Research proof index

g is invariant under S. Proof: g = (2H₂ + H₄)f₂ + (H₂ + 2H₄)f₄ Under S: H₂ ↦ -H₄, H₄ ↦ -H₂, f₂ ↦ -f₄, f₄ ↦ -f₂ g|S = (2(-H₄) + (-H₂))(-f₄) + ((-H₄) + 2(-H₂))(-f₂) = (2H₄ + H₂)f₄ + (H₄ + 2H₂)f₂ = g

g is invariant under T. Proof: Under T: H₂ ↦ -H₂, H₄ ↦ H₃, f₂ ↦ -f₂, f₄ ↦ f₃ = f₂ + f₄ g|T = (2(-H₂) + H₃)(-f₂) + ((-H₂) + 2H₃)(f₂ + f₄) Using Jacobi: H₃ = H₂ + H₄, simplifies to g.

h is invariant under S. Proof: h = f₂² + f₂f₄ + f₄² Under S: f₂|[4]S = -f₄, f₄|[4]S = -f₂ Using mul_slash_SL2: (f₂²)|[8]S = (f₂|[4]S)² = (-f₄)² = f₄² (f₂f₄)|[8]S = (f₂|[4]S)(f₄|[4]S) = (-f₄)(-f₂) = f₂f₄ (f₄²)|[8]S = (f₄|[4]S)² = (-f₂)² = f₂² So h|[8]S = f₄² + f₂f₄ + f₂² = f₂² + f₂f₄ + f₄² = h

h is invariant under T. Proof: Under T: f₂ ↦ -f₂, f₄ ↦ f₃ = f₂ + f₄ h|T = (-f₂)² + (-f₂)(f₂ + f₄) + (f₂ + f₄)² = f₂² - f₂² - f₂f₄ + f₂² + 2f₂f₄ + f₄² = f₂² + f₂f₄ + f₄² = h

For a finite group of odd order and a three-term-progression-free set $s$, the normalized inner product between $\mu_s*\mu_s$ and the uniform measure on $2s$ is exactly $|s|^{-2}$. The identity records the precise normalized count forced by the absence of nontrivial three-term progressions.

For $x\in G\setminus G'$, summing the localized Carleson contribution over all tiles is exactly the same as summing the corresponding oscillatory kernel integral over the integer scales from $\sigma_1(x)$ to $\sigma_2(x)$. This is the identity that converts the discrete tile model back into the finitary operator.