Research proof index

f₂ tends to 0 at infinity. Proof: f₂ = serre_D 2 H₂ - (1/6)H₂(H₂ + 2H₄) Since H₂ → 0, both serre_D 2 H₂ → 0 and H₂(H₂ + 2H₄) → 0, so f₂ → 0.

f₄ transforms under S as f₄|S = -f₂. Proof outline (symmetric to f₂_S_action): 1. (serre_D 2 H₄)|[4]S = serre_D 2 (H₄|[2]S) = serre_D 2 (-H₂) = -serre_D 2 H₂ 2. (H₄(2H₂ + H₄))|[4]S = (-H₂)(2(-H₄) + (-H₂)) = H₂(H₂ + 2H₄) 3. f₄|[4]S = -serre_D 2 H₂ + (1/6)H₂(H₂ + 2H₄) = -f₂

Key algebraic identity for proving f₂ = f₄ = 0. Given Af₂ + Bf₄ = 0, we have f₄² * (A² - AB + B²) = A² * (f₂² + f₂f₄ + f₄²).

f₄ transforms under T as f₄|T = f₃. Proof outline: 1. (serre_D 2 H₄)|[4]T = serre_D 2 (H₄|[2]T) = serre_D 2 H₃ 2. (H₄(2H₂ + H₄))|[4]T = H₃(2(-H₂) + H₃) = H₃(H₃ - 2H₂) Using Jacobi H₃ = H₂ + H₄: H₃ - 2H₂ = H₄ - H₂ 3. f₄|[4]T = serre_D 2 H₃ + (1/6)H₃(H₃ - 2H₂) But H₂² - H₄² = (H₂ - H₄)(H₂ + H₄) = (H₂ - H₄)H₃ So (1/6)(H₂² - H₄²) = -(1/6)H₃(H₄ - H₂) = -(1/6)H...

f₄ tends to 0 at infinity. Proof: f₄ = serre_D 2 H₄ + (1/6)H₄(2H₂ + H₄) serre_D 2 H₄ = D H₄ - (1/6)E₂ H₄ → 0 - (1/6)11 = -1/6 (since H₄ → 1, E₂ → 1) H₄(2H₂ + H₄) → 1*(0 + 1) = 1 So f₄ → -1/6 + (1/6)*1 = 0.

Let $G$ be a finite vector space over $\mathbb F_q$, where $q\ge3$ is prime. If a nonempty set $A\subseteq G$ contains no nontrivial three-term arithmetic progression, then $\dim_{\mathbb F_q}G \le 2^{148}\,\mathcal L(\alpha)^9$, where $\alpha=|A|/|G|$ and $\mathcal L$ is the project's capped logarithm.