more math, however, not satisfying!

diff --git a/physics_compact/math_app.tex b/physics_compact/math_app.tex
index f3277509004ed259528c292c279254bdfd5a8363..e3417bdadbfc503842a193833540682248b02294 100644 (file)
--- a/physics_compact/math_app.tex
+++ b/physics_compact/math_app.tex
@@ -69,7 +69,10 @@ $(\cdot,\cdot):V\times V \rightarrow K$, which satisfies
       (positive definite)
 \end{itemize}
 for $\vec{u},\vec{v}\in V$ and $\lambda\in K$.
-Taking the complex conjugate $(\cdot)^*$ is the map from $K\ni z=a+bi\mapsto a-bi=z^*\in K$.
+Taking the complex conjugate $(\cdot)^*$ is the map from
+\begin{equation}
+z=a+bi\mapsto z^*=a-bi \text{, } z,z^*\in K \text{.}
+\end{equation}
 \end{definition}
 
 \begin{remark}
@@ -82,7 +85,7 @@ Due to conjugate symmetry, linearity in the first argument results in conjugate
 This is called a sesquilinear form.
 If $K=\mathbb{R}$, conjugate symmetry reduces to symmetry and the sesquilinear form gets a bilinear form.
 
-The inner product $(\cdot,\cdot)$ provides a mapping
+Furtermore, the inner product $(\cdot,\cdot)$ provides a mapping
 \begin{equation}
 V\rightarrow V^{\dagger}:\vec{v}\mapsto \varphi_{\vec{v}}
 \quad
@@ -100,18 +103,21 @@ If the inner product is nondegenerate, i.e.\  $\forall\vec{u}\, (\vec{v},\vec{u}
 Since the dimension of $V$ and $V^{\dagger}$ is equal, it is additionally surjective.
 Then, $V$ is isomorphic to $V^{\dagger}$.
 Vector $\vec{v}^{\dagger}\equiv \varphi_{\vec{v}}\in V^{\dagger}$ is said to be the dual vector of $\vec{v}\in V$.
+The dual pairing $[\vec{v}^{\dagger},\vec{u}]=[\varphi_{\vec{v}},\vec{u}]=\varphi_{\vec{v}}(\vec{u})$ is associated with the inner product $(\vec{v},\vec{u})$.
 
-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 $V^{\dagger}$
+Now, 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 this case, the antilinearity property is assigned to element $\varphi_{\vec{v}}=\vec{v}^{\dagger}$ of dual space indicating an isomorphism of $V$ to the conjugate complex of its dual space.
 \begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}(\vec{u})\vec{v}
-\text{ CHECK ! ! !}
+[(\lambda\vec{v})^{\dagger},\vec{u}]=
+[\varphi_{\lambda\vec{v}},\vec{u}]=
+\varphi_{\lambda\vec{v}}(\vec{u})=
+\lambda^*\varphi_{\vec{v}}(\vec{u})=
+\lambda^*(\vec{v},\vec{u})
 \end{equation}
-or the conjugate transpose in matrix formalism
+According to this, in matrix formalism, the dual vector is associated with the conjugate transpose.
 \begin{equation}
-(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v} \text{ .}
+(\vec{u},\vec{v}) \rightarrow \vec{u}^{\dagger}\vec{v}
 \end{equation}
-In doing so, the conjugate transpose is associated with the dual vector.
 \end{remark}
 
 \begin{definition}[Outer product]