Visualise Gram sets of nonnegative polynomials in the cone of PSD matrices I am currently reading Harnessing Sparsity over the Continuum:
Atomic Norm Minimisation for Super Resolution by Yuejie Chi and Maxime Ferreira Da Costa.
In the box "From Bounded Polynomials to Linear Matrix Inequalities" on page 5 the following plot can be found: 
It aims to show that if a trigonometric polynomial $S$ is larger than another, $R$, everywhere on the unit circle, that for any pair of matrices in their respective Gram sets $G \in \mathcal{G}(R)$, $H \in \mathcal{G}(S)$, the one belonging to the dominating polynomial, $H$, is "more positive semidefinite than the other": $H \succeq G$.
Background
A (hermitian) trigonometric polynomial is
$$
R(z)
= \sum_{k = -n}^{n} r_k z^{-k},
$$
with $r_{-k} = r_{k}^*$.
Let $\psi(z) = [1, z, z^2, \ldots, z^{n - 1}]^{\mathsf{T}}$ and $\Theta_k$ be the Hermitian Toeplitz matrix whose $k$ diagonal is only ones.
A Hermitian matrix $G$ is a Gram matrix associated with $R$ (denoted by $G \in \mathcal{G}(R)$) if
$$
R(z) = \psi(z^{-1})^{\mathsf{T}} G \psi(z).
$$
If $G \in \mathcal{G}(R)$, then $r_k = \text{Tr}(\Theta_k G)$.
A trigonometric polynomial is nonnegative of the unit circle if and only if there exists a positive semidefinite matrix $G$ with $r_k = \text{Tr}(\Theta_k G)$.
My Question
I know that the set of Hermitian positive semidefinite matrices $C$ form a proper pointed (i.e. $C \cap (-C) = \{ 0 \}$) convex cone, so I understand the yellow cone to be an accurate representation of $C$, which is the set from which all Gram matrices come from. But why are the Gram sets $\mathcal{G}$ represented as ellipses (and thus connected sets)?
Is there a map, which projects $C$ onto a proper pointed convex cone in $\mathbb R^2$ such that this representation is valid? What about if we only consider $2 \times 2$ matrices?
 A: 
why are the Gram sets $\mathcal{G}$ represented as ellipses (and thus connected sets)?

I guess the Gram sets are represented as ovals because those are simple and natural representations of subsets of a set, like Venn diagrams, and they usually do not have a corresponding geometrical interpretation.
But since a space $\Bbb H_n$ of all $n\times n$ Hermitian matrices is (an $n^2$ dimensional) linear space over $\Bbb R$,  the sets $\mathcal G(R)$ have a natural geometrical representation. For each $G=\|g_{ij}\|\in\Bbb H_n$ put $$\mathcal R(G)(z)= \psi(z^{-1})^{\mathsf{T}} G \psi(z)=\sum_{1\le i,j\le n} g_{ij}z^{j-i},$$ which is a Hermitian trigonometric polynomial. I guess $\mathcal R(G)$ is non-negative on the unit circle iff $G$ is positive semidefinite, that is $G\in C$. Given a Hermitian trigonometric polynomial $R$, a set $\mathcal G(R)$ consists of Hermitian $n\times n$ matrices $H$ such that $\mathcal R(G)=R$. Since $\mathcal R$ is a linear mapping from $\Bbb H_n$ to a space $\Bbb{Tr}_{n-1}$ of all Hermitian trigonometric polynomials of “degree” $n-1$, a preimage $\mathcal R^{-1}(R)$ of each point $R\in \Bbb{Tr}_{n-1}$ is an affine subspace of $\Bbb H_n$. It has dimension of $\operatorname{dim}\operatorname{ker}\mathcal R=\operatorname{dim} \Bbb H_n-\operatorname{dim} \Bbb{Tr}_{n-1}=(n-1)^2$. Thus a set of all positive semidefinite matrices $G$ such that $\mathcal R(G)=R$ is an intersection of the affine subspace $\mathcal R^{-1}(R)$ with the cone $C$. How this cone looks like, what shape this intersection can have, and whether can it be elliptical? I guess the shape of the cone $C$ can be already studied. Sylvester’s criterion suggests that it can be complicated.

Is there a map, which projects $C$ onto a proper pointed convex cone in $\mathbb R^2$ such that this representation is valid?

Any linear map $P$ from $\Bbb H_n$ to $\Bbb R^2$ “projects” $C$ to a convex cone and keeps affine subspaces, so in this case the intersections are restricted to points, lines, rays, segments or to the whole cone.

What about if we only consider $2 \times 2$ matrices?

Already in this case $G\in C$ iff $g_{11}\ge 0$, $g_{22}\ge 0$, and $g_{11}g_{22}\ge (\operatorname{Re} g_{12})^2+(\operatorname{Im} g_{12})^2$. This is a four-dimensional shape, so it is hard to visualize it. But since $\operatorname{dim}\operatorname{ker}\mathcal R=1$, the intersections $\mathcal R^{-1}(R)\cap C$ are at most one-dimensional.
