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Consider a lattice $\mathcal{L} = \mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_l$ in $\mathbb{R}^n$ and let $S_0$ be the set of edges of the fundamental unit of $\mathcal{L}$. We call a region $X$ lattice path connected iff for every two lattice points $P$ and $Q$ there exists a path of lattice points $P = M_1, M_2, \ldots, M_{t - 1}, M_t = Q$ with $M_i = M_{i-1} + u_i$ and $u_i \in S_0$ and $M_i \in X$.

Let $P \subset \langle v_1, \ldots, v_l \rangle$ be a convex $k$ dimensional polytope. Suppose the supporting linear space of $P$ has spanning set $\mathbf{S}$ which is a subset of $S_0$ and moreover every codimension $1$ face has a spanning set which is a subset of $\mathbf{S}$.

Is it true that $P$ is lattice path connected ? This seems very intuitive as one would just follows the direction of the codimension $1$ faces to connect the two points.

Still, I'm curious to see if such a result is known.

Here are two examples in $\mathbb{R}^2$.

1. Take $\mathcal{L}$ to be the standard lattice. Then the polytopes in question are just horizontal segments, vertical segments and rectangles which are trivially $\mathcal{L}$ - connected.

2. Take $\mathcal{L}$ to be the triangular matrix whose fundamental unit is the equilateral triangle with vertices $(0,0), (1,0), e^{\frac{i\pi}{3}}$. Here there are more classes of polytopes whose edges are parallel to the edges of the fundamental unit. Even so, a case by case analysis shows that the result still holds true.

Consider a lattice $\mathcal{L} = \mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_l$ in $\mathbb{R}^n$ and let $S_0$ be the set of edges of the fundamental unit of $\mathcal{L}$. We call a region $X$ lattice path connected iff for every two lattice points $P$ and $Q$ there exists a path of lattice points $P = M_1, M_2, \ldots, M_{t - 1}, M_t = Q$ with $M_i = M_{i-1} + u_i$ and $u_i \in S_0$ and $M_i \in X$.

Let $P \subset \langle v_1, \ldots, v_l \rangle$ be a convex $k$ dimensional polytope. Suppose the supporting linear space of $P$ has spanning set $\mathbf{S}$ which is a subset of $S_0$ and moreover every codimension $1$ face has a spanning set which is a subset of $\mathbf{S}$.

Is it true that $P$ is lattice path connected ? This seems very intuitive as one would just follows the direction of the codimension $1$ faces to connect the two points.

Still, I'm curious to see if such a result is known.

Here are two examples in $\mathbb{R}^2$.

1. Take $\mathcal{L}$ to be the standard lattice. Then the polytopes in question are just horizontal segments, vertical segments and rectangles which are trivially $\mathcal{L}$ - connected.

2. Take $\mathcal{L}$ to be the triangular matrix whose fundamental unit is the equilateral triangle with vertices $(0,0), (1,0), e^{\frac{i\pi}{3}}$. Here there are more classes of polytopes whose edges are parallel to the edges of the fundamental unit. Even so, a case by case analysis shows that the result still holds true.

EDITED. (formalism of lattices and fundamental units)

I'll try a formal definition. First let's consider a lattice $\mathcal{L}$ just a set of points given by sum direct sum $\mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_n \subset \mathbb{R}^n$. So far $\mathcal{L}$ is just a set of points. There are "no directions" attached to it. There is no notion of "adjacency". There is no "repeating pattern".

One could of course define such notions by just using the $v_1, \ldots, v_n$ coordinates. Then the repeating pattern would be the convex polytope with vertices in $0$, the $v_i$s, the $v_i + v_j$s and so on. But then morally, one simply recovers the "standard" lattice $\mathbb{Z}^n$ with the "standard" directions.

However there are of courses lattices that have the same set of points $\mathcal{L}$ but the directions and repeating pattern distinct. How do we morally define them as "different lattices".

Well $v_1, \ldots, v_n$ are just lattice points that happen to form a basis for $\mathcal{L}$ but there are of courses other spanning sets of lattice points in $\mathcal{L}$.

A fundamental unit for $\mathcal{L}$ is a finite set of lattice points $\{u_1, \ldots, u_k\}$ of $\mathcal{L}$ that form a convex polytope $K$ with edges $k_1, \ldots, k_s$ such that $\mathcal{L} = \mathbb{Z}k_1 + \ldots + \mathbb{Z}k_s$.

So for me a lattice is really the initial data $\mathcal{L}, K$.

Consider a lattice $\mathcal{L} = \mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_l$ in $\mathbb{R}^n$. We call a region $X$ lattice path connected iff for every two lattice points $P$ and $Q$ there exists a path of lattice points $P = M_1, M_2, \ldots, M_{t - 1}, M_t = Q$ with $M_i = M_{i-1} + u_i$ and $u_i \in \{v_1,\ldots, v_l\}$ and $M_i \in X$.

Let $P \subset \langle v_1, \ldots, v_l \rangle$ be a convex $k$ dimensional polytope. Suppose the supporting linear space of $P$ has spanning set $\mathbf{S}$ which is a subset of $\{v_1, \ldots, v_l\}$ and moreover every codimension $1$ face has a spanning set which is a subset of $\mathbf{S}$.

Is it true that $P$ is lattice path connected ? This seems very intuitive as one would just follows the direction of the codimension $1$ faces to connect the two points.

Still, I'm curious to see if such a result is known.

Consider a lattice $\mathcal{L} = \mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_l$ in $\mathbb{R}^n$ and let $S_0$ be the set of edges of the fundamental unit of $\mathcal{L}$. We call a region $X$ lattice path connected iff for every two lattice points $P$ and $Q$ there exists a path of lattice points $P = M_1, M_2, \ldots, M_{t - 1}, M_t = Q$ with $M_i = M_{i-1} + u_i$ and $u_i \in S_0$ and $M_i \in X$.

Let $P \subset \langle v_1, \ldots, v_l \rangle$ be a convex $k$ dimensional polytope. Suppose the supporting linear space of $P$ has spanning set $\mathbf{S}$ which is a subset of $S_0$ and moreover every codimension $1$ face has a spanning set which is a subset of $\mathbf{S}$.

Is it true that $P$ is lattice path connected ? This seems very intuitive as one would just follows the direction of the codimension $1$ faces to connect the two points.

Still, I'm curious to see if such a result is known.

Here are two examples in $\mathbb{R}^2$.

1. Take $\mathcal{L}$ to be the standard lattice. Then the polytopes in question are just horizontal segments, vertical segments and rectangles which are trivially $\mathcal{L}$ - connected.

2. Take $\mathcal{L}$ to be the triangular matrix whose fundamental unit is the equilateral triangle with vertices $(0,0), (1,0), e^{\frac{i\pi}{3}}$. Here there are more classes of polytopes whose edges are parallel to the edges of the fundamental unit. Even so, a case by case analysis shows that the result still holds true.

Consider a lattice $\mathcal{L} = \mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_l$ in $\mathbb{R}^n$. We call a region $X$ lattice path connected iff for every two lattice points $P$ and $Q$ there exists a path of lattice points $P = M_1, M_2, \ldots, M_{t - 1}, M_t = Q$ with $M_i = M_{i-1} + u_i$ and $u_i \in \{v_1,\ldots, v_l\}$.

Let $P \subset \langle v_1, \ldots, v_l \rangle$ be a convex $k$ dimensional polytope. Suppose the supporting linear space of $P$ has spanning set $\mathbf{S}$ which is a subset of $\{v_1, \ldots, v_l\}$ and moreover every codimension $1$ face has a spanning set which is a subset of $\mathbf{S}$.

Is it true that $P$ is lattice path connected ? This seems very intuitive as one would just follows the direction of the codimension $1$ faces to connect the two points.

Still, I'm curious to see if such a result is known.

Consider a lattice $\mathcal{L} = \mathbb{Z}v_1 \oplus \ldots \oplus \mathbb{Z}v_l$ in $\mathbb{R}^n$. We call a region $X$ lattice path connected iff for every two lattice points $P$ and $Q$ there exists a path of lattice points $P = M_1, M_2, \ldots, M_{t - 1}, M_t = Q$ with $M_i = M_{i-1} + u_i$ and $u_i \in \{v_1,\ldots, v_l\}$ and $M_i \in X$.

Let $P \subset \langle v_1, \ldots, v_l \rangle$ be a convex $k$ dimensional polytope. Suppose the supporting linear space of $P$ has spanning set $\mathbf{S}$ which is a subset of $\{v_1, \ldots, v_l\}$ and moreover every codimension $1$ face has a spanning set which is a subset of $\mathbf{S}$.

Is it true that $P$ is lattice path connected ? This seems very intuitive as one would just follows the direction of the codimension $1$ faces to connect the two points.

Still, I'm curious to see if such a result is known.