EnsembleSpaces: integrated results from discussion with Jacky

Author: gcarcassi

pm_EnsembleSpaces.tex

@@ -1276,7 +1276,7 @@ \subsection{Affine combinations}
 	A convex space $X$ is \textbf{cancellative} if $p \ens[a] + \bar{p} \ens = p \ens[b] + \bar{p} \ens$ for some $p \in (0,1)$ implies $\ens[a] = \ens[b]$.
 \end{defn}
 
-\begin{thrm}[Ensemble spaces are cancellative]
+\begin{thrm}[Ensemble spaces are cancellative]\label{pm-es-ensembleSpaceCancellative}
 	Let $\Ens$ be an ensemble space. Let $\ens[a], \ens[b], \ens \in \Ens$ such that $p\ens[a] + \bar{p} \ens = p \ens[b] + \bar{p} \ens$ for some $p \in (0,1)$. Then $\ens[a] = \ens[b]$.
 \end{thrm}
 
@@ -1351,6 +1351,27 @@ \subsection{Affine combinations}
 	\end{equation}
 	Since $S(\lambda\ens[a] + \bar{\lambda}\ens[b]) = \lambda S(\lambda\ens[a] + \bar{\lambda}\ens[b]) + \bar{\lambda} S(\lambda\ens[a] + \bar{\lambda}\ens[b]) = \lambda S(\ens[a]) + \bar{\lambda} S(\ens[b])$, $\ens[a] = \ens[b]$ by strict concavity.
 \end{proof}
+
+TODO: review
+
+\begin{coro}\label{pm-es-uniqueMixtureCoefficient}
+	Let $\ens[a], \ens[b] \in \Ens$ and $p,\lambda \in [0,1]$. Then $p\ens[a] + \bar{p} \ens[b] = \lambda \ens[a] + \bar{\lambda} \ens[b]$ if and only if $\ens[a] = \ens[b]$ or $p=\lambda$.
+\end{coro}
+
+\begin{proof}
+	Suppose that $\ens[a] = \ens[b]$ then $p\ens[a] + \bar{p} \ens[b] = p\ens[a] + \bar{p} \ens[a] = \ens[a] = \lambda \ens[a] + \bar{\lambda} \ens[a] = \lambda \ens[a] + \bar{\lambda} \ens[b]$ for all $p,\lambda \in [0,1]$. Suppose $p=\lambda$. Then $p\ens[a] + \bar{p} \ens[b] = \lambda \ens[a] + \bar{\lambda} \ens[b]$.
+	
+	Now suppose $p\ens[a] + \bar{p} \ens[b] = \lambda \ens[a] + \bar{\lambda} \ens[b]$. If $p=\lambda$, the relationship is satisfied for all $\ens[a],\ens[b] \in \Ens$. Without loss of generality, suppose $p \geq \lambda$. We have
+	\begin{align}
+		\lambda \ens[a] + \bar{\lambda} \ens[b] &= p \ens[a] + \bar{p} \ens[b] = \lambda \ens[a] + (p-\lambda) \ens[a] + \bar{p} \ens[b] = \lambda \ens[a] + \bar{\lambda}\left(\frac{p-\lambda}{\bar{\lambda}} \ens[a] + \frac{\bar{p}}{\bar{\lambda}} \ens[b] \right).
+	\end{align}
+	Using \ref{pm-es-ensembleSpaceCancellative}, we have
+	\begin{align}
+		\frac{p-\lambda}{\bar{\lambda}} \ens[a] + \frac{\bar{p}}{\bar{\lambda}} \ens[b] &= \ens[b] = \frac{p-\lambda}{\bar{\lambda}} \ens[b] + \frac{\bar{p}}{\bar{\lambda}} \ens[b].
+	\end{align}
+	Using \ref{pm-es-ensembleSpaceCancellative} again, we have $\ens[a] = \ens[b]$.
+\end{proof}
+
 \end{mathSection}
 
 The fact that the convex space is cancellative essentially allows us to ``invert'' a convex combination, allowing affine combinations. An affine combination is a linear combination where the coefficients sum to one, like in a convex combination, but coefficients can be negative, unlike in a convex combination. Conceptually we can ``take part of an ensemble out'' from another ensemble. The ability in quantum mechanics to use negative pseudo-probability, for example in Wigner functions, stems from the fact that the space of ensembles allows affine combinations because it is cancellative.
@@ -2026,7 +2047,7 @@ \subsection{Mixing entropy as pseudo-distance}
 	Since both the entropy and the mixing operation are continuous, the mixing entropy is also continuous. An entropic open ball is the reverse image of an open set through a continuous function and it is therefore an open set.
 \end{proof}
 
-\begin{prop}
+\begin{prop}\label{pm-es-Hausdorff}
 	Ensemble spaces are Hausdorff topological spaces.
 \end{prop}
 
@@ -2111,21 +2132,32 @@ \subsection{Mixing entropy as pseudo-distance}
 	Let $\ens[a], \ens[b]_i\in \Ens$ and $\ens[c]_i = p \ens[a] +\bar{p} \ens[b]_i$ for some $p \in (0,1)$. Then $\ens[b]_i \to \ens[b]$ if and only if $\ens[c]_i \to \ens[c]$. 
 \end{conj}
 
-\begin{conj}[Affine combinations are continuous]
-	Let $\ens \in \Ens$ and let $p \in [0,1]$. Let $+_{p\ens[a]} : \Ens \to \Ens$ be the curried function $+_{p\ens[a]}(\ens[b]) = p\ens[a] + \bar{p}\ens[b]$. The function is a homeomorphism between $\Ens$ and $+_{p\ens[a]}(\Ens)$.
-\end{conj}
+TODO: reorganize
 
-\begin{conj}
-	Mixing is an open map.
+\begin{remark}
+	Note that the curried function $+_{p\ens[a]}(\ens[b]) = p\ens[a] + \bar{p}\ens[b]$ is continuous and bijective, but it's not an open map over $\Ens$. Consider, in fact, $\ens[0,1]$ and take $p = 0.5$ and $\ens[a] = 0.5$. Then $+_{p\ens[a]}(\Ens) = [0.25,0.75]$ which is not an open set in the topology of $\Ens$. The map effectively shrinks sets by half, and the isolated points at the algebraic boundary are no longer isolated points in the bulk.
+	
+	Note that $[0.25,0.75]$ is, however, an open set in the subspace topology. It may be, then, that the map is open on the subspace topology.
+\end{remark}
+
+\begin{conj}[Affine combinations are continuous]
+	Let $\ens \in \Ens$ and let $p \in [0,1]$. Let $+_{p\ens[a]} : \Ens \to \Ens$ be the curried function $+_{p\ens[a]}(\ens[b]) = p\ens[a] + \bar{p}\ens[b]$. The function is a homeomorphism between $\Ens$ and $+_{p\ens[a]}(\Ens)$. That is, $+_{p\ens[a]}$ is a topological embedding.
 \end{conj}
 
 \begin{conj}
 	An ensemble space embeds continuously into a topological vector space.
 \end{conj}
 
-\begin{conj}
-	Let $+_{\ens[a]\ens[b]} : [0,1] \to \Ens$ be the curried function $+_{\ens[a]\ens[b]}(p) = p\ens[a] + \bar{p}\ens[b]$. The function is a homeomorphism between $[0,1]$ and $+_{\ens[a]\ens[b]}([0,1])$.
-\end{conj}
+TODO: reorganize
+
+\begin{prop}
+	Let $\ens[a], \ens[b] \in \Ens$ be two distinct ensembles.	Let $+_{\ens[a]\ens[b]} : [0,1] \to \Ens$ be the curried function $+_{\ens[a]\ens[b]}(p) = p\ens[a] + \bar{p}\ens[b]$. The function is a homeomorphism between $[0,1]$ and $+_{\ens[a]\ens[b]}([0,1])$.
+\end{prop}
+
+\begin{proof}
+	The map $+_{\ens[a]\ens[b]}$ is continuous as it $+$ is continuous by the axiom of mixture. Since $\ens[a]$ and $\ens[b]$ are distinct, by \ref{pm-es-uniqueMixtureCoefficient}, $p\ens[a] + \bar{p} \ens[b] = \lambda \ens[a] + \bar{\lambda} \ens[b]$ if and only if $p=\lambda$, which means $+_{\ens[a]\ens[b]}$ is injective, and it is therefore bijective onto its image. The domain of $+_{\ens[a]\ens[b]}$ is $[0,1]$, which is a compact. The image $+_{\ens[a]\ens[b]}([0,1])$ is a Hausdorff space since $\Ens$ is Hausdorff by \ref{pm-es-Hausdorff}. Putting it all together, $+_{\ens[a]\ens[b]}$ is a continuous bijective function from a compact space to a Hausdorff space and is therefore a homeomorphism.
+\end{proof}
+
 
 \begin{remark}
 	These are all about showing that mixing allows a continuous inverse. It is easy to show that mixing gives us a continuous bijection. The problem is showing that the inverse is continuous.