Y_v\to0, \qquad 0\to U_S\to U_{S\cup\{v\}}\xrightarrow{\operatorname{ord}}Y_v\to0. \end{equation} The first surjection can balance its augmentation using the old places. The second becomes surjective over ${\mathbb{Q}}$ because the $S$-ideal class group is finite. Both split over ${\mathbb{Q}}[G]$ by Maschke's theorem. Choose $f'$ compatible with $f$ on the old summands and with the identity on the two $Y_v$ quotients.

The chosen normalized absolute values give $-\log|u|_w=\operatorname{ord}_w(u)\log Nw$ and $\log Nw=f_v\log Nv$. The regulator is consequently block triangular, and, with $r_\chi=\dim V_\chi^D$, its quotient determinant is $(f_v\log Nv)^{r_\chi}$. On the analytic side the removed Euler factor is \begin{equation} E_v(s,\chi)=\det(1-F_v(Nv)^{-s}\mid V_\chi). \end{equation} Because $F_v$ has finite order, it is semisimple, and \begin{equation} E_v^*(0,\chi)=(\log Nv)^{r_\chi} \det(1-F_v\mid V_\chi/V_\chi^D). \end{equation} It follows, with precisely the sign and dual conventions in the target, that \begin{equation}\label{wsref-0-33-30-30-30-36-37-36-34-3a-70-6c-61-63-65} \frac{A_{S\cup\{v\}}(\chi,f')}{A_S(\chi,f)} =\frac{f_v^{r_\chi}} {\det(1-F_v\mid V_\chi/V_\chi^D)}. \end{equation} The denominator is nonzero. Its characteristic coefficients are in ${\mathbb{Q}}(\chi)$ by the same power-trace argument, and they transform with $\chi$ under $\operatorname{Aut}({\mathbb{C}})$. The equality of invariant multiplicities for a character and its dual follows by averaging over $D$. Equation (30006764:place), together with change of $f$, reduces all allowed $S$ to the archimedean and ramified places. It reconstructs the relevant known reduction rather than assuming an Euler factor is rational just because it has finite-order eigenvalues [S1], Remark 1.2.

Attempt: an exact quadratic normalization test.

Take $L={\mathbb{Q}}(\sqrt5)$, $K={\mathbb{Q}}$, $S=\{\infty,5\}$, and $\varphi=(1+\sqrt5)/2$. Let $w_+,w_-$ be the real places and $w_5$ the single place over 5. Set \begin{equation} a=w_+-w_-,\qquad b=w_++w_--2w_5, \qquad f(a)=\varphi,\quad f(b)=5. \end{equation} Units are written multiplicatively on the right, as elements of the rationalized unit module. Conjugation sends $\varphi$ to $-\varphi^{-1}$, so $f$ is $G$-equivariant after torsion is killed. The two elements have independent logarithms and give the required isomorphism. The logarithm convention yields \begin{equation} \lambda(\varphi)=-(\log\varphi)a,\qquad \lambda(5)=-(\log5)b. \end{equation} In particular, $|5|_{w_5}=5^{-2}$; replacing this by $5^{-1}$ would break the product formula and the computation.

For completeness, the Minkowski ideal-class bound is $\sqrt5/2<2$, so the class number is one. The positive unit $\varphi$ is fundamental: a unit $1<u<\varphi$ of norm $+1$ would have integral trace $u+u^{-1}$ strictly between $2$ and $\sqrt5$; one of norm $-1$ would have trace $u-u^{-1}$ strictly between $0$ and $1$. Both are impossible. Multiplying any positive unit by a power of $\varphi$ reduces it to this interval or its endpoints.

The analytic class-number formula, in the normalization used in the Stark source, gives $\zeta_L^*(0)=-(\log\varphi)/2$ and $\zeta_{{\mathbb{Q}}}(0)=-1/2$. Hence \begin{equation} L_S^*(0,\chi_5)=\log\varphi, \qquad L_S^*(0,1)=-\tfrac12\log5, \qquad A_S(\chi_5,f)=-1,\quad A_S(1,f)=2. \end{equation} The ramified Euler factor for $\chi_5$ is one. This is a worked known case, compatible with [S1], Remark 1.4, not evidence that the higher-character descent problem has disappeared.

Stress test: products cannot replace characterwise descent.

Even all permutation-character products may lose essential information. For $G=C_3$, the transitive permutation characters are $1$ and $1+\chi+\bar\chi$. As a model of the information carried by such products, set \begin{equation} A(1)=1,\qquad A(\chi)=z,\qquad A(\bar\chi)=\bar z, \qquad |z|=1, \end{equation} where $z$ is transcendental. Such a $z$ exists because the unit circle is uncountable and the algebraic numbers are countable. Every product attached to a permutation character is one, but the individual nontrivial ratios are not algebraic. This is not a construction of Stark ratios for a number field and not a counterexample. It proves only that the indicated product data, even with complex-conjugation compatibility, do not logically force individual algebraicity.

Brauer induction with nontrivial one-dimensional characters is stronger than permutation-character information; invoking it would require the appropriate Stark information for those characters, not just analytic class-number formulas for fields. Similarly, the rational power traces in (30006764:trace) apply to $a\in\operatorname{End}_{{\mathbb{Q}}[G]}X$, not to the logarithmic map itself. Treating the regulator as such a rational endomorphism would assume the arithmetic descent that remains to be proved.

Outcome.

Outcome: Conditional reduction.

The argument explicitly reduces the target to one choice of $f$ and minimal $S$, without a Schur-index assumption, and verifies all signs and local norms in a quadratic test. These reductions are already known in the literature; their proofs here are normalization checks, not novelty claims. The class-number-product attack has a demonstrated information deficit. General characterwise algebraicity and full $\operatorname{Aut}({\mathbb{C}})$-equivariance remain the unresolved input; no claim about totally odd or abelian cases has been enlarged to the full catalogue assertion.

Sources

• S1 Andreas Nickel, The strong Stark conjecture for totally odd characters; arXiv2106.05619v1 June10,2021; live metadata lists ProcAMS152$2024$147–162. https://arxiv.org/pdf/2106.05619. Catalogue formulation source.
• E1 Samit Dasgupta, Mahesh Kakde, Jesse Silliman and Jiuya Wang, "The Brumer--Stark Conjecture over ${\mathbb{Z}}$," arXiv:2310.16399 $2023$. Theorem 1.1 and its CM/abelian hypotheses. https://arxiv.org/abs/2310.16399. An adjacent integral theorem; not a solution of arbitrary-character rational Stark.