## The Question

Let $V\simeq \mathbb{R}^r$ be an $r$-dimensional vector space with the usual Euclidean inner product.

Let $\mathcal K\subset V$ be a cone defined as
$$
\mathcal K=\Big\{x\in V\ \Big|\ \exists y\in\mathbb{R}^s~\mathrm{such~that}~\sum_{i=1}^r x_i X_i+\sum_{j=1}^s y_j Y_j\geq0\Big\},
$$
where $\{X_i\}_{1\leq i\leq r},\{Y_j\}_{1\leq j\leq s}\in B(H)^{sa}$ are self-adjoint operators acting on some finite-dimensional Hilbert space $H$. These cones are called *semidefinite representable* (SDR). They are projections of spectrahedra and have all the nice properties of the latter, plus some others. Some good reference on these beasts would be great.

The dual of $\mathcal K$ is defined as $\mathcal K^*=\{x'\in V\ |\ x\cdot x'\geq0~\forall x\in \mathcal K\}$.

Question:Is $\mathcal K^*$ an SDR cone? What are the corresponding operators?

In particular I am interested in an answer presented in the same form, *i.e.* some set of self-adjoint operators defining $\mathcal K^*$. However, but I'm not sure it's even possible.

Of course a general solution would be great, but I can settle by making a few assumptions.

**Assumption 0:** Without loss of generality, $\{Y_j\}$ can be assumed to be linearly independent.

**Assumption 1:** The cone $\mathcal K$ is *pointed* or *salient*; $\mathcal K\cap(-\mathcal K)=\{0\}$.

**Assumption 2:** The cone $\mathcal K$ is generating; $V=\mathcal K-\mathcal K$.

## Some partial insights

Assumption 1 already tells us some things about $\{X_i\}$ and $\{Y_j\}$. A few immediate consequences are

**Fact 1:**$\{X_i\}$ are linearly independent. If they weren't $\mathcal K$ would contain some entire subspace of $V$, thus wouldn't be pointed.**Fact 2:**$\mathrm{span}\{Y_i\}$ does not contain an order unit for $\mathrm{span}\{X_i\}$ or any subspace thereof. Same reason as above.

Fact 3:$\{X_i,Y_j\}$ are linearly independent.

*Proof*: Suppose they are not. Take
$$\sum_i \alpha_i X_i+\sum_j \beta_j Y_j=0$$
with some nonzero coefficients, so that $\sum_i \alpha_i X_i=-\sum_j \beta_j Y_j$ is nonzero because $\{X_i\}$ and $\{Y_j\}$ are linearly independent. Then $\alpha\neq0$. For any $\lambda\in\mathbb{R}$, $\lambda\alpha\in\mathcal K$. Thus $\mathcal K$ is not pointed. $\blacksquare$

From **Fact 3** we can complete the set of operators to $\{X_i,Y_j,Z_k\}_{(1\leq i\leq r, 1\leq j\leq s, 1\leq k\leq t)}$ to form a basis of $B(H)^{sa}$. In addition, define the conjugate basis with respect to the Hilbert-Schmidt inner product
\begin{align}
\begin{array}{ccc}
\mathrm{tr}[X_i \tilde X_{i'}]=\delta_{ii'}, &\mathrm{tr}[X_i \tilde Y_{j'}]=0&\mathrm{tr}[X_i \tilde Z_{k'}]=0\\\
\mathrm{tr}[Y_j \tilde X_{i'}]=0, &\mathrm{tr}[Y_j \tilde Y_{j'}]=\delta_{jj'}&\mathrm{tr}[Y_j \tilde Z_{k'}]=0\\\
\mathrm{tr}[Z_k \tilde X_{i'}]=0, &\mathrm{tr}[Z_k \tilde Y_{j'}]=0&\mathrm{tr}[Z_k \tilde Z_{k'}]=\delta_{kk'}\\\
\end{array}
\end{align}

Partial answer:With the conjugate basis one can define $$ \mathcal C=\Big\{a\in V\ \Big|\ \exists c\in\mathbb{R}^t~\mathrm{such~that}~\sum_{i=1}^r a_i \tilde X_i+\sum_{k=1}^t c_k \tilde Z_k\geq0\Big\}. $$ and show that $\mathcal C\subseteq\mathcal K^*$.

*Proof:* Let $a\in\mathcal C$. Then there is $c\in\mathbb{R}^t$ such that
$$
\mathcal A=\sum_{i}a_{i} \tilde X_{i}+\sum_{k}c_{k} \tilde Z_k\geq0.
$$
For any $x\in\mathcal K$, there is $y\in\mathbb{R}^s$ such that
$$
\mathcal X=\sum_{i}x_i X_i+\sum_{j}y_j Y_j\geq0
$$
thus the inner product $x\cdot a=\mathrm{tr}[\mathcal X\mathcal A]\geq0$. Therefore,
$$
a\in\mathcal C~~\Rightarrow~~ x\cdot a\geq0~\forall x\in\mathcal K~~\Rightarrow~~ a\in\mathcal K^*.
$$

Alternative Question:Under what conditions it is true that $\mathcal K^*=\mathcal C$ ?

[Edit] Partial answer: A sufficient condition for equality is that $\mathrm{span}\{X_i,Y_j\}$ intersects the interior of the positive semidefinite cone, as Noah explains in his answer. Whether Assumption 2 guarantees this is an open question (for me). Interestingly, for spectrahedra, this is always the case.