2026-09-18

We know that the existence of two disjoint Mobius strip results in absolute determinant $\ge 4$. We are unsure of the existence of two overlapping Mobius strips. Let’s explore $\mathbb{R}{P}^2$ which contains multiple overalapping Mobius strips, but is constructed form a single Mobius strip.

Below is the triangulation of $\mathbb{R}{P}^2$ we will look at:

There are three Mobius strips:

$$\begin{array}{rl}{M}_1& =\{[0,1,3],[0,3,5],[3,4,5],[1,4,5],[0,1,4]\}\\ M_2& =\{[1,2,3],[2,3,4],[3,4,5],[1,4,5],[1,2,5]\}\\ M_3& =\{[0,2,4],[2,3,4],[3,4,5],[0,3,5],[0,2,5]\}.\end{array}$$

We start by looking at the submatrix corresponding to $L=M_1\cup M_2$.

$$B(M_1\cup M_2)=\begin{array}{ccccccccc}& [0,1,3]& [0,3,5]& [3,4,5]& [1,4,5]& [0,1,4]& [1,2,3]& [2,3,4]& [1,2,5]\\ [0,1]& 1& 0& 0& 0& 1& 0& 0& 0\\ [0,3]& -1& 1& 0& 0& 0& 0& 0& 0\\ [3,5]& 0& 1& -1& 0& 0& 0& 0& 0\\ [4,5]& 0& 0& 1& 1& 0& 0& 0& 0\\ [1,4]& 0& 0& 0& 1& 1& 0& 0& 0\\ [1,2]& 0& 0& 0& 0& 0& 1& 0& 1\\ [2,3]& 0& 0& 0& 0& 0& 1& 1& 0\\ [3,4]& 0& 0& 1& 0& 0& 0& 1& 0\\ [4,5]& 0& 0& 1& 1& 0& 0& 0& 0\\ [1,5]& 0& 0& 0& -1& 0& 0& 0& -1\\ [1,2]& 0& 0& 0& 0& 0& 1& 0& 1\end{array}$$

All $n\times n$ minors, where $n=|L|$, have determinant 0 or $\pm 2$.

Below is the full boundary matrix:

$$B=\begin{array}{ccccccccccc}& [0,1,3]& [0,1,4]& [0,2,4],& [0,2,5]& [0,3,5]& [1,2,3],& [1,2,5]& [1,4,5]& [2,3,4]& [3,4,5]\\ [0,1]& 1& 1& 0& 0& 0& 0& 0& 0& 0& 0\\ [0,2]& 0& 0& 1& 1& 0& 0& 0& 0& 0& 0\\ [0,3]& -1& 0& 0& 0& 1& 0& 0& 0& 0& 0\\ [0,4]& 0& -1& -1& 0& 0& 0& 0& 0& 0& 0\\ [0,5]& 0& 0& 0& -1& -1& 0& 0& 0& 0& 0\\ [1,2]& 0& 0& 0& 0& 0& 1& 1& 0& 0& 0\\ [1,3]& 1& 0& 0& 0& 0& -1& 0& 0& 0& 0\\ [1,4]& 0& 1& 0& 0& 0& 0& 0& 1& 0& 0\\ [1,5]& 0& 0& 0& 0& 0& 0& -1& -1& 0& 0\\ [2,3]& 0& 0& 0& 0& 0& 1& 0& 0& 1& 0\\ [2,4]& 0& 0& 1& 0& 0& 0& 0& 0& -1& 0\\ [2,5]& 0& 0& 0& 1& 0& 0& 1& 0& 0& 0\\ [3,4]& 0& 0& 0& 0& 0& 0& 0& 0& 1& 1\\ [3,5]& 0& 0& 0& 0& 1& 0& 0& 0& 0& -1\\ [4,5]& 0& 0& 0& 0& 0& 0& 0& 1& 0& 1\end{array}$$

I wrote code (see below) which computed every $k\times k$ minor for $1\le k\le 10$ and found that all minors had absolute determinant at most 2, i.e., $B$ is totally bimodular!

m, n = A.shape
num_bad_minors = 0
 
for k in range(1, min(m, n) + 1):
    print("Testing all {}x{} minors...".format(k, k))
    for rows in combinations(range(m), k):
        for cols in combinations(range(n), k):
            submatrix = A[np.ix_(rows, cols)]
            det = np.linalg.det(submatrix)
            if np.abs(det) > 2:
                num_bad_minors += 1
                print("{}".format(det))
 
print("Found {} minors with absolute determinant greater than 2.".format(num_bad_minors))

Subdividing this triangulation causes the number of minors to blow up. Restricting to just two Mobius strips and including all triangles in our submatrix resulted in determiant 0, $\pm 1$, or $\pm 2$.

---

Theorem: Let $[{\partial }_2]$ be the $m\times n$ boundary matrix for a triangulation of $\mathbb{R}{P}^2$. Then every $n\times n$ submatrix of $[{\partial }_2]$ has absolute determinant at most 2. Moreover, any square submatrix of $[{\partial }_2]$ has absolute determinant at most $2^k$ for some $k$.

Proof. Let $A$ be a subcomplex of $\mathbb{R}{P}^2$. The snake lemma gives that

$$H_1(A)\to H_1(\mathbb{R}{P}^2)\to H_1(\mathbb{R}{P}^2,A)\to H_0(A)\to H_0(\mathbb{R}{P}^2)$$

is an exact sequence. Recall that

$$H_k(\mathbb{R}{P}^2)=\{\begin{array}{ll}\mathbb{Z}& k=0\\ \mathbb{Z}/2\mathbb{Z}& k=1\\ 0& k\ge 2\end{array}$$

so

$$H_1(A)\stackrel{i_{\ast }}{\to }\mathbb{Z}/2\mathbb{Z}\stackrel{{\pi }_{\ast }}{\to }{H}_1(\mathbb{R}{P}^2,A)\stackrel{\partial }{\to }{H}_0(A)\stackrel{j_{\ast }}{\to }\mathbb{Z}.$$

Because the sequence is exact, $\text{im}\partial =\text{ker}{j}_{\ast }$. Note that $j_{\ast }$ is induced by the inclusion $j:A\hookrightarrow \mathbb{R}{P}^2$ so it follows that for a vertex $a_i\in A$, $j_{\ast }[a_i]=[a_i]$. Assume that we have two vertices $a_1$ and $a_2$ which get mapped to $[a_1]$ and $[a_2]$ in $H_0(\mathbb{R}{P}^2)$. Because $\mathbb{R}{P}^2$ is connected, it follows that there is a 1-chain $\gamma$ such that ${\partial }_1\gamma =a_2-a_1$ in $\mathbb{R}{P}^2$. In other words, $j_{\ast }(a_2-a_1)=0$ implying that $[a_1]=[a_2]$. Thus, every class in $H_0(A)$ gets mapped to the same generator. This implies that $\text{rank}(j_{\ast })=1$ which results in $\text{ker}(j_{\ast })\cong {\mathbb{Z}}^{{\beta }_0-1}$ where ${\beta }_0$ is the number of connected components of $A$. Hence, we can reduce our exact sequence to

$$H_1(A)\stackrel{i_{\ast }}{\to }\mathbb{Z}/2\mathbb{Z}\stackrel{{\pi }_{\ast }}{\to }{H}_1(\mathbb{R}{P}^2,A)\stackrel{\partial }{\to }{\mathbb{Z}}^{{\beta }_0-1}\to 0.$$

Due to exactness at $\mathbb{Z}/2\mathbb{Z}$, it follows that $\text{ker}{\pi }_{\ast }=\text{im}{i}_{\ast }$. By the first isomorphism theorem,

$$\text{im}{\pi }_{\ast }\cong (\mathbb{Z}/2\mathbb{Z})/\text{ker}({\pi }_{\ast })=(\mathbb{Z}/2\mathbb{Z})/\text{im}(i_{\ast })=\text{coker}(i_{\ast }).$$

We also have exactness at $H_1(\mathbb{R}{P}^2,A)$ which implies that $\text{ker}\partial =\text{im}{\pi }_{\ast }\cong \text{coker}(i_{\ast })$. So we can reduce our exact sequence further to

$$0\to \text{coker}(i_{\ast })\to H_1(\mathbb{R}{P}^2,A)\stackrel{\partial }{\to }{\mathbb{Z}}^{{\beta }_0-1}\to 0.$$

This sequence splits as ${\mathbb{Z}}^{{\beta }_0-1}$ is a free abelian group, thus it follows that

$$H_1(\mathbb{R}{P}^2,A)\cong \text{coker}(i_{\ast })\oplus {\mathbb{Z}}^{c-1},$$

i.e., either $H_1(\mathbb{R}{P}^2,A)$ is torsion free, or its only torsion subgroup is $\mathbb{Z}/2\mathbb{Z}$. Indeed, because $i_{\ast }:H_1(A)\to \mathbb{Z}/2\mathbb{Z}$ it follows that $i_{\ast }$ is onto or $i_{\ast }$ is trivial. These two cases give $\text{coker}(i_{\ast })=0$ and $\text{coker}(i_{\ast })=\mathbb{Z}/2\mathbb{Z}$, respectively. Therefore, every $n\times n$ minor, where $n$ is the number of columns, has absolute determinant at most 2.

Now let $L_0\subset L\subset \mathbb{R}{P}^2$ be a pair of pure subcomplexes of dimension 1 and 2, respectively. Let $p:S^2\to \mathbb{R}{P}^2$ be the usual sphere construction where a point $p$ is identified with its antipodal point $-p$. Set $L^{\prime }=p^{-1}(L)$ and $L_0^{\prime }=p^{-1}(L_0)$. Because $L\ne \mathbb{R}{P}^2$, it follows that $L^{\prime }\subseteq S^2\setminus \{p\}$ for some point $p$. Since $S^2\setminus \{p\}\cong {\mathbb{R}}^2$, it follows that $L^{\prime }$ is embeddable in ${\mathbb{R}}^2$. Thus it follows that $H_1(L^{\prime },L_0)$ must be torsion-free as the boundary map $[{\partial }_2]$ for any complex embedded in ${\mathbb{R}}^2$ is TU. Notice that our map $p$ gives a lifted triangulation of $S^2$. Define $t:C_2(L)\to C_2(L^{\prime })$ by

$$t(\sigma )=\tau +{\tau }^{\prime }$$

where $\tau$ and ${\tau }^{\prime }$ are the lifted triangles of $p^{-1}(\sigma )$. Then it follows that $t(C_2(L_0))=C_2(L_0^{\prime })$ implying that $t$ induces a homomorphism $t_{\ast }:H_1(L,L_0)\to H_1(L^{\prime },L_0^{\prime })$. Because $p$ induces a map $p_{\#}:C_1(L^{\prime })\to C_1(L)$ given by

$$p_{\#}\left(\sum _i{c}_i{\sigma }_i\right)=\sum _i{c}_{i}p({\sigma }_i)$$

it follows that

$$p_{\#}(t(\sigma ))=p_{\#}(\tau +{\tau }^{\prime })=p_{\#}(\tau )+p_{\#}({\tau }^{\prime })=2\sigma$$

implying that $p_{\ast }\circ t_{\ast }=2$ where $p_{\ast }:H_1(L^{\prime },L_0^{\prime })\to H_1(L,L_0)$ is induced by $p_{\#}$. Let $x\in H_1(L,L_0)$ be such that $kx=0$ for some $k\in \mathbb{Z}$. Notice that

$$k{t}_{\ast }(x)=t_{\ast }(kx)=t_{\ast }(0)=0$$

implying that $t_{\ast }(x)$ is torsion in $H_1(L^{\prime },L_0^{\prime })$. But $H_1(L^{\prime },L_0^{\prime })$ is torsion free, so we get that $t_{\ast }(x)=0$ which implies that

$$2x=p_{\ast }(t_{\ast }(x))=p_{\ast }(0)=0.$$

Therefore, the torsion subgroup of $H_1(L,L_0)$ is $(Z/2\mathbb{Z}{)}^k$ for some $k$. $\square$