From: hackbard Date: Fri, 10 Feb 2012 08:29:18 +0000 (+0100) Subject: more math - finish isomorphism + CHECK X-Git-Url: https://hackdaworld.org/gitweb/?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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,