Recovering a set from its projections in varying coordinate systems - a projection hull?

Question

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.

Answer 1

$\newcommand{\R}{\mathbb R}\newcommand{\tU}{\tilde U}$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 $\tilde{U} \supsetneqq U$.

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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, $\tU\subseteq C\subseteq\tU$. $\quad\Box$

Answer 2

Your definition is a special case of the following. Choose some set $S$ whose elements are subsets of $V$. Say that a subset of $V$ is $S$-convex if it is the intersection of some elements of $S$. The $S$-hull of a set $U$ is the smallest $S$-convex set containing $U$, i.e. the intersection of all the elements of $S$ that contain $U$.

For you the elements of $S$ are the sets $\varphi^{-1}(\mathbb R\backslash \{a\})$ where $\varphi$ is a non-zero element of $V^\ast$ and $a\in \mathbb R$, i.e. the complements of the affine hyperplanes in $V$.

If instead we choose $S$ to consist of the closed half-spaces $\varphi^{-1}([a,\infty))$, then $S$-convex means convex and closed. Or if we choose $S$ to consist of the open half-spaces then $S$-convex means something else, stronger than convex; let's call it "strongly convex".

As observed by Pinelis, if $U$ is connected and closed and bounded then your $\tilde U$ is the (closed) convex hull. More generally if $U$ is connected then $\tilde U$ is the strongly convex hull.

More generally, if $U$ cannot be separated into two nonempty pieces by any hyperplane then $\tilde U$ is the strongly convex hull.

$U$ is equal to $\tilde U$ if $U$ is empty, or if $U$ is a nonempty strongly convex set, or if $U$ is the union of two sets of that kind that can be separated by some hyperplane, or if $U$ is the union of three sets of that kind such that each of them can be separated from the other two by a hyperplane, "and so on".