fixed outer product, redo inner product!

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 9ff2b4b39b6a303405994ee0b29d6375491c5197..79e4ec996d602cf18cf629e0cd0ce6ee4bcfbbc8 100644 (file)
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -74,14 +74,14 @@ In doing so, conjugacy is associated with duality.
 \end{remark}
 
 \begin{definition}
-If $\vec{u}\in U$, $\vec{v}\in V$ and $\vec{v}^{\dagger}\in V^{\dagger}$  are vectors within the respective vector spaces and $V^{\dagger}$ is the dual space of $V$,
-the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{v}^{\dagger}$ and $\vec{u}$,
+If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{y}^{\dagger}\in V^{\dagger}$  is a linear functional of the dual space $V^{\dagger}$ of $V$,
+the outer product $\vec{u}\otimes\vec{v}$ is defined as the tensor product of $\vec{y}^{\dagger}$ and $\vec{u}$,
 which constitutes a map $A:V\rightarrow U$ by
 \begin{equation}
-\vec{v}\mapsto\vec{v}^{\dagger}(\vec{v})\vec{u}
+\vec{v}\mapsto\vec{y}^{\dagger}(\vec{v})\vec{u}
 \text{ ,}
 \end{equation}
-where $\vec{v}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{v}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
+where $\vec{y}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{y}^{\dagger}\in V^{\dagger}$ on $V$ when evaluated at $\vec{v}\in V$, a scalar that in turn is multiplied with $\vec{u}\in U$.
 
 In matrix formalism, with respect to a given basis ${\vec{e}_i}$ of $\vec{u}$ and ${\vec{e}'_i}$ of $\vec{v}$,
 if $\vec{u}=\sum_i^m \vec{e}_iu_i$ and $\vec{v}=\sum_i^n\vec{e}'_iv_i$,