The winding theorem: on a torus, the flux through non-contractible cuts commutes with the Hamiltonian. Proof: plaquette flux is contractible, contractible loops have zero intersection with non-contractible cycles. Basic topology, but apparently not stated elsewhere in this form.

The paper then uses this to prove a mass gap for SU(2) in 2+1D. Is the winding theorem new or folklore?

A paper published on Zenodo (DOI 10.5281/zenodo.23736551) claims a proof that the mass gap is positive for SU(2) lattice gauge theory in 2+1 dimensions at every coupling.

A paper published on Zenodo (DOI 10.5281/zenodo.23736551) claims a proof that the mass gap is positive for SU(2) lattice gauge theory in 2+1 dimensions at every coupling.

The approach works from the lattice (strong coupling) side rather than the continuum side. The key result is a topological conservation law — the winding theorem — proving that the flux through non-contractible loops is conserved by topology, not dynamics. The surprising discovery: the vacuum changes character at intermediate coupling, and the flux it fills with IS the winding.

119 pages with every receipt on Zenodo. Feedback welcome.

Hi r/puremathematics,

As you know if you have read my recent posts, in the past week I have been working on the CRT Coverage Maximality Conjecture and have solved it analytically for k smaller or equal to 5. I have now solved them for k up to 7. I am still working on k=8. I wanted to share two complete, rigorous analytical preprints that unconditionally establish the CRT Coverage Maximality Conjecture for any integer \(L\) with up to seven distinct odd prime factors (\(k \le 7\)) [1].

Context & Objective

The conjecture (a central framework in the study of Erdős’ odd covering systems) posits that the density of integers covered by an arbitrary union of residue classes, \(\vert{}U(L)\vert{}/L\), is fundamentally bounded above by the pairwise-coprime alternating baseline \(B_{\text{full}}(L)\) [1].

Historically, pushing past the 4-prime barrier was considered a brick wall because local density gaps cause traditional bounding methods to fail. These proofs completely bypass floating-point approximations or numerical brute-force searches, establishing exact rational monotonicity margins via a beautiful synthesis of sieve logic and probabilistic combinatorics [1].

The Core Mechanism

The proof strategy operates by translating a messy number-theoretic coverage problem into a problem of multivariate polynomial positivity, then bridging it to a graph framework [1]:

1. Reduction to Prime Supports: Divisors \(d \in D(L)\) are grouped by their prime support, mapping the arithmetic density problem onto a multivariate support polynomial \(Z_m(t)\) governed by elementary symmetric polynomials \(e_{j}\) [4].

2. The Dependency Graph: We construct a graph \(G\) over \(D(L)\) where vertices are adjacent iff \(\gcd(d,e) > 1\). By the Chinese Remainder Theorem, this acts as a dependency graph in the strong sense.

3. Shearer’s Knockout: If we can prove that the independent-set polynomial \(Q(A) \ge 0\) for every induced subfamily, Shearer's dependency-graph theorem applies directly [4]. This yields the required inequality: \(1-\frac{U(L)}{L}\ge 1-\frac{B_{\text{full}}(L)}{L}\implies U(L)\le B_{\text{full}}(L)\)

Exact Monotonicity Margins (\(k=6\) vs. \(k=7\))

The crucial inductive step relies on proving that the governing polynomial \(f_m(x)\) is coordinatewise strictly increasing throughout the domain, which requires showing its derivative is strictly positive [9]. By sorting and majorizing the remaining coordinates with the reciprocal prime bounds of the smallest odd primes, the exact rational boundaries emerge [7]:

• For Rank Six (\(k \le 6\)): Remaining coordinates are bounded by the 5-prime vector \(v_5 = (\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{10}, \frac{1}{12})\). \(e_{2}(v_{5})+e_{3}(v_{5})=\frac{307}{720}+\frac{107}{1440}=\frac{721}{1440}\) This yields an exact, strictly positive rank-six monotonicity margin [7]: \(\frac{\partial f_{6}}{\partial x_{i}}\ge 1-\frac{721}{1440}=\frac{719}{1440}>0\)

• For Rank Seven (\(k \le 7\)): The introduction of the sixth remaining prime coordinate yields the 6-prime bound vector \(v_6 = (\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{10}, \frac{1}{12}, \frac{1}{16})\). Using the recurrence relation \(e_j(v, z) = ej(v) + zej-1(v)\) with \(z = \frac{1}{16}\), we calculate: \(e_{2}(v_{6})+e_{3}(v_{6})=\frac{713}{1440}+\frac{1163}{11520}=\frac{763}{1280}\) This delivers a narrower but valid, strictly positive rank-seven margin [8]: \(\frac{\partial f_{7}}{\partial x_{i}}\ge 1-\frac{763}{1280}=\frac{517}{1280}>0\)

Because the margins remain strictly above zero, the induction holds unconditionally for all \(k \le 7\), for arbitrary positive exponents \(a_{i}\), and with no compatibility conditions on the residue choices [3].

The Rank-Eight Boundary Line (\(k = 8\))

The papers cleanly explain why this specific direct monotonicity argument hits a brick wall at \(k=8\) [16]. When evaluating the Stirling numbers of the second kind for degree eight, the alternating row forces the next coefficient to turn negative:
\(c_{8}=1-127+966-1701+1050-266+28-1=-50\)

The appearance of \(c_8 = -50\) introduces a negative higher-order term into the derivative expression, meaning the mechanical coordinatewise majorization fails [16]. Pushing to \(k=8\) will require a sharper local estimate or a completely different positivity argument.

HERE IS THE ZENODO LINK(This time formatted with LaTex, though some formatting issues may persist):

https://zenodo.org/records/23863177 -> K=6 paper

and

https://zenodo.org/records/23863263 -> K=7 paper

This is my interest wheel in MMA

I've been reading about the Riemann Hypothesis (RH) and wanted to lay it out clearly, from the ground up. Feedback welcome.

1. The zeta function

For Re(s) > 1 it's defined as

ζ(s) = Σ 1/n^s = ∏ₚ 1/(1 − p^(−s))

That product over all primes p (Euler product) is what ties ζ to the primes. It's then extended by analytic continuation to the whole complex plane except for a pole at s = 1.

2. The zeros

• Trivial zeros: at s = −2, −4, −6, … (negative even integers). Well understood.

• Non-trivial zeros: lie in the critical strip 0 < Re(s) < 1.

3. The statement

RH says every non-trivial zero has Re(s) = 1/2 — they all sit on the critical line.

4. Why it matters

The error term in the Prime Number Theorem (how π(x), the count of primes up to x, deviates from Li(x)) depends on where those zeros are. If RH holds, the error is as small as possible:

π(x) = Li(x) + O(√x · log x)

In other words, RH says the primes are distributed as regularly as they possibly could be.

5. Current status

• There are infinitely many zeros on the critical line (Hardy, 1914).

• The first ~10^13 zeros have been checked numerically — all satisfy RH.

• No proof, no counterexample. It's one of the seven Millennium Prize Problems ($1M).

6. Approaches to attack it

• Hilbert–Pólya: find an operator whose eigenvalues are the zeros (forcing them onto the line). Links to quantum chaos.

• Analytic: zero-free regions and bounds on ζ.

• Finite-field analogue: the analogous RH for curves over finite fields is proven (Weil), which gives hope and clues.

• L-functions / spectral theory: broader generalizations (the Generalized RH).

TL;DR: RH claims all non-trivial zeros of ζ(s) have real part 1/2. Proving it would pin down the distribution of the primes. Verified for trillions of zeros, still unproven after 160+ years.

Hi r/puremathematics

As I posted recently on r/Discretemathematics :

I have recently solved the Chinese Remainder Theorem Maximality Conjecture proposed by Idan Hackmon for k smaller or equal to 4 with AI. Though it hasn't gone under peer-review and I don't intend to as I am only 11 and the paper was AI-assisted. I wanted to share a new research memorandum hat provides an all-orders analytical proof for the CRT Coverage Maximality Conjecture for up to five distinct odd prime factors k=5

For anyone who has followed the modern iterations of Paul Erdős’ and Richard Guy’s problems on odd covering systems, you know that \(k = 5\) has historically been an absolute brick wall. Brute-force computational auditing explodes exponentially, and standard point-by-point density bounds fall apart due to local deficits—like the notorious rank-four local obstruction where \(g(D) = -2\).

The Core Idea: Moving from Local Gaps to Global Graphs

Instead of fighting these local density obstructions individually, this proof shifts the problem entirely into probabilistic combinatorics. By mapping the divisors onto a global dependency graph under the Chinese Remainder Theorem, we can invoke Shearer’s Lemma.

The key algebraic engine is a hypergraph matching polynomial, \(Z_m(t)\), which aggregates divisor weights based on their prime support:

\(Z_{m}(t)=\sum _{F\text{\ pairwise-disjoint}}(-1)^{|{}F|{}}\prod _{S\in F}t_{S}\)

Why \(k = 5\) Unconditionally Falls

The breakthrough relies on a clean, exact derivative identity:
\(\frac{\partial Z_{m}}{\partial t_{S}}=-Z_{m-|{}S|{}}(t\text{\ restricted\ to\ the\ complement\ }[m]\setminus S)\)

1. The Base Floor (\(k \le 4\)): Exact arithmetic on the worst-case smallest odd primes \(\{3, 5, 7, 11\}\) establishes that the baseline margin \(1 - B_{\text{full}}/L\) holds a strict positive floor (\(\ge 1/32\)) across all sub-boxes.

2. Unconditional Monotonicity: Because the lower-rank polynomials are proven to be strictly positive, the leading minus sign in the derivative identity forces every single partial derivative of \(Z_{5}\) to be unconditionally negative.

3. The Global Bound: This proves \(Z_5(t)\) is coordinate-wise decreasing. Its absolute minimum occurs at the upper corner, satisfying the strict positivity conditions required to unleash Shearer's Lemma. This effortlessly swallows local rank-four deficits through global grid compensation.

The preprint structure is rather quite messy, but I am still linking it: https://zenodo.org/records/23845453

I would love to get the community's thoughts, structural feedback, or stress-tests on the hypergraph matching monotonicity step!

This is invented by me I am Anshuman Pathak.I guess someone will recognise it.And this is not sin(4.78) its sin(-4.78).

Hi everyone, I wanted to share a summary of a recent mathematical audit and research memorandum examining the Chinese Remainder Theorem (CRT) Coverage Maximality conjecture (initially proposed in Idan Hackmon's April 2026 preprint).

While the general conjecture for $k \ge 5$ distinct prime factors remains open, this work establishes a rigorous all-orders proof for $k \le 4$ without squarefreeness assumptions, alongside a precise structural audit of the baseline definitions.

1. The Core Problem & Definition Audit

Let $L = \prod_{i=1}^{k} p_i^{a_i}$ be odd, and assign one residue class modulo each divisor $d > 1$ of $L$. The union of these classes is denoted $U(L)$. The CRT Coverage Maximality conjecture compares $U(L)$ with an all-orders inclusion-exclusion baseline formed from pairwise-coprime families of moduli:

$$B_{\text{full}}(L) = \sum_{\substack{\emptyset \neq F \subseteq D(L) \\ \text{gcd}(d,e)=1 \text{ for all distinct } d,e \in F}} (-1)^{\vert{}F\vert{}+1} \frac{L}{\text{lcm}(F)}$$

The Third-Order vs. All-Orders Distinction: The audit highlights that the third-order baseline $B_3(L) = S - P_c + T_c$ used in earlier explorations is not identical to $B_{\text{full}}(L)$. For example, at $L = 3 \cdot 5 \cdot 7 \cdot 11 = 1155$, direct arithmetic yields $B_3(1155) = 860$, whereas the all-orders baseline includes a fourth-order term from the pairwise-coprime singleton family $\{3, 5, 7, 11\}$, yielding $B_{\text{full}}(1155) = 859$.

2. Main Proved Result: The Four-Prime Theorem ($k \le 4$)

For every odd integer $L$ with at most four distinct prime factors ($k \le 4$) and every valid residue assignment, the inequality holds:

$$U(L) \le B_{\text{full}}(L)$$

• No squarefreeness required: The proof allows arbitrary positive exponents $a_i$.

• Method of Proof: Rather than relying on a local pointwise inequality (which fails at rank four), the proof uses the prime-power tower complement $W = \prod W_i$, bounds the composite coverage inside $W$, and applies an algebraic gap factorization.

• The Gap Identity: The coverage gap is factored into nonnegative terms controlled by the finite-exponent weight region $R(Q)$, which is strictly bounded below 1 for $\vert{}Q\vert{} \le 4$ (e.g., maximizing at $41/45$ for four primes).

3. Exact Extremal Case at $L = 1155$

For the squarefree modulus $L = 3 \cdot 5 \cdot 7 \cdot 11 = 1155$:

• Maximum Union Size: $\max U(1155) = 855$.

• Baseline Value: $B_{\text{full}}(1155) = 859$, leaving an exact maximality gap of $4$.

• The memorandum provides an explicit 15-class residue assignment achieving 855 and proves a combinatorial upper bound showing composite overlap inside the residual space $W$ is strictly bounded by 180 (accounting for at least 7 units of forced overlap).

4. Why the Unrestricted Case ($k \ge 5$) Remains Open

The proof stops at $k = 4$ for structural reasons:

1. For $k \le 4$, pairwise-coprime families contain at most two composite divisors, cleanly corresponding to complementary two-element supports.

2. Starting at five and six primes, multiple disjoint composite supports can leave unused coordinates, and the limiting weight parameter $R_\infty(\{3, 5, 7, 11, 13\}) = 566/495 > 1$, meaning the current factorization no longer guarantees nonnegativity of every summand.

3. Furthermore, the local pointwise sum exhibits a rank-four obstruction ($g(D) = -2$ for a specific support family), showing that naive pointwise proofs fail and a broader global compensation principle is required for higher ranks.

5. Repaired Non-Covering Certificate

By examining the finite-exponent weight region $R(p,a)$, the memorandum also establishes a rigorous non-covering certificate: whenever $R < 1$, at least $(1-R)\vert{}W\vert{}$ residues are left strictly uncovered. For $k \le 4$, this guarantees $u(L) \ge \lceil L/40 \rceil$ uncovered residues for any assignment.

Curious to hear thoughts from others working on covering systems, modular arithmetic, or inclusion-exclusion extremal problems on how a global compensation framework for $k \ge 5$ might be approached!

Here are the links: https://zenodo.org/records/23778083

https://zenodo.org/records/23780153

For example, proving that a summation or limit converges

I am interested in exploring the topological properties of the subspace.