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[lectures/latex.git] / physics_compact / math_app.tex

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 9ff2b4b..687f928 100644 (file)
--- 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}
 
 
 \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}
 \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 $K=\mathbb{R}$)
 \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 $\lambda\in K$.
       (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}
 \end{definition}
 
 \begin{remark}


@@ -61,27 +83,45 @@ 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}
 \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 \varphi_{\vec{v}}
+\quad
+\text{ defined by }
+\quad
+\varphi_{\vec{v}}(\vec{u})=(\vec{v},\vec{u}) \text{ .}
+\end{equation}
+Since the inner product is linear in the first argument, the same is true for the defined mapping.
+\begin{equation}
+\lambda(\vec{u}+\vec{v}) \mapsto
+\varphi_{\lambda(\vec{u}+\vec{v})}=
+\lambda\varphi_{\vec{u}}+\lambda\varphi_{\vec{v}}\\
+\end{equation}
+The kernel is $\vec{v}=0$, structural identity (isomorphism) of $V$ and $V^{\dagger}$ is .
+

 In physics and matrix algebra, the inner product is often defined with linearity in the second argument and conjugate linearity in the first argument.
 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}
 \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}
 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
 \end{equation}
 \end{equation}
 or the conjugate transpose in matrix formalism
 \begin{equation}
 (\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
 \end{equation}

-In doing so, conjugacy is associated with duality.
+In doing so, the conjugate transpose is associated with the dual vector.

 \end{remark}
 
 \begin{definition}
 \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 $\vec{\varphi}^{\dagger}\in V^{\dagger}$  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}
 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{\varphi}^{\dagger}(\vec{v})\vec{u}

 \text{ ,}
 \end{equation}
 \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 $\vec{\varphi}^{\dagger}(\vec{v})$ 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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
 
 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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,