Chapter 10 — Worked Examples

Prerequisites: Chapters 2--8 (Part I).

10.1 The $U(1)$ Theory in Full#

Setup#

• Gauge group: $G=U(1)=\{e^{i\alpha}:\alpha\in[0,2\pi)\}$.
• Relation transit: $g_t(i,j)=e^{i\alpha_t(i,j)}$.
• Gauge transformation: $h(i)=e^{i\varphi_i}$.

Gauge action#

Holonomy and curvature#

For triangle $(i,j,k)$:

Discrete curvature:

Since $U(1)$ is abelian, the holonomy itself (not just its conjugacy class) is gauge-invariant.

Flattening energy#

Optimal gauge#

Critical-point equation: $\sum_{j\sim i}W_t(i,j)\sin(\alpha_{ij}+\varphi_i-\varphi_j)=0$ for all $i\in F^\circ$.

At a global minimum with small residual curvature ($|\tilde\alpha_{ij}|<\pi$ for all edges), the Hessian is a weighted graph Laplacian with positive weights (Theorem 7.11), so the minimum is unique up to a constant shift $\varphi\mapsto\varphi+c$.

In the small-curvature regime, the normal equations give:

where $L=B\,\mathrm{diag}(W)B^T$ is the weighted graph Laplacian and $L^\dagger$ its pseudoinverse.

Canonical connection and residual curvature#

The constant $c$ cancels in $\tilde\alpha_{ij}$, confirming Conjecture 7.10 for $U(1)$.

10.2 Five-Node Example ($G=U(1)$)#

Graph#

  1 ─── 2
  │ ╲ ╱ │
  │  3   │
  │ ╱ ╲ │
  4 ─── 5

$V=\{1,2,3,4,5\}$, $G=U(1)$, $\theta=0.15$, $\tau=0.8$.

Weights:

• Internal to $\{1,2,3\}$: $W(1,2)=W(1,3)=W(2,3)=10$.
• Internal to $\{4,5\}$: $W(4,5)=8$.
• Bridge: $W(3,4)=1$, $W(2,5)=0.5$.
• All others: $0$.

Step 1: Degrees#

Step 2: Fruit identification#

Candidate $F=\{4,5\}$:

• $\mathrm{vol}(\{4,5\})=9+8.5=17.5$.
• $\mathrm{vol}(V)=20+20.5+21+9+8.5=79$.
• (F1): $17.5\le 39.5=\frac{1}{2}\cdot 79$. Yes.
• $\mathrm{cut}(\{4,5\},\{1,2,3\})=W(4,3)+W(5,2)=1+0.5=1.5$.
• $\phi(\{4,5\})=1.5/17.5\approx 0.086\le 0.15$. Yes.

$\{4,5\}\in\mathfrak{F}_t$.

Candidate $F=\{1,2,3\}$: $\mathrm{vol}(\{1,2,3\})=61.5>39.5$. Violates (F1). Not a fruit.

Step 3: Door detection#

For $F=\{4,5\}$:

$\Sigma=\{4\}$, $e_4=1$.

Step 4: Existence#

$F^\circ=\{4,5\}\setminus\{4\}=\{5\}$. Single node, no internal edges.

$[A_\infty]$ is trivial (empty connection). $\mathrm{Existence}(\{4,5\},t)=([\emptyset],\{4\},\{1\})$.

Verification of theorems#

• Theorem A: Internal energy $=2\cdot 8=16$; $\mathrm{vol}=17.5$; ratio $=16/17.5\approx 0.914\ge 1-0.15=0.85$. Satisfied.
• Theorem B: $|\Sigma|=1\le 0.15\cdot 17.5/0.8\approx 3.28$. Satisfied.
• Theorem D: Expected escape time $\ge 1/(2\cdot 0.15)\approx 3.3$ steps. Reasonable for a 2-node fruit.

10.3 Ten-Node Example ($G=U(1)$, Complete Calculation)#

Graph structure#

Fruit A: {1,2,3,4} — complete graph, internal weight 10
Fruit B: {7,8,9,10} — complete graph, internal weight 8
Stem: {5,6}
 
Bridges:
  3 -- 5 (weight 0.5)
  4 -- 5 (weight 0.3)
  5 -- 6 (weight 2.0)
  6 -- 7 (weight 0.4)
  6 -- 8 (weight 0.6)

Parameters: $\theta=0.1$, $\tau=0.5$.

Step 1: Degrees#

Fruit A internal edges (complete $K_4$, weight 10): each node has 3 internal edges, so internal degree $=3\times 10=30$.

$\mathrm{vol}(V)=30+30+30.5+30.3+2.8+3.0+24.4+24.6+24+24=223.6$.

Step 2: Fruit identification#

Fruit A $=\{1,2,3,4\}$:

• $\mathrm{vol}(A)=30+30+30.5+30.3=120.8$.
• (F1): $120.8>111.8=\frac{1}{2}\cdot 223.6$. Violates (F1).

So $\{1,2,3,4\}$ is too large (majority). Check complement:

Fruit B $=\{7,8,9,10\}$:

• $\mathrm{vol}(B)=24.4+24.6+24+24=97.0$.
• (F1): $97.0\le 111.8$. Yes.
• $\mathrm{cut}(B,\bar B)=W(7,6)+W(8,6)=0.4+0.6=1.0$.
• $\phi(B)=1.0/97.0\approx 0.0103\le 0.1$. Yes.

$\{7,8,9,10\}\in\mathfrak{F}_t$.

Checking smaller subsets: $\{5,6\}$ has $\mathrm{vol}=5.8$, $\mathrm{cut}=0.5+0.3+0.4+0.6=1.8$, $\phi=1.8/5.8\approx 0.31>0.1$. Not a fruit.

Summary of fruits: $\mathfrak{F}_t=\{\{7,8,9,10\}\}$ (only Fruit B qualifies under the strict (F1) condition).

Remark. If we relax (F1) or embed this graph in a larger graph so that $\mathrm{vol}(A)\le\frac{1}{2}\mathrm{vol}(V)$, then $A$ would also be a fruit with $\phi\approx 0.007$.

Step 3: Door detection for Fruit B#

$\Sigma=\{8\}$, $e_8=0.6$.

Step 4: Kernel and existence#

$B^\circ=\{7,9,10\}$.

Internal edges of $B^\circ$: $(7,9),(7,10),(9,10)$, all with weight 8. This is a complete triangle $K_3$.

Transit angles (assign): $\alpha_{79}=0.3$, $\alpha_{7,10}=0.1$, $\alpha_{9,10}=0.2$.

Holonomy of triangle $(7,9,10)$: $\Omega=e^{i(0.3+0.2-0.1)}=e^{0.4i}$. Non-trivial curvature.

Optimal gauge (small-curvature approximation):

Incidence matrix for $K_3$ on $\{7,9,10\}$ (edges ordered as $e_1=(7,9),e_2=(7,10),e_3=(9,10)$):

$L=BWB^T=8\begin{pmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{pmatrix}.$

$\alpha=(0.3,0.1,0.2)^T$.

$B\,\mathrm{diag}(W)\alpha=8B\alpha=8\begin{pmatrix}-0.3+0.1\\-0.3+0.2\\-0.1+0.2\end{pmatrix}=8\begin{pmatrix}-0.2\\-0.1\\0.1\end{pmatrix}=\begin{pmatrix}-1.6\\-0.8\\0.8\end{pmatrix}.$

Normal equation: $L\varphi^*=\begin{pmatrix}1.6\\0.8\\-0.8\end{pmatrix}$ (note sign).

$L$ has eigenvalues $0,24,24$ with $\ker L=\mathrm{span}((1,1,1)^T)$. Projecting out the kernel:

$\varphi^*=L^\dagger\begin{pmatrix}1.6\\0.8\\-0.8\end{pmatrix}+c\begin{pmatrix}1\\1\\1\end{pmatrix}.$

The RHS has zero mean already: $1.6+0.8-0.8=1.6\ne 0$. Project: $\bar v=1.6/3\approx 0.533$. Centered: $(1.067, 0.267, -1.333)$.

$L^\dagger=\frac{1}{24}\begin{pmatrix}2/3&-1/3&-1/3\\-1/3&2/3&-1/3\\-1/3&-1/3&2/3\end{pmatrix}$ (pseudoinverse of $L$ on the orthogonal complement of $\mathbf{1}$).

Actually, for a $3\times 3$ Laplacian $L=8\cdot 3(I-\frac{1}{3}\mathbf{1}\mathbf{1}^T)=24(I-\frac{1}{3}\mathbf{1}\mathbf{1}^T)$, the pseudoinverse is $L^\dagger=\frac{1}{24}(I-\frac{1}{3}\mathbf{1}\mathbf{1}^T)$.

$\varphi^*=\frac{1}{24}(I-\frac{1}{3}\mathbf{1}\mathbf{1}^T)\begin{pmatrix}1.6\\0.8\\-0.8\end{pmatrix}=\frac{1}{24}\begin{pmatrix}1.6-0.533\\0.8-0.533\\-0.8-0.533\end{pmatrix}=\frac{1}{24}\begin{pmatrix}1.067\\0.267\\-1.333\end{pmatrix}\approx\begin{pmatrix}0.0444\\0.0111\\-0.0556\end{pmatrix}.$

(Taking $c=0$.)

Residual angles:

• $\tilde\alpha_{79}=0.3+(0.0444-0.0111)=0.3+0.0333=0.333$.
• $\tilde\alpha_{7,10}=0.1+(0.0444-(-0.0556))=0.1+0.1=0.2$.
• $\tilde\alpha_{9,10}=0.2+(0.0111-(-0.0556))=0.2+0.0667=0.267$.

Check: holonomy $=0.333+0.267-0.2=0.4$ (unchanged, as expected---gauge transformation preserves holonomy).

Residual curvature:

• $\rho(7)=8(0.333^2+0.2^2)=8(0.111+0.04)=1.208$.
• $\rho(9)=8(0.333^2+0.267^2)=8(0.111+0.071)=1.456$.
• $\rho(10)=8(0.2^2+0.267^2)=8(0.04+0.071)=0.888$.

Total energy: $\mathcal{E}=8(0.333^2+0.2^2+0.267^2)=8(0.111+0.04+0.071)=1.776$.

Existence of Fruit B:

where $\tilde g$ encodes residual angles $(0.333, 0.2, 0.267)$ on the triangle $\{7,9,10\}$.

Verification#

• Theorem A: Internal energy $=6\times 2\times 8=96$ (6 directed pairs in $K_4$, each weight 8). $\mathrm{vol}(B)=97$. Ratio $=96/97\approx 0.990\ge 1-0.1=0.9$. Satisfied.
• Theorem B: $|\Sigma|=1\le 0.1\times 97/0.5=19.4$. Satisfied.
• Theorem E: Curvature is distributed over the $K_3$ kernel. Node 10 (farthest from door 8) has the smallest $\rho$, consistent with localisation.

10.4 $SU(2)$ Example (Sketch)#

Setup#

$G=SU(2)$, with elements $g=\begin{pmatrix}a&-\bar b\\b&\bar a\end{pmatrix}$, $|a|^2+|b|^2=1$.

Non-abelian features#

• Holonomy conjugacy class $\ne$ holonomy itself (base-point dependence remains).
• Scalar curvature $\omega_t(\triangle)=d_{SU(2)}(\Omega,I)^2$ is still gauge-invariant.
• Energy landscape $\mathcal{E}_{F^\circ}(h)$ is non-convex: multiple local minima possible.

Triangle example#

$\{1,2,3\}$ with:

Holonomy: $\Omega=e^{i\pi\sigma_x/4}\cdot e^{i\pi\sigma_y/4}\cdot e^{i\pi\sigma_z/4}\ne I$. Non-zero curvature that cannot be removed by gauge transformation (topological obstruction from holonomy).

This illustrates that for non-abelian $G$, complete flattening may be impossible, and the residual curvature reflects genuine topological content.