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diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index 079d8d9..687f928 100644
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -36,15 +36,120 @@ 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 ...
+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 $K=\mathbb{R}$)
+\item $(\lambda\vec{u},\vec{v})=\lambda(\vec{u},\vec{v})$ and
+      $(\vec{u}'+\vec{u}'',\vec{v})=(\vec{u}',\vec{v})+(\vec{u}'',\vec{v})$
+      (linearity in first argument)
+\item $(\vec{u},\vec{u})\geq 0 \text{, } ``=" \Leftrightarrow \vec{u}=0$
+      (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}
+Due to conjugate symmetry, linearity in the first argument results in conjugate linearity (also termed antilinearity) in the second argument.
+This is called a sesquilinear form.
+\begin{equation}
+(\vec{u},\lambda(\vec{v}'+\vec{v}''))=(\lambda(\vec{v}'+\vec{v}''),\vec{u})^*=
+\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.
+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}
+(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
+\end{equation}
+In doing so, the conjugate transpose is associated with the dual vector.
+\end{remark}
+
 \begin{definition}
-The outer product ...
+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{\varphi}^{\dagger}(\vec{v})\vec{u}
+\text{ ,}
+\end{equation}
+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 $\vec{v}=\sum_i^n\vec{e}'_iv_i$,
+the outer product can be written as matrix $A$ as
+\begin{equation}
+\vec{u}\otimes\vec{v}=A=\left(
+\begin{array}{c c c c}
+u_1v_1 & u_1v_2 & \cdots & u_1v_n\\
+u_2v_1 & u_2v_2 & \cdots & u_2v_n\\
+\vdots & \vdots & \ddots & \vdots\\
+u_mv_1 & u_mv_2 & \cdots & u_mv_n\\
+\end{array}
+\right)
+\text{ .}
+\end{equation}
 \end{definition}
+\begin{remark}
+The matrix can be equivalently obtained by matrix multiplication:
+\begin{equation}
+\vec{u}\otimes\vec{v}=\vec{u}\vec{v}^{\dagger} \text{ ,}
+\end{equation}
+if $\vec{u}$ and $\vec{v}$ are represented as $m\times 1$ and $n\times 1$ column vectors, respectively.
+Here, $\vec{v}^{\dagger}$ represents the conjugate transpose of $\vec{v}$.
+By definition, and as can be easily seen in the matrix representation, the following identity holds:
+\begin{equation}
+(\vec{u}\otimes\vec{v})\vec{w}=\vec{u}(\vec{v},\vec{w})
+\end{equation}
+\end{remark}
 
 \section{Spherical coordinates}