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Phys-45 · physics · mathematics · Physics

Exactly one threshold bound state in every relative SU(N) BFSS model

Status OpenLLM status Open unverified

Statement

Definitions and mathematical statement

Statement recorded in the supplied source.
For every integer $N\geq 2$, does the gauge-invariant relative SU(N) BFSS Hamiltonian, with its source-defined self-adjoint domain and fermion grading, have a one-dimensional zero-energy kernel consisting of Spin(9)-invariant even states?

Short English statement

For every integer $N\geq 2$, does the gauge-invariant relative SU(N) BFSS Hamiltonian, with its source-defined self-adjoint domain and fermion grading, have a one-dimensional zero-energy kernel consisting of Spin(9)-invariant even states?

Sources

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Phys-45-A1 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Phys-45

For every finite $N\geq2$, the relative undeformed $SU(N)$ BFSS Hamiltonian is claimed to have exactly one square-integrable zero-energy state. The question retains these same parameters and conventions. Within these assumptions, this is a claimed affirmative answer to the stated question.

Exactly one threshold bound state in every relative SU(N) BFSS model

OpenAI's claim

OpenAI presents a proof in the selected manuscript. For every finite $N\geq2$, the relative undeformed $SU(N)$ BFSS Hamiltonian is claimed to have exactly one square-integrable zero-energy state.

Mathematical statement of the claimed result

Result 1.
For each $N\ge2$, use the gauge-invariant space $\mathcal H_N=[L^2(\mathbb R^{9(N^2-1)})\otimes\mathcal F_N]^{\operatorname{SU}(N)}$, with $\operatorname{Tr}(T_AT_B)=\delta_{AB}$, $[T_A,T_B]=if_{ABC}T_C$, and $\{\theta_\alpha^A,\theta_\beta^B\}=\delta_{\alpha\beta}\delta_{AB}$. The relative Hamiltonian closes the smooth invariant-core form \[ q_N(\Psi)=\frac1{16}\sum_\alpha\|Q_\alpha\Psi\|^2,\quad Q_\alpha=-i\partial_{iA}\gamma^i_{\alpha\beta}\theta_\beta^A +\tfrac12 f_{ABC}x_i^Bx_j^C\gamma^{ij}_{\alpha\beta}\theta_\beta^A. \] The asserted threshold-state count is \[ \dim\ker H_N=1. \] That kernel is $\operatorname{Spin}(9)$-invariant and even for the exterior grading whose polarization is $\gamma^1\gamma^2\gamma^3$, with the specified empty vector chosen even. The nine $\gamma^i$ are real symmetric $16\times16$ matrices with $\{\gamma^i,\gamma^j\}=2\delta_{ij}I$ and $\gamma^{ij}=\tfrac12[\gamma^i,\gamma^j]$. The space $\mathcal F_N$ is the irreducible complex Clifford module for $16(N^2-1)$ self-adjoint fermions, of dimension $2^{8(N^2-1)}$. The polarization's empty vector is the minimizer of $\tfrac i2\theta^{\mathsf T}(\gamma^1\gamma^2\gamma^3)\theta$; it is declared even, fixing the grading's sign.

Outline of the argument

A massive deformation has a computable Fredholm index. Corrected profiles near separated centers and exterior estimates control the undeformed limit, while massive tightness prevents loss of states and the profile exhaustion fixes the kernel count after center-of-mass removal.

Formal verification

Family 270 has no Lean scope page in this release. Its selected result is not presented as formalized. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

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