I think the following statement on flat-tori is true, but I feel uncertain and am thus posting as a sanity check:
First the convetions I am using:
The $1$-torus is the quotient space $\mathbb{T}^1 := \mathbb{R}/_\mathbb{Z}$. We consider this as a topological space equipped with the quotient topology induced by the absolute value as a norm in $\mathbb{R}$.
For $d \in \mathbb{N}^+$ the $d$-torus is the Cartesian product $\mathbb{T}^d := (\mathbb{T}^1)^d$.
We call a subset $A \subset \mathbb{T}^d$ rectangular if it can canonically be identified with a rectangle in $\mathbb{R}^d$, i.e., if for some $S \subset \mathbb{R}^d$ that is a product of half-open intervals of length $1$ (this is a set of representatives of $\mathbb{T}^d$) the set of representatives $R \subset S$ of representatives of $A$ is the Cartesian product of $d$ intervals.
Now the statement that I want to verify:
Let $A \subset \mathbb{T}^d$. Then $A$ is rectangular if and only if it is the Cartesian product of connected subsets of $\mathbb{T}^1$.
The intuition here is clear from the fact that in $(\mathbb{R}, |\cdot |)$ sets are connected if and only if they are intervals. My statement for the torus seems fine, right?
I have not worked with quotient topologies before, so I feel uneasy with the connectedness in $\mathbb{T}^1$ and how that relates to a "square" choice of representatives. The validity if my statement should boil down to the validity in the case $d=1$, right? Thanks in advance.
It is true.
Let $p : \mathbb R \to T^1$ denote the quotient map. This induces a map $$p^d : \mathbb R^d \to T^d, p^d(x_1, \ldots, x_d) = (p(x_1),\ldots, p(x_n)). $$
Define a rectangular subset of $\mathbb R^d$ to be one of the form $$R =\prod_{i=1}^d J_i$$ with intervals $J_i$ of length $\le 1$ which are required to be half-open if the length is $1$.
Clearly each such $R$ is contained in some set $$S =\prod_{i=1}^d J'_i$$ with half open intervals $J'_i$ of length $1$. Each such $S$ is mapped by $p^d$ bijectively onto $T^d$. But note that $p^d : S \to T^n$ is not a homeomorphism.
Then a rectangular susbet of $T^d$ is one form the form $A = p^d(R)$ with a rectangular $R \subset \mathbb R^d$.
A rectangular $A \subset T^d$ is the Cartesian product of connected subsets of $T^1$.
This is true because $A = \prod_{i=1}^d p(J_i)$. Now use the fact that continuous images of connected sets are connected.
The Cartesian product $\prod_{i=1}^n C_i$ of connected subsets $C_i \subset T^1$ is a rectangular subset of $T^d$.
To see this, it suffices to know that the connected subsets $C_i \subset T^1$ have the form $C_i = P(J_i)$ with intervals $J_i$ occurring as factors of rectangular subsets of $\mathbb R^d$. But this is well-known.