gpt 6 astra ultra attempt on Math-429

Lück's Determinant Conjecture for Discrete Groups

Definitions and mathematical statement

For a discrete group $G$, the involution on ${\mathbb{Z}}[G]$ sends $g$ to $g^{-1}$. A matrix $A\in M_n({\mathbb{Z}}[G])$ acts by convolution on $\ell^2(G)^n$, and $A^*A$ is bounded positive. Let $\mu$ be its spectral measure for the canonical group trace combined with normalized matrix trace. The selected determinant inequality is \[ \int_{(0,\infty)}\log t{\,\mathrm{d}}\mu(t)\ge0. \] The atom at zero is omitted, not assigned $\log0$; integrability of the negative part is part of the conjectured conclusion, since the integral is a priori allowed to be $-\infty$. Equivalently, the corresponding regularized Fuglede--Kadison determinant is at least one, with the conventional square-root factor for passing from $A^*A$ to $A$. This asserts neither invertibility nor trivial kernel of $A$.

Formulation and concept references: [S1].

Short English statement

Does every square integer group-ring matrix satisfy the nonnegative logarithmic spectral-integral bound after its zero eigenspace is omitted?

Research attempt

GPT-6 Astra Ultra; 27 September 2026. Status: UNRESOLVED.

The logarithm is discontinuous at zero, and the prescribed convention omits the zero atom. I therefore focus on proving a limit argument with that convention intact. The result recovers the inequality for residually finite groups from integer matrices and gives an explicit sufficient spectral-approximation criterion. It does not establish the criterion for an arbitrary group.

The exact finite-dimensional inequality.

Let $C$ be an $m\times m$ integer matrix and put $D=C^{\mathsf T}C$. If its nonzero eigenvalues are $\lambda_1,\ldots,\lambda_r$, its characteristic polynomial is \[ \det(tI-D)=t^{m-r}\prod_{i=1}^r(t-\lambda_i). \] The coefficient of $t^{m-r}$ is $(-1)^r\prod_i\lambda_i$, a nonzero integer. Since the eigenvalues are positive, their product is at least one. Thus the normalized spectral measure satisfies \[ \int_{(0,\infty)}\log t\,d\mu_D(t) =\frac1m\log\prod_{i=1}^r\lambda_i\ge0. \tag{1} \] If $C=0$, the empty product is one and the integral is zero. This case would fail under an unmodified $\log0=-\infty$ convention. For a finite group $Q$, convolution by an integer group-ring matrix is an integer matrix of size $n|Q|$, and the normalized matrix trace agrees with normalized group-and-matrix trace. Hence $1$ proves the catalogue inequality for finite groups, including matrices with kernels.

A limit lemma retaining the zero atom.

Suppose probability measures $\mu_j$ converge weakly to $\mu$, all are supported in $[0,K]$ with $K\ge e$, and each has finite negative logarithmic part off zero and satisfies \[ I(\mu_j):=\int_{(0,K]}\log t\,d\mu_j(t)\ge0. \] Then the same integrability and inequality hold for $\mu$. Here is a proof, including the kernel mass which weak convergence alone need not control.

The positive logarithmic contribution is at most $\log K$. Therefore \[ \int_{(0,1)}-\log t\,d\mu_j(t)\le\log K, \quad \mu_j((0,\varepsilon])\le \frac{\log K}{|\log\varepsilon|}\quad(0<\varepsilon<1). \tag{2} \] Choose $\varepsilon$ which is not an atom of $\mu$. Weak convergence gives convergence of the masses on $[0,\varepsilon]$, in the relative space $[0,K]$. Hence \[ \liminf_j\mu_j(\{0\})\ge \mu([0,\varepsilon])-\frac{\log K}{|\log\varepsilon|}. \] Letting such $\varepsilon$ tend to zero gives a lower bound $\mu(\{0\})$. The opposite upper bound follows from the closed-set part of the Portmanteau theorem. Thus \[ \mu_j(\{0\})\longrightarrow\mu(\{0\}). \tag{3} \]

For fixed $0<\varepsilon<1$, define \[ I_\varepsilon(\eta)= \int_{[0,K]}\log\max(t,\varepsilon)\,d\eta(t) -\eta(\{0\})\log\varepsilon. \] This truncates only the nonzero small eigenvalues. Since it increases their logarithms, $I_\varepsilon(\mu_j)\ge I(\mu_j)\ge0$. The integrand is bounded continuous, so weak convergence and $3$ give $I_\varepsilon(\mu)\ge0$. As $\varepsilon\downarrow0$, the positive part stays fixed and the truncated negative part increases monotonically. The inequalities bound that negative part by $\log K$. Monotone convergence proves its integrability and $I(\mu)\ge0$. This establishes the lemma without assuming uniform integrability of the logarithm in advance.

Application to residually finite groups.

Fix $A\in M_n(\mathbb ZG)$ and replace $G$ by the finitely generated subgroup containing all its supports; the canonical traces of all powers of $A^*A$ are unchanged. If this subgroup is residually finite, choose finite quotients $Q_j$ which separate all nonidentity words of length at most $j$ in a fixed finite generating set. Let $A_j$ be the image of $A$. There is a uniform norm bound \[ \|A_j\|\le C,\qquad\|A\|\le C, \] where one may take $C$ to be the sum of the absolute values of all coefficients in all entries. Each convolution summand is a scalar multiple of a unitary between coordinate summands, so the triangle inequality supplies this admittedly coarse bound.

For each fixed $k$, the normalized trace of $(A_j^*A_j)^k$ eventually equals the canonical trace of $(A^*A)^k$. Indeed its coefficient expansion uses only finitely many bounded-length words, and the quotient eventually creates no additional identity words among them. All spectral measures lie in $[0,C^2]$; convergence of every polynomial moment therefore implies weak convergence by uniform polynomial approximation on that compact interval. The finite-group inequality $1$ and the limit lemma prove the target inequality for $G$. No claim of invertibility has entered the proof.

Why determinant convergence would be too strong.

Even uniformly bounded supports and nonnegative logarithmic integrals do not force those integrals to converge. For $j\ge2$, set \[ \eta_j=(1-1/j)\delta_{\exp(j/(j-1))} +(1/j)\delta_{\exp(-j)}. \] Then $I(\eta_j)=1-1=0$, while $\eta_j$ converges weakly to $\delta_e$, whose integral is one. All supports lie in $[0,e^2]$. The limit lemma proves an inequality, not an equality of determinants. The small mass at an exponentially small positive eigenvalue carries a fixed logarithmic contribution which disappears from the weak limit. These measures are a countertest to an analytic inference, not asserted spectral measures of integer group-ring matrices.

Verification, scope, and first gap.

Exact checks of small integer matrices include the zero matrix and singular matrices and use characteristic coefficients, not rounded eigenvalues. The normalization factor is $1/(n|Q|)$ in the finite quotient case; passing from $A^*A$ to the determinant convention for $A$ introduces a factor $1/2$, which leaves the sign unchanged. [S1] studies invariant random subgroups and constructs objects satisfying an analogous determinant property outside a particular approximation class. It does not announce the universal group theorem.

For the general target, the first missing step is a suitable spectral approximation by measures with the integer determinant bound and a uniform norm bound, or another source of the small-eigenvalue estimate $2$. Arbitrary weak approximations do not suffice. The attempt proves the finite and residually finite cases and the precise limit lemma; the assertion for all discrete groups remains UNRESOLVED.

Sources

• S1 Aareyan Manzoor, Invariant Random Subgroups, Soficity, and Lück's determinant conjecture, arXiv:2508.15154v2 (21 September 2025). https://arxiv.org/html/2508.15154v2. Formulation source.
• E1 Aareyan Manzoor, Invariant Random Subgroups, Soficity, and Lück's determinant conjecture, arXiv:2508.15154v2, Introduction and Section 2.1. https://arxiv.org/html/2508.15154v2. Version-specific scope check of [S1]. The IRS theorem is not a proof or disproof for all groups. The zero-atom limit argument is written out independently above. Consulted 27 September 2026.