2026-05-26

According to Claude, the example complex I went through on 2026-05-14 cannot be embedded in ${\mathbb{R}}^3$. I repeated the run on a modified complex that should be a triangulation of $M\vee S^2$ where $M$ is the Mobius strip.

Iteration 1-19: Phase 1 searching for initial basis

Iteration 20:

$$\begin{array}{rl}z& =6\\ x& =[1,4]+[0,1]+[2,3]+[4,5]-[0,3]-[2,5]\\ y& =0\end{array}$$

Iteration 21:

$$\begin{array}{rl}z& =5\\ x& =[2,3]+[4,5]-[0,4]-[0,3]-[2,5]\\ y& =-[0,1,4]\end{array}$$

Iteration 22:

$$\begin{array}{rl}z& =4\\ x& =[3,4]+[2,3]+[4,5]-[2,5]\\ y& =[0,3,4]-[0,1,4]\end{array}$$

The remaining 7 iterations are the same solution $(z,x,y)$ but with equivalent bases.

The final basis matrix:

Which has Smith normal form

$$UBV=\text{diag}(I_{17},2)$$

where

$$U=\left[\begin{array}{cccccccccccccccccc}& 1& & & & & & & & & & & & & & & & \\ & & 1& & & & & & & & & & & & & & & \\ & & & 1& & & & & & & & & & & & & & \\ & & & & 1& & & & & & & & & & & & & \\ & & & & & 1& & & & & & & & & & & & \\ & & & & & & 1& & & & & & & & & & & \\ & & & & & & & 1& & & & & & & & & & \\ & & & & & & & & 1& & & & & & & & & \\ & & & & & & & & & 1& & & & & & & & \\ & & & & & & & & & & 1& & & & & & & \\ & & & & & & & & & & & 1& & & & & & \\ & & & & & & & & & & & & 1& & & & & \\ 1& & & & & & & & & & & & & 1& -1& & & -1\\ & & & & & & & & & & & & & & 1& & & \\ & & & & & & & & & & & & & & & 1& & \\ & & & & & & & & & & & & & & & & 1& \\ & & & & & & & & & & & & & & & & & 1\\ & & & & & & & & & & & & & -1& 1& & & 1\end{array}\right].$$