PhyslibAlpha.ProbabilisticTheory.Representation.Covariance.Basic

For a probabilistic theory $E$, this definition provides the action of a symmetry $\phi \in \text{Symmetry}(E)$ on an effect $e \in \text{Effect}(E)$. The transformed effect $\phi \cdot e$ is obtained by applying the symmetry's underlying linear transformation to the effect $e$.

For a probabilistic theory $E$, a symmetry $\phi \in \text{Symmetry}(E)$, and an effect $e \in \text{Effect}(E)$, the action of the symmetry on the effect, denoted as $\phi \cdot e$, is equal to the application of the symmetry's underlying transformation $\phi$ to the effect $e$ as an element of $E$. In mathematical notation, this is expressed as $(\phi \cdot e) = \phi(e)$.

For a probabilistic theory $E$, the group of symmetries $\text{Symmetry}(E)$ forms a group action on the set of effects $\text{Effect}(E)$. This structure ensures that the identity symmetry acts as the identity transformation on effects and that the action is compatible with the group multiplication (composition) of symmetries, such that $(\phi \cdot \psi) \cdot e = \phi \cdot (\psi \cdot e)$ for all $\phi, \psi \in \text{Symmetry}(E)$ and $e \in \text{Effect}(E)$.

Let $\Omega$ be a measurable space and $G$ a group acting on $\Omega$. If $S \subseteq \Omega$ is a measurable set, then for any group element $g \in G$, the set $g \cdot S$ (the image of $S$ under the action of $g$) is also a measurable set. This follows from the fact that $g \cdot S$ can be viewed as the preimage of $S$ under the measurable action of $g^{-1}$.

Let $G$ be a group acting on a measurable space $\Omega$ and $E$ be a physical system. Let $\rho : G \to \text{Symmetry}(E)$ be a group homomorphism that maps group elements to symmetries of the system $E$. An effect-valued measure $\mu$ (the abstract form of a POVM) is **covariant** with respect to $\rho$ if for every group element $g \in G$ and every measurable set $S \subseteq \Omega$, the effect assigned to the translated set $g \cdot S$ is equal to the transformation of the effect $\mu(S)$ under the symmetry $\rho(g)$: $$\mu(g \cdot S) = \rho(g) \cdot \mu(S)$$ where the right-hand side denotes the action of the symmetry group on the space of effects.

Let $G$ be a group, and let $\rho_1: G \to \text{Symmetry}(E_1)$ and $\rho_2: G \to \text{Symmetry}(E_2)$ be representations of $G$ as symmetries acting on the systems $E_1$ and $E_2$, respectively. A channel $\phi : E_1 \to E_2$ is **covariant** if for all $g \in G$, the channel intertwines the actions of the group, which is expressed by the condition $\phi \circ \rho_1(g) = \rho_2(g) \circ \phi$.

Let $G$ be a group and $E_1$ be a physical system. Let $\rho: G \to \text{Symmetry}(E_1)$ be a representation of $G$ as symmetries acting on $E_1$. The identity channel on $E_1$, denoted $\text{id}_{E_1}$, is covariant with respect to the action $\rho$ on both the input and the output. That is, for all $g \in G$, the identity channel intertwines the action with itself: $\text{id}_{E_1} \circ \rho(g) = \rho(g) \circ \text{id}_{E_1}$.

Let $G$ be a group, and let $E_1, E_2$, and $E_3$ be physical systems. Suppose $\rho_1: G \to \text{Symmetry}(E_1)$, $\rho_2: G \to \text{Symmetry}(E_2)$, and $\rho_3: G \to \text{Symmetry}(E_3)$ are representations of $G$ as symmetry actions on these systems. If a channel $\phi : E_1 \to E_2$ is covariant with respect to $\rho_1$ and $\rho_2$, and a channel $\psi : E_2 \to E_3$ is covariant with respect to $\rho_2$ and $\rho_3$, then their composition $\psi \circ \phi$ is a channel that is covariant with respect to $\rho_1$ and $\rho_3$.