$7$-location, weak systolicity and isoperimetry

Let $W=W_{1}\cup W_{2}$ be a $(k,l)$-dwheel subcomplex of $X$ of boundary length at most $7$ such that $W_{1}$ and $W_{2}$ are full subcomplexes. By $5$-largeness of $X$, we have $k,l\geq 5$. Then, since $W$ has boundary length at most $7$, either $W$ is a planar or nonplanar $(5,5)$-dwheel or $W$ is a planar $(5,6)$-dwheel. By Remark 2.2 it suffices to show that $W$ is contained in the $1$-ball centered at a vertex of $X$.

The above three claims together imply that $X$ is $7$-located. ∎

Since $D$ is a simplicial complex and $v$ is an interior vertex, it is the center of a full wheel $W_{v}$. Similarly, we have a full wheel $W_{w}$ in $D$ with center $w$ and $W_{v}\cup W_{w}$ is a planar dwheel of boundary length $\deg(v)+\deg(w)-4$. Since $D$ is planar, the dwheel $W_{v}\cup W_{w}$ cannot be contained in a $1$-ball. Thus $\deg(v)+\deg(w)-4\geq 8$. ∎

If $W_{1}$ and $W_{2}$ intersect in their interiors then they must share a triangle $\sigma$. Thus the central vertices $v_{1}$ and $v_{2}$ of $W_{1}$ and $W_{2}$ are adjacent. But then, by Lemma 4.1, we have $|\partial W_{1}|+|\partial W_{2}|=\deg(v_{1})+\deg(v_{2})\geq 12$ so that one of the wheels must have boundary length at least $6$. ∎

Since $D$ is simply connected, it suffices to show that the sum of the angles of the corners of triangles incident to any interior vertex $v$ of $D$ is at least $2\pi$ [4, Theorem II.5.4 and Lemma II.5.6]. Because $D$ is flag, it has no interior vertices of degree $3$. Any interior vertex of degree $4$ or $5$ is the central vertex of a flattened wheel and so has angle sum exactly $2\pi$. Away from flattened wheel centers, corner angles of triangles are either $\frac{\pi}{4}$, $\frac{3\pi}{10}$ or $\frac{\pi}{3}$. Thus any interior vertex of degree at least $8$ has angle sum at least $2\pi$.

Thus we may assume that $v$ is an interior vertex of degree $6$ or $7$. If $v$ is not incident to any flattened wheel then its incident triangles are all regular Euclidean and so have corner angle $\frac{\pi}{3}$ so that $v$ has angle sum $\frac{6\pi}{3}=2\pi$ or $\frac{7\pi}{3}>2\pi$. So we may assume that $v$ is incident to a flattened wheel. Since $\deg(v)\leq 7$, by Lemma 4.1, any central vertex of a flattened wheel $W$ incident to $v$ has degree $5$. Then $v$ is incident to a vertex of degree $5$ and any triangle corner incident to $v$ has angle at least $\frac{3\pi}{10}$. So, by Lemma 4.1, we have $\deg(v)=7$ and so $v$ has angle sum at least $\frac{21\pi}{10}>2\pi$. ∎

The area of a minimal disc diagram for a loop of length $n$ in a $\operatorname{CAT}(0)$ space is bounded by the area of a circle of circumference $n$ in the Euclidean plane [4, Theorem III.2.17]. It follows that if $D$ is a $7$-located disc of boundary length $n$, it has area at most $\frac{n^{2}}{4\pi}$ when endowed with the flattened wheel metric. Each triangle of $D$ has area at least $\frac{1}{4}$ so the number of triangles of $D$ is at most $f(n)=\frac{n^{2}}{\pi}$. ∎

The subdisc bounded by $(u,a,v,b)$ has at least two triangles. We delete this subdisc. We glue $u$ to $v$, the edge $\langle a,u\rangle$ to the edge $\langle a,v\rangle$, and the edge $\langle b,u\rangle$ to the edge $\langle b,v\rangle$. Thus we get a new disc $D^{\prime}$. See Figure 7. Let $f^{\prime}$ be the map induced by the gluing. This is well defined since $f(u)=f(v)$, $f(\langle a,u\rangle)=f(\langle a,v\rangle)$ and $f(\langle b,u\rangle)=f(\langle b,v\rangle)$. Thus $(D^{\prime},f^{\prime})$ is a disc diagram for $\gamma$ of lesser area than $D$.

∎

The subdisc bounded by $\alpha$ has at least three $2$-simplices. We delete this subdisc. We obtain a new disc diagram $D^{\prime}$ by glueing $u$ to $v$, gluing the edge $\langle a,u\rangle$ to the edge $\langle a,v\rangle$ and gluing a $2$-simplex $\langle u,b,c\rangle$. See Figure 8. Let $f^{\prime}$ be the map induced by the gluing. This is well defined since $f(u)=f(v)$, $f(\langle a,u\rangle)=f(\langle a,v\rangle)$ and, by flagness, $\langle f(u),f(b),f(c)\rangle$ is a triangle in $X$. Thus $(D^{\prime},f^{\prime})$ is a disc diagram for $\gamma$ of lesser area than $D$.

∎

The subdisc bounded by $\alpha$ has at least four $2$-simplices. We obtain a new disc diagram $D^{\prime}$ by deleting the subdisc bounded by $\alpha$, gluing $u$ to $v$ and then gluing in two triangles $\langle a,b,u\rangle$ and $\langle u,c,d\rangle$. See Figure 9. Let $f^{\prime}$ be the map induced by gluing. This is well defined since $f(u)=f(v)$ and, by flagness, $\langle f(a),f(b),f(u)\rangle$, $\langle f(u),f(c),f(d)\rangle$ are triangles in $X$. Hence $(D^{\prime},f^{\prime})$ is a disc diagram for $\gamma$ of lesser area than $D$.

∎

Suppose $W$ is a $4$-wheel in $D$. If the restriction $f|_{W}$ is not injective then some pair of antipodal vertices $u,v\in\partial W$ have a common image $f(u)=f(v)$. This contradicts minimality by Lemma 5.2.

If the restriction $f|_{W}$ is injective then, since $X$ is locally $5$-large, a pair of antipodal vertices $u,v\in\partial W$ have images $f(u)$ and $f(v)$ that are joined by an edge $e$ in $X$. Since $X$ is flag, the image $f(\partial W)$ along with $e$ span two triangles so that $f(\partial W)$ has a disc diagram of area $2$. Thus we may remove the interior of $W$ in $D$ and replace it with an edge joining $u$ and $v$ and a pair of triangles to obtain a disc diagram for $\gamma$ of lesser area, contradicting minimality of $D$. ∎

It follows from Lemma 5.3 that $f|_{W}$ is injective. Then $f(W)$ is a $5$-wheel in $X$. We need only verify that it is full. If not then some pair of vertices $u,v\in\partial W$ that are not adjacent in $\partial W$ have images $f(u)$ and $f(v)$ that are joined by an edge $e$. Since $X$ is flag, the edge $e$ spans a triangle with each of $f(u)$ and $f(v)$. Let $a$ be the unique vertex of $\partial W$ adjacent to both $u$ and $v$. Let $c$ be the central vertex of $W$. Then $(u,a,v,c)$ is a $4$-cycle bounding a subdisc of $D$ with two triangles. We delete this subdisc, join $u$ and $v$ by an edge $e^{\prime}$ and add triangles $\langle a,u,v\rangle$ and $\langle c,u,v\rangle$ to obtain a new disc diagram $D^{\prime}$ for $\gamma$ of the same (minimal) area. In particular, $D^{\prime}$ is a minimal area disc diagram. But the triangle $\langle c,u,v\rangle$ along with the three remaining original triangles of $W$ form a $4$-wheel, which contradicts minimality by Lemma 5.5. ∎

For the sake of deriving a contradiction, we assume that $f|_{W}$ is injective. Since $D$ is planar, the dwheel $W$ is also planar. We consider first the case where $W$ is a planar $(5,5)$-dwheel. By Lemma 5.6, the $(5,5)$-dwheel $f(W)$ has full wheels. Then, since $X$ is $7$-located, the image $f(W)$ is contained in the $1$-ball centered at a vertex $v$ of $X$. Since $X$ is flag, it follows that $f(\partial W)$ has a disc diagram with at most $6$ triangles, contradicting minimality of $D$.

We now consider the case where $W$ is a planar $(5,6)$-dwheel. Let $W=W_{1}\cup W_{2}=(w_{6};v_{1},v_{2},v_{3},v_{4},v_{5})\cup(v_{2};w_{1}=v_{1},w_{2},w_{3},$ $w_{4},w_{5}=v_{3},w_{6})$; see Figure 10. By Lemma 5.6, the image $f(W_{1})$ is a full $5$-wheel. If the $6$-wheel $f(W_{2})$ is also a full then we can argue as in the $(5,5)$-dwheel case that $f(\partial W)$ has a disc diagram with at most $7$ triangles, which leads to a contradiction with the minimality of $D$. Thus we need only consider the case where $f(W_{2})$ is not full. Then some pair of vertices $u,v\in\partial W_{2}$ that are not adjacent in $\partial W_{2}$ have images $f(u)$ and $f(v)$ that are adjacent.

Thus the additional edges in the full subcomplex induced on the $6$-wheel $f(W_{2})$ are either $\langle f(w_{6}),f(w_{2})\rangle$ or $\langle f(w_{6}),f(w_{4})\rangle$ or both. If just one of these is present then $f(W)$ spans a non-planar $(5,5)$-dwheel with full wheels so that $\partial W$ has a disc diagram with at most $7$ triangles, contradicting minimality of $(D,f)$. If, on the other hand, both edges are present then we perform two disc diagram surgeries as in the proof of Lemma 5.5 transforming $W$ from a $(5,6)$-dwheel to a $(7,4)$-dwheel. The resulting disc diagram has the same (minimal) area and yet contains a $4$-wheel, contradicting Lemma 5.5. ∎