an update of various things

docs/mathematics/differential-geometry/topological-notions.md

@@ -1,4 +1,6 @@
-# Fiber bundles
+# Topological notions
+
+## Fiber bundles
 
 Let $X$ be a manifold over a field $F$. 
 
@@ -54,4 +56,4 @@ with $T^{ij}_k \in \mathbb{K}$ holors of $\mathbf{T}$. Furthermore, we have a ba
 >
 > of the tensor fiber bundle $\mathscr{B}$. 
 
-Therefore, a tensor field assigns a tensor fiber (or tensor) to each point on a section of the  manifold. These tensors may vary smoothly along the section of the manifold.
+Therefore, a tensor field assigns a tensor fiber (or tensor) to each point on a section of the  manifold. These tensors may vary smoothly along the section of the manifold.

docs/mathematics/linear-algebra/tensors/tensor-formalism.md

@@ -45,7 +45,7 @@ $$
 
 ## Outer product
 
-> *Definition 3*: the outer product $f \otimes g: X \times Y \to F$ of two scalar functions $f: X \to F$ and $g: Y \to F$ is defined as
+> *Definition 3*: the **outer product** $f \otimes g: X \times Y \to F$ of two scalar functions $f: X \to F$ and $g: Y \to F$ is defined as
 >
 > $$
 >   (f \otimes g)(x,y) = f(x) g(y),
@@ -133,9 +133,45 @@ By definition tensors are basis independent. Holors are basis dependent.
 
 We have from theorem 2 that the outer product of two tensors yields another tensor, with ranks adding up.
 
+## Interior product
+
+> *Definition 5*: the **left interior product** $f \llcorner \alpha: Y \to F$ of a scalar function $f:X \times Y \to F$ and a scalar $\alpha \in X$ is defined as
+>
+>   $$
+>       (f \llcorner \alpha)(y) = f(\alpha,y) \qquad \forall y \in Y,
+>   $$
+>
+> and the **right interior product** $f \lrcorner \alpha: X \to F$ for a scalar $\alpha \in Y$ is defined as
+>
+> $$
+>       (f \lrcorner \alpha)(x) = f(x,\alpha) \qquad \forall x \in X.
+> $$
+
+Note that neither interior product is associative, commutatitive and only distributive in the field addition. 
+
+With the interior product we can partially contract a tensor (i.e. reduce its basis) with a (co)vector. Consider $\mathbf{T} \in \mathscr{T}^2_0(V)$ then
+
+$$
+    \begin{align*}
+        \mathbf{T} \llcorner \mathbf{\hat v} &= T^{ij} \mathbf{k}(\mathbf{\hat v}, \mathbf{e}_i) \mathbf{e}_j,\\
+        &= v_i T^{ij} \mathbf{e}_j.
+    \end{align*}
+$$
+
+or similarly
+
+$$
+    \begin{align*}
+        \mathbf{T} \lrcorner \mathbf{\hat v} &= T^{ij} \mathbf{e}_i \mathbf{k}(\mathbf{\hat v}, \mathbf{e}_j),\\
+        &= v_j T^{ij} \mathbf{e}_i.
+    \end{align*}
+$$
+
+for all $\mathbf{\hat v} \in V^*$. 
+
 ## Inner product
 
-> *Definition 5*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to F$ which satisfies
+> *Definition 6*: an **inner product** on $V$ is a bilinear mapping $\bm{g}: V \times V \to F$ which satisfies
 >
 > 1. for all $\mathbf{u}, \mathbf{v} \in V: \; \bm{g}(\mathbf{u}, \mathbf{v}) = \overline{\bm{g}}(\mathbf{v}, \mathbf{u}),$ 
 > 2. for all $\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$ and $\lambda, \mu \in F: \;\bm{g}(\mathbf{u}, \lambda \mathbf{v} + \mu \mathbf{w}) = \lambda \bm{g}(\mathbf{u}, \mathbf{v}) + \mu \bm{g}(\mathbf{u}, \mathbf{w}),$
@@ -144,7 +180,7 @@ We have from theorem 2 that the outer product of two tensors yields another tens
 
 It may be observed that $\bm{g} \in \mathscr{T}_2^0$. Unlike the Kronecker tensor, the existence of an inner product is never implied. 
 
-> *Definition 6*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as 
+> *Definition 7*: let $G$ be the Gram matrix with its components $G \overset{\text{def}}= (g_{ij})$ defined as 
 > 
 > $$
 >   g_{ij} = \bm{g}(\mathbf{e}_i, \mathbf{e}_j).
@@ -227,7 +263,7 @@ $$
 
 with $u^j = g^{ij} u_i$. 
 
-> *Definition 7*: the basis $\{\mathbf{e}_i\}$ of $V$ induces a **reciprocal basis** $\{\mathbf{g}^{-1}(\mathbf{\hat e}^i)\}$ of $V$ given by 
+> *Definition 8*: the basis $\{\mathbf{e}_i\}$ of $V$ induces a **reciprocal basis** $\{\mathbf{g}^{-1}(\mathbf{\hat e}^i)\}$ of $V$ given by 
 > 
 > $$
 > \mathbf{g}^{-1}(\mathbf{\hat e}^i) = g^{ij} \mathbf{e}_j.

docs/mathematics/linear-algebra/tensors/tensor-symmetries.md

@@ -1,6 +1,6 @@
 # Tensor symmetries
 
-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 $\{\mathbf{\hat e}^i\}$, a field F 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 $\{\mathbf{\hat e}^i\}$, a field $F$ and a pseudo inner product $\bm{g}$ on $V$.
 
 ## Symmetric tensors
 

docs/mathematics/linear-algebra/tensors/tensor-transformations.md

@@ -12,7 +12,7 @@ so that $\mathbf{\hat e}^i = A^i_j \mathbf{\hat f}^j$.
 
 ## Transformation of tensors
 
-Recall from the section of [tensor-formalism]() that a holor depends on the chosen basis, but the corresponding tensor itself does not. This implies that holors transform in a particular way under a change of basis, which is characteristic for tensors. 
+Recall from the section of [tensor-formalism](tensor-formalism.md) that a holor depends on the chosen basis, but the corresponding tensor itself does not. This implies that holors transform in a particular way under a change of basis, which is characteristic for tensors. 
 
 > *Theorem 1*: let $\mathbf{T} \in \mathscr{T}^p_q(V)$ be a tensor with $p=q=1$ without loss of generality and $B = A^{-1}$. Then $\mathbf{T}$ may be decomposed into
 >
@@ -26,12 +26,20 @@ Recall from the section of [tensor-formalism]() that a holor depends on the chos
 > with the holors related by
 >
 > $$
->   \overline T^i_j = B^i_k A^j_l T^k_l.
+>   \overline T^i_j = B^i_k A^l_j T^k_l.
 > $$
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    We have
+
+    $$
+        \begin{align*}
+            \mathbf{T} &= T^i_j B^k_i \mathbf{f}_k \otimes A^j_l \mathbf{\hat f}^l,\\
+            &= B^k_i A^j_l T^i_j \mathbf{f}_k \otimes \mathbf{\hat f}^l,\\
+            &= \overline T^i_j \mathbf{f}_i \otimes \mathbf{\hat f}^j.
+        \end{align*}
+    $$
 
 The homogeneous nature of the tensor transformation implies that a holor equation of the form $T^i_j = 0$ holds relative to any basis if it holds relative to a particular one.
 
@@ -54,7 +62,17 @@ The homogeneous nature of the tensor transformation implies that a holor equatio
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    We have
+
+    $$
+        \begin{align*}
+            \bm{\mu} &= \mu_{i_1\dots i_n} A^{i_1}_{j_1} \mathbf{\hat f}^{j_1} \otimes \dots \otimes A^{i_n}_{j_n} \mathbf{\hat f}^{j_n},\\
+            &= A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} \mu_{i_1\dots i_n} \mathbf{\hat f}^{j_1} \otimes \dots \otimes \mathbf{\hat f}^{j_n},\\
+            &= A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} [i_1,\dots,i_n] \mathbf{\hat f}^{j_1} \otimes \dots \otimes \mathbf{\hat f}^{j_n},\\
+            &= \det(A) [j_1,\dots,j_n] \mathbf{\hat f}^{j_1} \otimes \dots \otimes \mathbf{\hat f}^{j_n},\\
+            &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \dots \otimes \mathbf{\hat f}^{i_1}.
+        \end{align*}
+    $$
 
 Then $\det(A)$ is the volume scaling factor of the transformation with $A$. So that if $\bm{\mu}(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1$, then $\bm{\mu}(\mathbf{f}_1, \dots, \mathbf{f}_n) = \det(A).$
 
@@ -75,10 +93,6 @@ Then $\det(A)$ is the volume scaling factor of the transformation with $A$. So t
 >
 > then $\mu_{i_1 \dots i_n} = \overline \mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$ is an invariant holor. 
 
-??? note "*Proof*:"
-
-    Will be added later.
-
 ## Transformation of Levi-Civita form
 
 > *Theorem 3*: let $\bm{\epsilon} \in \bigwedge_n(V)$ be the Levi-Civita tensor with
@@ -94,4 +108,4 @@ Then $\det(A)$ is the volume scaling factor of the transformation with $A$. So t
 
 ??? note "*Proof*:"
 
-    Will be added later.
+    Follows directly from the definition $\bm{\epsilon} = \sqrt{g} \bm{\mu}$.

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 $\epsilon^{|i_1 \dots i_n|} = \frac{1}{\sqrt{g}}$.