Flip cycles in plabic graphs

Abstract

Planar bicolored (plabic) graphs are combinatorial objects introduced by Postnikov to give parameterizations of the positroid cells of the totally nonnegative Grassmannian $\text {Gr}^{\ge 0}(n,k)$. Any two plabic graphs for the same positroid cell can be related by a sequence of certain moves. The flip graph has plabic graphs as vertices and has edges connecting the plabic graphs which are related by a single move. A recent result of Galashin shows that plabic graphs can be seen as cross-sections of zonotopal tilings for the cyclic zonotope Z(n, 3). Taking this perspective, we show that the fundamental group of the flip graph is generated by cycles of length 4, 5, and 10, and use this result to prove a related conjecture of Dylan Thurston about triple crossing diagrams. We also apply our result to make progress on an instance of the generalized Baues problem.

Appendix

The proof of Theorem 4.7 in Sect. 4 relies on the base case, which can be reformulated as follows. Consider the CW-complex $\mathbb {U}_{n}$, defined as follows

• The vertices of $\mathbb {U}_n$ are the triangulations of a convex n-gon.

• The edges of $\mathbb {U}_n$ are the flips between triangulations differing inside a quadrilateral.

• Every five triangulations differing inside a pentagon give rise to a 5-cycle of flips, along which we glue a 2-cell in $\mathbb {U}_n$.

• Every two commuting triangulations occurring in non-overlapping quadrilaterals give rise to a 4-cycle of flips, along which we glue a 2-cell in $\mathbb {U}_n$.

Fact A.1

The complex $\mathbb {U}_{n}$ is simply connected.

This is equivalent to the case $k=1$ of Theorem 4.7. Fact A.1 is well-known, and goes back at least to Stasheff’s work [19], where he constructs his celebrated associahedron. The formulation as above (but in greater generality) could be found, for example, in [11, Theorem 7]. An independent proof of this fact follows from the argument below.

We give a different proof of Theorem 4.7 for the case of cyclic connectivity $\pi ^: = \pi (n,k)$. An advantage of this proof that it doesn’t rely on the base case $k=1$, unlike the one in Sect. 4.

Our strategy is to consider a cycle of plabic moves as a cycle of zonotopal tiling flips, and apply Theorem 2.6. We would perform each plabic move by doing flips in tiling containing the graph as a cross-section. Unfortunately the appropriate flip isn’t always available, but luckily we can set it up without changing the relevant layer. Let $\Delta$ be a zonotopal tiling and $G_k$ be a plabic graph formed by a cross-section of $\Delta$.

Lemma A.2

Suppose M is a possible black (resp. white) trivalent move in $G_k$. Then there exists a finite sequence of flips $(S_1,S_2,\ldots ,S_m)$ in $\Delta$, such that $G_\ell$ is unchanged by each of the first $m-1$ mutations for any $\ell$ at least (resp. at most) k, but the move M occurs on the last mutation.

Proof

Complementing all of the labels of the vertices doesn’t change the structure of the available flips but does change the colors of all of the regions, so it suffices to prove the result when M is a black trivalent move. We proceed by induction on k. When $k \le 2$, there are no legal black trivalent moves, so the claim holds vacuously. Now, the black trivalent move corresponds to two black triangles in $\Sigma _k$, which by Lemma 3.6 creates two white triangles in $\Sigma _{k-1}$, which are forced to border two black regions. If the black regions are triangulated such that a square move is legal using the white triangles, then perform the corresponding flip and we’re done. Otherwise, there exists a sequence of triangulation flips in the black regions which would make the square move legal. By the inductive hypothesis, each of these flips can be done through a finite sequence of mutations, each of which (except the last) leave $G_\ell$ unchanged for all $\ell \ge k-1$. The last flip in each sequence performs a black trivalent move in $\Sigma _{k-1}$, so also leaves $S_k$ unchanged. Therefore we can set up the square move in $\Sigma _{k-1}$ without changing $\Sigma _k$ at all, so the induction is complete. $\square$

In order to properly embed cycles as cyclic zonotopal flips, we also need to match up tilings which share a cross-section. Once we’ve done that, we are ready to prove the result for cyclic connectivities.

Lemma A.3

Let $\Delta$ and $\Delta '$ be two fine zonotopal tilings of Z(n, 3) which are identical on $G_k$ for some fixed k. Then there exists a series of flips, none of which alter $G_k$, which transform $\Delta$ into $\Delta '$.

Proof

It suffices to find such a sequence of moves which make $\Delta$ match $\Delta '$ on $\Sigma _{k+1}$ (without ever changing $G_k$ or any lower layer) and $\Sigma _{k-1}$ (without ever changing $G_k$ or any higher layer). Once this is done we can recursively match all of the layers to transform $\Delta$ into $\Delta '$. By Lemma 3.6, $\Delta$ and $\Delta '$ already agree up to white (resp. black) triangulation on $\Sigma _{k+1}$ (resp. $\Sigma _{k-1})$. By the flip connectivity of triangulations, there exists a sequence of white (resp. black) trivalent flips in $G_{k+1}$ (resp. $G_{k-1}$) which transform $\Delta$ to completely match $\Delta '$ on $\Sigma _{k+1}$ (resp. $\Sigma _{k-1}$). By Lemma A.2, for each of these flips there exists a finite sequence of flips which perform only this move in $G_{k+1}$ (resp. $G_{k-1}$, none of which change $G_\ell$ for any $\ell \le k$ (resp. $\ell \ge k$). Therefore all of these triangulation moves can be performed without ever changing $G_k$ or any lower (resp. higher) layer, as desired. $\square$

Proof of Theorem 4.7 for cyclic permutations

Fix any cyclic permutation $\pi ^: = \pi (n,k)$. Let $\gamma = M_1M_2\cdots M_{m}$ be a loop in $\mathbb {X}_\pi ^:$ connecting plabic graphs $G_k^1,G_k^2,\ldots ,G_k^{m+1} = G_k^1$ with connectivity $\pi (n,k)$. We will construct a loop $Z(\gamma )$ in $\mathbb {Z}_{n,3}$ such that the flips in $Z(\gamma )$ cause exactly the moves $M_1,M_2,\ldots ,M_m$ to occur in $G_k$, in that order. For $i = 0,1,\ldots ,m-1$, there exists $\Delta _i$ whose cross-section at height k is exactly $G_k^{i+1}$. By Lemma 3.5 (if $M_{i+1}$ is a square move) and Lemma A.2 (if $M_{i+1}$ is a black or white trivalent move), there exists a sequence of moves starting from $\Delta _i$ which performs only the move $M_{i+1}$ in $G_k^{i+1}$. The resulting tiling $\Delta _i'$ from this sequence of moves is identical to $\Delta _{i+1}$ at height k, so by Lemma A.3 there exists another sequence flips, none of which cause a move in $G_k$, which turns $\Delta _i'$ into $\Delta _{i+1}$, where $i+1$ is considered modulo m. Concatenating all these sequences of moves results in our loop $Z(\gamma )$ with the desired properties.

By Theorem 2.6, the loop $Z(\gamma )$ is contractible, by moving it across the 2-cells in $\mathbb {Z}_{n,3}$. We will show that these 2-cells correspond to 2-cells in $\mathbb {X}_{\pi (n,k)}$ nicely, in order to contract $\gamma$.

The quadrilaterals in $\mathbb {Z}_{n,3}$ are formed by two commuting flips in Z(n, k), which result in either two moves in separate parts of $G_k$ (a quadrilateral in $\mathbb {X}_{\pi (n,k)})$, one move being performed twice in $G_k$ (an edge in $\mathbb {X}_{\pi (n,k)}$), or no moves in $G_k$ (a point in $\mathbb {X}_{\pi (n,k)}$). In all cases, when $Z(\gamma )$ is moved across a quadrilateral, the image of the quadrilateral in $\mathbb {X}_{\pi (n,k)}$ is a vertex, edge, or 2-cell which $\gamma$ can also be moved across.

The only other 2-cells in $\mathbb {Z}_{n,3}$ are decagons whose vertices correspond to the ten refinements of an instance of Z(5, 3) inside Z(n, 3). Depending on where the plane $x=k$ intersects the copy of Z(5, 3), one of five things could happen in $G_k$ as the ten flips in the decagon are performed (see Fig. 5).

• (1) If $x=k$ does not intersect the copy of Z(5, 3) or only touches the top or bottom vertex, no moves occur in $G_k$ and the image of the decagon in $\mathbb {X}_{\pi (n,k)}$ is a vertex.

• (2) If $x=k$ intersects the copy of Z(5, 3) at relative height 1, then five white trivalent moves occur in a subgraph of $G_k$ with connectivity $\pi {(5,1)}$. The image of the decagon in $\mathbb {X}_{\pi (n,k)}$ is a pentagon.

• (3) If $x=k$ intersects the copy of Z(5, 3) at relative height 2, then five square moves and five white trivalent moves occur in a subgraph of $G_k$ with connectivity $\pi {(5,2)}$. The image of the decagon in $\mathbb {X}_{\pi (n,k)}$ is another decagon.

• (4) If $x=k$ intersects the copy of Z(5, 3) at relative height 3, then five square moves and five black trivalent moves occur in a subgraph of $G_k$ with connectivity $\pi {(5,3)}$. The image of the decagon in $\mathbb {X}_{\pi (n,k)}$ is another decagon.

• (5) If $x=k$ intersects the copy of Z(5, 3) at relative height 4, then five black trivalent moves occur in a subgraph of $G_k$ with connectivity $\pi {(5,4)}$. The image of the decagon in $\mathbb {X}_{\pi (n,k)}$ is a pentagon.

In all cases, when $Z(\gamma )$ is moved across the decagon, the image of the decagon is a vertex or 2-cell in $\mathbb {X}_{\pi (n,k)}$ which $\gamma$ can be moved across.

Finally, let $Z(\gamma )'$ be a deformation of $Z(\gamma )$ by moving it across a 2-cell. We have considered all possible 2-cells in $\mathbb {Z}_{n,3}$ and shown that there always exists a cell in $\mathbb {X}_{\pi (n,k)}$ which $\gamma$ can be moved across to create $\gamma '$ such that $Z(\gamma ') = Z(\gamma )'$. Therefore by contracting $Z(\gamma )$ to a point in $\mathbb {Z}_{n,3}$ step-by-step while adjusting $\gamma$ along the way, $\gamma$ is also contracted to a point.

We note that that the procedure of moving across the 2-cells in fact builds a cellular contraction, cf. Definition 4.8. $\square$

This method of proof cannot be straightforwardly applied to the more general statement of Theorem 4.7. Although a loop of flips for any connectivity $\pi ^:$ could still be included in a loop of zonotopal tiling flips, the contraction of the loop might not stay inside the graph with connectivity $\pi ^:$ (see the second question in Sect. 7).