From: hackbard <hackbard@hackdaworld.org>
Date: Fri, 10 Feb 2012 08:29:18 +0000 (+0100)
Subject: more math - finish isomorphism + CHECK
X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=31582cb1db0cbd9f53573e31f2eb30c9905bc655;p=lectures%2Flatex.git

more math - finish isomorphism + CHECK
---

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 79e4ec9..d097e70 100644
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -36,11 +36,32 @@ The addition of two vectors is called vector addition.
 
 \subsection{Dual space}
 
+\begin{definition}
+The dual space $V^{\dagger}$ of vector space $V$ over field $K$ is defined as the set of all linear maps from the vector space $V$ into its field $K$
+\begin{equation}
+\varphi:V\rightarrow K \text{ .}
+\end{equation}
+These type of linear maps are termed linear functionals.
+The dual space $V^{\dagger}$ becomes a vector space over $K$ itself by the following additional definitions
+\begin{eqnarray}
+(\varphi+\psi)(\vec{v}) & = & \varphi(\vec{v})+\psi(\vec{v}) \\
+(\lambda\varphi)(\vec{v}) & = & \lambda\varphi(\vec{v})
+\end{eqnarray}
+for all $\vec{v}\in V$, $\varphi,\psi\in V^{\dagger}$ and $\lambda\in K$.
+
+The map $V^{\dagger}\times V \rightarrow K: [\varphi,\vec{v}]=\varphi(\vec{v})$ is termed dual pairing of a functional $\varphi\in V^{\dagger}$ and an elemnt $\vec{v}\in V$.
+\end{definition}
+
 \subsection{Inner and outer product}
 \label{math_app:product}
 
 \begin{definition}
-The inner product on a vector space $V$ over $K$ is a map $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
+The inner product on a vector space $V$ over $K$ is a map
+\begin{equation}
+(\cdot,\cdot):V\times V \rightarrow K
+\text{ ,}
+\end{equation}
+which satisfies
 \begin{itemize}
 \item $(\vec{u},\vec{v})=(\vec{v},\vec{u})^*$
       (conjugate symmetry, symmetric for $K=\mathbb{R}$)
@@ -51,6 +72,7 @@ The inner product on a vector space $V$ over $K$ is a map $(\cdot,\cdot):V\times
       (positive definite)
 \end{itemize}
 for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
+Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
 \end{definition}
 
 \begin{remark}
@@ -61,10 +83,22 @@ This is called a sesquilinear form.
 \lambda^*(\vec{v}',\vec{u})^*+\lambda^*(\vec{v}'',\vec{u})^*=
 \lambda^*(\vec{u},\vec{v}')+\lambda^*(\vec{u},\vec{v}'')
 \end{equation}
+
+The inner product $(\cdot,\cdot)$ provides a mapping
+\begin{equation}
+V\rightarrow V^{\dagger}:\vec{v}\mapsto \vec{v}^{\dagger}
+\end{equation}
+given by 
+\begin{equation}
+v^{\dagger}()
+\end{equation}
+indicating structural identity (isomorphism) of $V$ and $V^{\dagger}$.
+
 In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
-This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with the dual vector or linear functional of dual space $V^{\dagger}$
+This allows to express the inner product $(\vec{u},\vec{v})$ as a product of vector $\vec{v}$ with a dual vector or linear functional of dual space $V^{\dagger}$
 \begin{equation}
 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
+\text{ CHECK ! ! !}
 \end{equation}
 or the conjugate transpose in matrix formalism
 \begin{equation}
@@ -74,14 +108,14 @@ In doing so, conjugacy is associated with duality.
 \end{remark}
 
 \begin{definition}
-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 $\vec{u}$,
+If $\vec{u}\in U$, $\vec{v}\in V$ are vectors within the respective vector spaces and $\vec{\varphi}^{\dagger}\in V^{\dagger}$  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 $\vec{\varphi}^{\dagger}$ and $\vec{u}$,
 which constitutes a map $A:V\rightarrow U$ by
 \begin{equation}
-\vec{v}\mapsto\vec{y}^{\dagger}(\vec{v})\vec{u}
+\vec{v}\mapsto\vec{\varphi}^{\dagger}(\vec{v})\vec{u}
 \text{ ,}
 \end{equation}
-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 $\vec{u}\in U$.
+where $\vec{\varphi}^{\dagger}(\vec{v})$ denotes the linear functional $\vec{\varphi}^{\dagger}\in V^{\dagger}$ 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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,