Suppose my domain is $\mathbb{R}^3$ where I have:
- a set of points $A = \left\lbrace x_i \right\rbrace_{i=1}^{n_1} $ and $n_1 > 3$.
- $A$ is contained in a plane (2-dimensional) $P_1$. i.e. $A \subset P_1 \subset \mathbb{R}^3$.
- There is no basis $ \left( v_1, v_2, v_3 \right)$ for $A$ $\textit{and}$ a lattice $\Lambda \subset P_1$
I want to construct a mapping, $f: \mathbb{R}^3 \mapsto B$, such that:
- There exists a basis $ \left( w_1, w_2, w_3 \right)$ for $f(A)$ $\textit{and}$ $f(A*) \subset f(P_1)$ where $f(A) \cup f(A*)$ is a lattice in $B$. That is to say $\left(f(A) \cup f(A*) \right) \subset f(P_1) \subset B$
- I can derive/define $(w_1, w_2, w_3)$ algebraically. i.e. $\begin{array}{rcl} w_1&=& \text{equation}_1 \in B\\ w_2&=& \text{equation}_2 \in B\\ w_3&=& \text{equation}_3 \in B\\ \end{array}$
- $f$ is bijective.
How do I do this? I figure $f$ would need to be a piece-wise shearing function where it takes $A$ as input and then $f$ finds some other set in $\mathbb{R}^3$ and the the mapped union of those sets forms a lattice in $B$. Although I am not sure.