Theorem E — Curvature Localisation

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1. #1

   (ii) Case . The optimal gauge satisfies the linear system (Theorem 7.11). The residual angles are . These solve:

   $$\tilde\alpha = (I-B^T L^\dagger B\,\mathrm{diag}(W))\,\alpha =: \Pi\,\alpha.$$
2. #2

   The operator is the projection onto the cycle space of the graph. The residual curvature at node is:

   $$\rho(i) = \sum_{j\sim i}W_t(i,j)\,\tilde\alpha_{ij}^2.$$
3. #3

   The source of non-zero is the curvature (holonomy around cycles). Near doors, the cycles passing through the exterior create non-trivial holonomy. The Green's function of the graph Laplacian on has the well-known decay property: for a graph with spectral gap ,

   $$|L^\dagger(i,j)|\le\frac{C_0}{\mathrm{vol}(F^\circ)}\cdot e^{-\sqrt{\lambda_2}\,d_{\mathrm{graph}}(i,j)}$$
4. #4

   At the optimal gauge , the Euler–Lagrange equation on each node reads:

   $$\sum_{j\sim i}W_t(i,j)\,\nabla_{h(i)}d_G(g_t^h(i,j),e)^2=0.$$
5. #5

   The energy on is then controlled by a discrete maximum principle: interior values of cannot exceed boundary values (door-adjacent). Summing:

   $$\sum_{i\in F_{\mathrm{deep}}}\rho(i)\le\frac{|F_{\mathrm{deep}}|}{|F_{\mathrm{near}}|}\cdot\mu\cdot\sum_{i\in F_{\mathrm{near}}}\rho(i)$$