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

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

Exploration of SNF

At any feasible basis, $B$ is of the form

$$B=\left[\begin{array}{cccccc}\pm e_{i_1}& \cdots & \pm e_{i_j}& \pm {\partial }_{i_{j+1}}& \cdots & \pm {\partial }_{i_m}\end{array}\right]$$

where each $e_{i_k}$ is a standard basis vector. Since order does not matter for basic variables, we assume WLOG that

$$i_1<i_2<\cdots <i_j.$$

Perform the following row swap operations

$$R_{i_k}\leftrightarrow R_k\forall k=1,\dots ,j$$

and

$$R_k\leftarrow -R_k\forall k\text{ where eik is negative}$$

at this point

$$UBV=\left[\begin{array}{cc}{I}_j& {\partial }_T\\ 0& {\partial }_B\end{array}\right]$$

where $V=I$ and $U=D{R}_{\pi }$ where

$$D_{k,\ell }=\{\begin{array}{ll}0& k\ne \ell \\ -1& k=\ell ,\text{ and eik is negative}\\ 1& \text{otherwise}\end{array}$$

and $R_{\pi }$ is the permutation matrix corresponding to

$$\pi :\left(\begin{array}{ccccccc}1& 2& \cdots & j& j+1& \cdots & m\\ i_1& i_2& \cdots & i_j& j+1& \cdots & m\end{array}\right)$$

We then cancel ${\partial }_T$ with column operations, affecting $V$ which we are not interested in. This leaves us with

$$(D{R}_{\pi })BV=\left[\begin{array}{cc}{I}_j& 0\\ 0& {\partial }^{\prime }\end{array}\right].$$

The remaining operations only affect, ${\partial }^{\prime }$. Let $U^{\prime }$ and $V^{\prime }$ such that $U^{\prime }{\partial }^{\prime }{V}^{\prime }=SNF({\partial }^{\prime })$. Then we get

$$\left[\begin{array}{cc}{I}_j& 0\\ 0& U^{\prime }\end{array}\right](D{R}_{\pi })B{V}^{\prime }\left[\begin{array}{cc}{I}_j& 0\\ 0& V^{\prime }\end{array}\right]=\left[\begin{array}{cc}{I}_j& 0\\ 0& SNF({\partial }^{\prime })\end{array}\right].$$

Note here that $|\det U|=|\det U^{\prime }|$ where $U$ is the matrices to the left of $B$. Note that $D$ and $R_{\pi }$ only affect the first $j$ columns. So

$$\left[\begin{array}{cc}{I}_j& 0\\ 0& U^{\prime }\end{array}\right](D{R}_{\pi })=\left[\begin{array}{cc}{D}_j{R}_j& 0\\ 0& U^{\prime }\end{array}\right]$$

where $D_j$ and $R_j$ are the top left $j\times j$ submatrices of $D$ and $R_{\pi }$, respectively.

Therefore, our conditions only depend on ${\partial }^{\prime }$, the $(m-j)\times (m-j)$ submatrix of $[\partial ]$.