# Convex subcomplexes of CAT(0) cubical complexes

Is the following statement true? If so, can anyone provide a reference?

Let $X$ be a CAT(0) cubical complex, and let $Y$ be a connected subcomplex of $X$. Then the following are equivalent:

1. $Y$ is convex in $X$.
2. For every cube $C$ in $X$, the intersection $C\cap Y$ is a face of $C$.

Here the empty set is considered a face of every cube, and each cube is a face of itself.

Clearly (1) implies (2), and in all the examples I can think of (2) also implies (1).

Yes, it is true.

You condition (2) implies that $X$ is locally convex; this can be proved the same way as the flag condition for $\mathrm{CAT}$-ness.

It remains to note that for $\mathrm{CAT}$-spaces local convexity + connectedness implies convexity.

In addition to Anton Petrunin's answer, I would like to mention that a more combinatorial argument is possible. Indeed, in a CAT(0) cube complex $$X$$, a full subcomplex $$Y$$ (i.e. a subcomplex which contains a cube if it already contains its vertices) is convex with respect to the CAT(0) metric if and only if it is convex with respect to the combinatorial metric; see Theorem 2.13 in Haglund's article Finite index subgroups of graph products. In other words, it suffices to show that the one-skeleton of $$Y$$ is convex in $$X$$ endowed with the graph metric. Now, the proposition is a straightforward consequence of the following two observations:

Proposition 1: Let $$X$$ be a CAT(0) cube complex and $$\alpha,\beta$$ two combinatorial paths with the same endpoints. Assume $$\beta$$ is a geodesic. There exists a sequence of combinatorial paths $$\gamma_1=\alpha, \gamma_2,\ldots, \gamma_{n-1},\gamma_n=\beta$$ such that, for every $$1 \leq i \leq n-1$$, $$\gamma_{i+1}$$ is obtained from $$\gamma_i$$ by removing a backtrack or flipping a square.

Here, by flipping a square, I mean replacing two consecutive edges of a square by the other two consecutive edges.

Proposition 2: Let $$X$$ be a CAT(0) cube complex and $$\alpha,\beta$$ two combinatorial geodesics with the same endpoints. There exists a sequence of combinatorial paths $$\gamma_1=\alpha, \gamma_2,\ldots, \gamma_{n-1},\gamma_n=\beta$$ such that, for every $$1 \leq i \leq n-1$$, $$\gamma_{i+1}$$ is obtained from $$\gamma_i$$ by flipping a square.

Proposition 2 is Theorem 4.6 in Sageev's thesis Ends of group pairs and non-positively curved cube complexes, and Proposition 1 can be proved similarly. We deduce that:

Let $$X$$ be a CAT(0) cube complex and $$Y \subset X$$ a full subcomplex. Then $$Y$$ is convex if and only if it is connected and every square containing two consecutive sides in $$Y$$ lies entirely in $$Y$$.

Fix two vertices $$a,b \in Y$$. Because $$Y$$ is connected, there exists a combinatorial path in $$Y$$ between $$a,b$$. As a consequence of Proposition 1, $$\alpha$$ can be turned into a combinatorial geodesic $$\gamma$$ in $$Y$$. As a consequence of Proposition 2, every geodesic between $$a$$ and $$b$$ can be obtained from $$\gamma$$ by flipping squares, and so must lie in $$Y$$. We conclude that $$Y$$ is (combinatorially) convex.