On Polynomial Characterization of Projection area of semidefinite matrices Suppose $m,n$ are positive integers.
$D$ denotes the set of $n\times n$ complex semidefinite positive matrices with unit trace.
$A_1,\cdots,A_m$ are $n\times n$ Hermitians.
We are interested in the projection of $D$ onto $A_1,\cdots,A_m$ in the following way,
$$
S=\{(tr(A_1N),\cdots,tr(A_m N)):N\in D\}
$$
It is clear $S$ is convex compact set. How to characterize its boundary.
Can it be characterized by polynomial inequalities of $\mathbb{R}^m$. What is the minimal degree of those polynomials? Is $m$ enough? How many polynomials are needed?
In particular, for the case $m=4$, what do we know? Can we compute the boundary efficiently?
 A: One can construct a "dual" $S^*$ to $S$, for which a lot is known. 
Indeed, this is related to  "convex algebraic geometry", namely, an approach to describe convex semialgebraic sets as sets $S^*$ of solutions of the linear matrix inequality in $y:=(y_1,\dots, y_m)$: 
$$
y_1A_1+\dots+y_mA_m\succeq A_0,\quad A_k=A_k^\top\in \mathbb{R}^{n\times n}, \tag{*}
$$
where $T\succeq U$ means $T-U$ is positive semidefinite. There is a wealth of material on this, see e.g. Victor Vinnikov's survey.
In loc.cit. you can find footnote on p.1, explaining that there is not much generality added by considering Hermitian $A_k$ rather than real symmetric $A_k$. 
On (*) one can consider a semindefinite optimisation problem (SDP):
$$
\max b^\top y, \quad y \text{ satisfies (*)}, \quad b:=(b_1,\dots,b_m)\in\mathbb{R}^m.\tag{P}
$$
It has a naturally defined dual:
$$
\min_{N\succeq 0}\ tr(A_0 N), \quad tr(A_kN)=b_k,\quad 1\leq k\leq m.\tag{D}
$$
This is related to your set $S$ by setting $A_0=0$ and $A_1=I$.
That is, a point $b$ is in $S$ if and only if (D) has a solution.
The duality of (P) and (D) comes from the computation 
$$
tr(A_0N)=tr((-T+\sum_k y_k A_k)N)=\sum_k y_k tr(A_k N)- tr(TN)=b^\top y-tr(TN), \quad T\succeq 0.
$$
Save some pathological cases, at the optimum of (D) one has $TN=0$, and the optima of (D) and (P) are equal.
Building upon this, one can relate $S$ and $S^*$: for each $b\in S$ there exists unique $y\in S^*$ such that $b^\top y=0$.

A lot is understood about the boundary of $S^*$, see Theorem 2.1.
Basically, it is a component in an algebraic set defined by $\det(-A_0+\sum_k y_kA_k)=0$. 
An interesting nontrivial example comes from determinantal representations of certain cubic curves, for which one has one compact convex component---this is the boundary of $S^*$, and one unbounded branch.
