an update of various things

docs/mathematics/linear-algebra/tensors/volume-forms.md

@@ -1,6 +1,6 @@
 # Volume forms
 
-We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n,$ a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}_{i=1}^n$ and a pseudo inner product $\bm{g}$ on $V.$ 
+We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim V = n$, with a basis $\{\mathbf{e}_i\}_{i=1}^n$, a corresponding dual space $V^*$ with a basis $\{\mathbf{\hat e}^i\}_{i=1}^n$, a field $F$ and a pseudo inner product $\bm{g}$ on $V$.
 
 ## n-forms
 
@@ -12,7 +12,7 @@ We have a $n \in \mathbb{N}$ finite dimensional vector space $V$ such that $\dim
 >
 > then $\bm{\mu}$ is the **unit volume form** with respect to the basis $\{\mathbf{e}_i\}$. 
 
-Note that $\dim \bigwedge_n(V) = 1$ and consequently if $\bm{\mu}_1, \bm{\mu}_2 \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$, then $\bm{\mu}_1 = \lambda \bm{\mu}_2$ with $\lambda \in \mathbb{K}$. 
+Note that $\dim \bigwedge_n(V) = 1$ and consequently if $\bm{\mu}_1, \bm{\mu}_2 \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$, then $\bm{\mu}_1 = \lambda \bm{\mu}_2$ with $\lambda \in F$. 
 
 > *Proposition 1*: the unit volume form $\bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$ may be given by
 >
@@ -27,7 +27,31 @@ Note that $\dim \bigwedge_n(V) = 1$ and consequently if $\bm{\mu}_1, \bm{\mu}_2
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    Let $\pi = [\pi(1),\dots,\pi(n)]$ be any permutation of the set $\{1,\dots,n\}$, the unit volume form $\bm{\mu}$ is defined as 
+
+    $$
+        \bm{\mu}(\mathbf{e}_{\pi(1)},\dots,\mathbf{e}_{\pi(2)}) = \mathrm{sign}(\pi),
+    $$
+
+    thus
+
+    $$
+        \bm{\mu} = \mu_{i_1\dots i_n} \mathbf{\hat e}^{i_1} \otimes \dots \otimes \mathbf{\hat e}^{i_n}.
+    $$
+
+    Furthermore $\mathscr{A}(\bm{\mu}) = \bm{\mu}$. Then
+
+    $$
+        \bm{\mu} = \mu_{i_1\dots i_n} \frac{1}{n!} \mathbf{\hat e}^{i_1} \wedge \dots \wedge \mathbf{\hat e}^{i_n},
+    $$
+
+    and going back to the definition only requires us to consider
+
+    $$
+        \bm{\mu} = \mathbf{\hat e}^{1} \wedge \dots \wedge \mathbf{\hat e}^{n},
+    $$
+
+    such that $\mu_{i_1\dots i_n} = [i_1,\dots,i_n]$.
 
 The normalisation of the unit volume form $\bm{\mu}$ requires a basis. Consequently, the identification $\mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$ holds only relative to the basis.
 
@@ -47,7 +71,7 @@ $$
 
 by convention, to resolve ambiguity with respect to the meaning of $\mu_{i_1 \dots i_n}$ without using another symbol or extra accents.
 
-Using theorem 2 in the section of [tensor symmetries]() we may state the following.
+Using theorem 2 in the section of [tensor symmetries](tensor-symmetries.md) we may state the following.
 
 > *Proposition 2*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form, then we have
 >
@@ -59,7 +83,15 @@ Using theorem 2 in the section of [tensor symmetries]() we may state the followi
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    We have
+
+    $$
+        \begin{align*}
+            \bm{\mu}(\mathbf{v}_1,\dots,\mathbf{v}_n) &= \mu_{i_1\dots i_n} \mathbf{k}(\mathbf{\hat e}^{i_1},\mathbf{v}_1) \cdots \mathbf{k}(\mathbf{\hat e}^{i_n},\mathbf{v}_n),\\
+            &= [i_1,\dots,i_n] \mathbf{k}(\mathbf{\hat e}^{i_1},\mathbf{v}_1) \cdots \mathbf{k}(\mathbf{\hat e}^{i_n},\mathbf{v}_n),\\
+            &= \det\big(\mathbf{k}(\mathbf{\hat e}^i,\mathbf{v}_j)\big).
+        \end{align*}
+    $$
 
 Which reveals the role of the Kronecker tensor and thus the role of the dual space in the definition of $\bm{\mu}$. We may also conclude that an oriented volume $\bm{\mu} \in \bigwedge_n(V)$ on a vector space $V$ does not require an inner product. 
 
@@ -73,9 +105,9 @@ From proposition 2 it may also be observed that within a geometrical context the
 >   \bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k(\mathbf{v}_{k+1}, \dots, \mathbf{v}_n) = \bm{\mu}(\mathbf{u}_1, \dots, \mathbf{u}_k, \mathbf{v}_{k+1}, \dots, \mathbf{v}_n),
 > $$
 >
-> for all $\mathbf{v}_{k+1}, \dots, \mathbf{v}_n \in V$ with $\lrcorner$ the insert operator. 
+> for all $\mathbf{v}_{k+1}, \dots, \mathbf{v}_n \in V$. 
 
-It follows that $(n-k)$-form $\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k \in \bigwedge_{n-k}(V)$ can be written as
+It follows that the $(n-k)$-form $\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k \in \bigwedge_{n-k}(V)$ can be written as
 
 $$
 \begin{align*}
@@ -120,7 +152,16 @@ then we have $\epsilon_{i_1 \dots i_n} = \sqrt{g} \mu_{i_1 \dots i_n}$ and $\eps
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    The reciprocal Levi-Civita tensor may be written as
+
+    $$
+        \begin{align*}
+            \mathbf{g}(\bm{\epsilon}) &= \sqrt{g} \mathbf{g}(\mathbf{\hat e}^1) \wedge \dots \wedge \mathbf{g}(\mathbf{\hat e}_n),\\
+            &= \sqrt{g} g^{1i_1} \mathbf{e}_{i_1} \wedge \dots \wedge g^{ni_n} \mathbf{e}_{i_n},\\
+            &= \sqrt{g} \det (G^{-1}) \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n,\\
+            &= \frac{1}{\sqrt{g}} \mathbf{e}_1 \wedge \dots \wedge \mathbf{e}_n.
+        \end{align*}
+    $$
 
 We may decompose the reciprocal Levi-Civita tensor by
 
@@ -128,4 +169,4 @@ $$
     \mathbf{g}(\bm{\epsilon}) = \epsilon^{i_1 \dots i_n} \mathbf{e}_{i_1} \otimes \cdots \otimes \mathbf{e}_{i_n} = \epsilon^{|i_1 \dots i_n|} \mathbf{e}_{i_1} \wedge \cdots \wedge \mathbf{e}_{i_n},
 $$
 
-then we have $\epsilon^{i_1 \dots i_n} = \frac{1}{\sqrt{g}} \mu^{i_1 \dots i_n}$ and $\epsilon^{|i_1 \dots i_n|} = \frac{1}{\sqrt{g}}$. 
+then we have $\epsilon^{i_1 \dots i_n} = \frac{1}{\sqrt{g}} \mu^{i_1 \dots i_n}$ and $\epsilon^{|i_1 \dots i_n|} = \frac{1}{\sqrt{g}}$.