Orbits in a Graph for Transitive Action

Question

Let $X$ and $Y$ be bipartite graphs over the same set of vertices $V = V_1\cup V_2$.

Let $E(X)$ and $E(Y )$ denote the set of edges of $X$ and $Y$, respectively.

Isomorphisms of $X$ and $Y$ $\textbf{preserve the parts $V_i, i \in \{1, 2\}$ by definition,}$ and, $G \leq \text{Sym}(V_1) \times \text{Sym}(V_2)$.

Consider the case where $G$ acts transitively on each part $V_i, i \in \{1, 2\}$. I am having hard time to comprehend the following line from a document:

Note that to be precise, in this case $B_i$ will be the set of all pairs $(v_1,v_2) \in V$ such that $v_1 \in V_1$ is in the $i^{th}$ orbit.

Questions:

1. Why does $B_i$ depend on $V_1$ only but include vertices of $V_2$ as the pair $(v_1,v_2)$?

2. Does $G$-orbits $B_i$ in $V_1$ created by $G$ induces or create automatically $G$-orbit in $V_2$?

3. What are the $B_i$ in the following graph?

This graph is bipartite,

with parts $V_1=\{1,2,3,4,5,6\}$ and $V_2=\{7,8,9,10,11,12\}$,

and has two orbits, $\{1,2,5,6,7,8,11,12\}$ and $\{3,4,9,10\}$.

Is $B_1\{= (1,7), (2,8), (5,11), (6,12)\}$ or $B_1=\{ (1,8), (2,7), (5,12), (6,11)\}$ or both?

Note that there is no ambiguity in $B_2=\{(3,10),(4,9)\}$.

Answer 1

Here $G \leq \def\Sym{\mathrm{Sym}}\Sym(V_1) \times \Sym(V_2)$, which means that $G$ has an action on $V_1$, some other action on $V_2$, and hence we have an action on $V_1 \times V_2$. Note that if we chop $V_1$ into orbits $E_i$ and $V_2$ into orbits $F_i$ then the orbits of $G$ on $V_1 \times V_2$ will be contained in the products $E_i \times F_j$ (but they could be smaller). Here we are simply letting $B_i = E_i \times V_2$. This is a $G$-invariant set obtained by looking at the action of $G$ on $V_1$ and ignoring $V_2$. Obviously $B_i$ tells you nothing about $V_2$ orbits, because we ignored them completely.

Your example is confused. The bold part of your question says that $V_1$ and $V_2$ are considered to be segregated permanently. Hence it is better to say that the orbits of $G$ (the automorphism group) on $V_1$ are $\{1, 2, 5, 6\}$ and $\{3, 4\}$ (while its orbits on $V_2$ are... similar, but who cares). Hence $B_1 = \{1, 2, 5, 6\} \times V_2$ and $B_2 = \{3, 4\} \times V_2$.