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: $h(i)=e^{i{\phi }_i}$.

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: ${\sum }_{j\sim i}{W}_t(i,j)\sin ({\alpha }_{ij}+{\phi }_i-{\phi }_j)=0$ 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 $\phi \mapsto \phi +c$.

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

where $L=B\mathrm{diag}(W)B^T$ is the weighted graph Laplacian and $L^{†}$ 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.
• $\mathrm{cut}(\{4,5\},\{1,2,3\})=W(4,3)+W(5,2)=1+0.5=1.5$.
• $\varphi (\{4,5\})=1.5/17.5\approx 0.086\le 0.15$. 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). $\mathrm{Existence}(\{4,5\},t)=([\varnothing ],\{4\},\{1\})$.

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.
• $\mathrm{cut}(B,\bar{B})=W(7,6)+W(8,6)=0.4+0.6=1.0$.
• $\varphi (B)=1.0/97.0\approx 0.0103\le 0.1$. Yes.

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

Checking smaller subsets: $\{5,6\}$ has $\mathrm{vol}=5.8$, $\mathrm{cut}=0.5+0.3+0.4+0.6=1.8$, $\varphi =1.8/5.8\approx 0.31>0.1$. 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 $\varphi \approx 0.007$.

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 $e_1=(7,9),e_2=(7,10),e_3=(9,10)$):

$L=BW{B}^T=8\left(\begin{array}{ccc}2& -1& -1\\ -1& 2& -1\\ -1& -1& 2\end{array}\right).$

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

$B\mathrm{diag}(W)\alpha =8B\alpha =8\left(\begin{array}{c}-0.3+0.1\\ -0.3+0.2\\ -0.1+0.2\end{array}\right)=8\left(\begin{array}{c}-0.2\\ -0.1\\ 0.1\end{array}\right)=\left(\begin{array}{c}-1.6\\ -0.8\\ 0.8\end{array}\right).$

Normal equation: $L{\phi }^{\ast }=\left(\begin{array}{c}1.6\\ 0.8\\ -0.8\end{array}\right)$ (note sign).

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

${\phi }^{\ast }=L^{†}\left(\begin{array}{c}1.6\\ 0.8\\ -0.8\end{array}\right)+c\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right).$

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: $(1.067,0.267,-1.333)$.

$L^{†}=\frac{1}{24}\left(\begin{array}{ccc}2/3& -1/3& -1/3\\ -1/3& 2/3& -1/3\\ -1/3& -1/3& 2/3\end{array}\right)$ (pseudoinverse of $L$ on the orthogonal complement of $1$).

Actually, for a $3\times 3$ Laplacian $L=8\cdot 3(I-\frac{1}{3}11^T)=24(I-\frac{1}{3}11^T)$, the pseudoinverse is $L^{†}=\frac{1}{24}(I-\frac{1}{3}11^T)$.

${\phi }^{\ast }=\frac{1}{24}(I-\frac{1}{3}11^T)\left(\begin{array}{c}1.6\\ 0.8\\ -0.8\end{array}\right)=\frac{1}{24}\left(\begin{array}{c}1.6-0.533\\ 0.8-0.533\\ -0.8-0.533\end{array}\right)=\frac{1}{24}\left(\begin{array}{c}1.067\\ 0.267\\ -1.333\end{array}\right)\approx \left(\begin{array}{c}0.0444\\ 0.0111\\ -0.0556\end{array}\right).$

(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 $g=\left(\begin{array}{cc}a& -\bar{b}\\ b& \bar{a}\end{array}\right)$, $|a{|}^2+|b{|}^2=1$.

Non-abelian features#

• Holonomy conjugacy class $\ne$ holonomy itself (base-point dependence remains).
• Scalar curvature ${\omega }_t(△)=d_{SU(2)}(\Omega ,I{)}^2$ 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.