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##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 $$ \mathcal K=\Big\{x\in V\Big|\exists y\in\mathbb{R}^s~\mathrm{such~that}~\sum_{i=1}^r x_i X_i+\sum_{j=1}^s y_j Y_j\geq0\Big\}, $$$$ \mathcal K=\Big\{x\in V\ \Big|\ \exists y\in\mathbb{R}^s~\mathrm{such~that}~\sum_{i=1}^r x_i X_i+\sum_{j=1}^s y_j Y_j\geq0\Big\}, $$ 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\}$$\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 $$ \mathcal C=\Big\{a\in V\Big|\exists c\in\mathbb{R}^t~\mathrm{such~that}~\sum_{i=1}^r a_i \tilde X_i+\sum_{k=1}^t c_k \tilde Z_k\geq0\Big\}. $$$$ \mathcal C=\Big\{a\in V\ \Big|\ \exists c\in\mathbb{R}^t~\mathrm{such~that}~\sum_{i=1}^r a_i \tilde X_i+\sum_{k=1}^t c_k \tilde Z_k\geq0\Big\}. $$ 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.

##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.

##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.

added partial answer to include new insights and highlight the remaining unknowns
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##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 $\mathcal K^*=\mathcal C$?

Incidentally, I do not know what to make[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.

##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 $\mathcal K^*=\mathcal C$?

Incidentally, I do not know what to make of Assumption 2.

##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.

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What is the dual of an semidefinitely representable (SDR) cone?

##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 $\mathcal K^*=\mathcal C$?

Incidentally, I do not know what to make of Assumption 2.