Let me describe the simplest non-trivial case of what I have in mind. Let $V$ be a 2-dimensional $\mathbb{R}$-vector space and fix an isomorphism $V \cong \mathbb{R}^2$, where $\mathbb{R}^2$ is equipped with the standard basis. This will be our starting frame of reference, so to speak. Let $S \subseteq V$ be an arbitrary subset. I am mostly interested in the cases where $S$ is a path, or a loop, or an open subset, but **for the purpose of this question let me give a formulation for the case when $S := U$ is an open subset of the plane.**


Let $I \in \operatorname{M}\_2(\mathbb{R}) \cong \operatorname{End}(V)$ denote the identity matrix, let $p\_I : U \twoheadrightarrow \mathbb{R}$ be the projection onto, say, the *first* coordinate with respect to the standard basis (hence the $I$-notation), and define
$$
U\_I := p\_I(U) \times \mathbb{R}.
$$
(It won't matter if we choose the first or the second projection.) Clearly, $U\_I \supseteq U$ is the best we can recover from only knowing $p\_I(U)$. Now change the basis (coordinate system) via $(A^T)^{-1} \in \operatorname{GL}\_2^+(\mathbb{R})$ and put similarly
$$
U\_A := p\_A(U) \times \mathbb{R},
$$
where $p\_A : U \twoheadrightarrow \mathbb{R}$ is the projection onto the *first* coordinate with respect to the new basis.
Define the *projection hull* (for a lack of a better term) of $U$ to be
$$
\tilde{U} = \bigcap\_{A \in \operatorname{GL}\_2^+(\mathbb{R})} U\_A
$$
Then $\tilde{U} \supseteq U$ is the best we can recover from knowing the 1-dimensional projection of $U$ in every coordinate system.


Of course, one can do a similar construction with $\operatorname{GL}\_2(\mathbb{R})$ and both projections, which produces the same result by rotation invariance of the problem, or one may consider more special types of coordinate changes, i.e. other subgroups of $\operatorname{GL}\_2(\mathbb{R})$.


Here is an explicit description of the set $U\_A$ with respect to the original frame of reference:
$$
U\_A = \bigg\{ \frac{1}{\det(A)} (a\_{11} a\_{22} x\_1 + a\_{12} a\_{22} x\_2 - a\_{12}y, -a\_{11} a\_{21} x\_1 - a\_{12} a\_{21} x\_2 + a\_{11} y) \mid (x\_1,x\_2) \in U, y \in \mathbb{R} \bigg\}
$$
where $a\_{11},a\_{12},a\_{21},a\_{22}$ are the coefficients of $A$. (I sincerely hope I haven't miscalculated or mistyped.)


Here are my questions:


**(Q1)** Is there an accepted name for $\tilde{U}$ in the literature?


**(Q2)** Assuming $U$ is a **bounded simply connected domain**, can we have $\tilde{U} \supsetneqq U$? If yes, is there a different, more tangible characterization of $\tilde{U}$?


*Please feel free to add more appropriate tags.*



Recovering a set from its projections in varying coordinate systems - a projection hull?
$\newcommand{\R}{\mathbb R}\newcommand{\tU}{\tilde U}$Suppose that $U$ is connected. Then all its projections are connected. So, all one-dimensional projections of $U$ will be convex. So, $\tilde U$ will be convex.


Now take any non-convex connected $U$. Then $\tilde U\ne U$ and hence $\tilde{U} \supsetneqq U$.




---


Suppose now that a connected subset $U$ of $\R^2$ is closed and bounded. Then
\begin{equation}
 \tU=\bigcap\_{l\in(\R^2)'}l^{-1}([m\_l,M\_l]),
\end{equation}
where $(\R^2)'$ is the space of all linear functionals on $\R^2$, $m\_l:=\min l(U)$, and $M\_l:=\max l(U)$.


Let us show that
$$\tU=C,$$
the closed convex hull of $U$. (So, is then a new name for $\tU$ needed?)


Clearly $\tU$ contains $U$ and is closed and convex, as the intersection of closed convex sets. So, $C\subseteq\tU$.


To obtain a contradiction, assume that $\tU\not\subseteq C$. So, we have some $x\in\tU\setminus C$. Then $x$ can be separated from $C$ by some $l\_x\in(\R^2)'$ so that $l\_x(x)<\min l\_x(C)\le\min l\_x(U)=m\_{l\_x}$. Then $x\notin l\_x^{-1}([m\_{l\_x},M\_{l\_x}])$, which contradicts the condition $x\in\tU$. So, $\tU\subseteq C\subseteq\tU$. $\quad\Box$


