Added volume-forms and tensor transformation and therefore finished the section of tensors.
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4 changed files with 234 additions and 3 deletions
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@ -96,6 +96,7 @@ nav:
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- 'Tensor formalism': mathematics/linear-algebra/tensors/tensor-formalism.md
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- 'Tensor symmetries': mathematics/linear-algebra/tensors/tensor-symmetries.md
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- 'Volume forms': mathematics/linear-algebra/tensors/volume-forms.md
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- 'Tensor transformations': mathematics/linear-algebra/tensors/tensor-transformations.md
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- 'Calculus':
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- 'Limits': mathematics/calculus/limits.md
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- 'Continuity': mathematics/calculus/continuity.md
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@ -168,7 +168,7 @@ $$
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Will be added later.
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An interesting result of the definition of the (anti)symmetric product is given in the theorem below.
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An interesting result of the definition of the symmetric and antisymmetric product is given in the theorem below.
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> *Theorem 2*: let $\mathbf{\hat u}_{1,2} \in V^*$ be covectors, the symmetric product of $\mathbf{\hat u}_1$ and $\mathbf{\hat u}_2$ may be given by
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>
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@ -184,8 +184,10 @@ An interesting result of the definition of the (anti)symmetric product is given
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> (\mathbf{\hat u}_1 \wedge \mathbf{\hat u}_2)(\mathbf{v}_1, \mathbf{v}_2) = \det \big(\mathbf{k}(\mathbf{\hat u}_i, \mathbf{v}_j) \big),
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> $$
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>
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> for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(j,k)$ denoting the entry of the matrix over which the determinant is taken.
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> for all $(\mathbf{v}_1, \mathbf{v}_2) \in V \times V$ with $(i,j)$ denoting the entry of the matrix over which the determinant is taken.
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??? note "*Proof*:"
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Will be added later.
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Will be added later.
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In some literature theorem 2 is used as definition for the symmetric and antisymmetric product from which the relation with the symmetrisation maps can be proven. Either method is valid, however it has been chosen that defining the products in terms of the symmetrisation maps is more general.
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@ -0,0 +1,97 @@
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# Tensor transformations
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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 $V.$
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Let us introduce a different basis $\{\mathbf{f}_i\}_{i=1}^n$ of $V$ with a corresponding dual basis $\{\mathbf{\hat f}^i\}_{i=1}^n$ of $V^*$ which are related to the former basis $\{\mathbf{e}_i\}_{i=1}^n$ by
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$$
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\mathbf{f}_j = A^i_j \mathbf{e}_i,
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$$
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so that $\mathbf{\hat e}^i = A^i_j \mathbf{\hat f}^j$.
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## Transformation of tensors
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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.
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> *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
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>
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> $$
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> \begin{align*}
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> \mathbf{T} &= T^i_j \mathbf{e}_i \otimes \mathbf{\hat e}^j, \\
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> &= \overline T^i_j \mathbf{f}_i \otimes \mathbf{\hat f}^j,
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> \end{align*}
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> $$
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>
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> with the holors related by
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>
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> $$
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> \overline T^i_j = B^i_k A^j_l T^k_l.
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> $$
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??? note "*Proof*:"
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Will be added later.
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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.
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## Transformation of volume forms
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> *Lemma 1*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form with
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>
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> $$
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> \begin{align*}
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> \bm{\mu} &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\
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> &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n},
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> \end{align*}
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> $$
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>
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> then we have
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>
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> $$
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> \overline \mu_{j_1 \dots j_n} = A^{i_1}_{j_1} \cdots A^{i_n}_{j_n} \mu_{i_1 \dots i_n} = \mu_{j_1 \dots j_n} \det (A).
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> $$
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??? note "*Proof*:"
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Will be added later.
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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 $\bm{\mu}(\mathbf{f}_1, \dots, \mathbf{f}_n) = \det(A).$
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> *Theorem 2*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form with
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>
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> $$
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> \begin{align*}
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> \bm{\mu} &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\
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> &= \overline \mu_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n},
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> \end{align*}
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> $$
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>
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> and if we define
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>
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> $$
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> \overline \mu_{i_1 \dots i_n} \overset{\text{def}}{=} \frac{1}{\det (A)} A^{j_1}_{i_1} \cdots A^{j_n}_{i_n} \mu_{j_1 \dots j_n},
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> $$
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>
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> then $\mu_{i_1 \dots i_n} = \overline \mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$ is an invariant holor.
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??? note "*Proof*:"
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Will be added later.
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## Transformation of Levi-Civita form
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> *Theorem 3*: let $\bm{\epsilon} \in \bigwedge_n(V)$ be the Levi-Civita tensor with
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>
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> $$
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> \begin{align*}
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> \bm{\epsilon} &= \epsilon_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n}, \\
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> &= \overline \epsilon_{i_1 \dots i_n} \mathbf{\hat f}^{i_1} \otimes \cdots \otimes \mathbf{\hat f}^{i_n},
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> \end{align*}
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> $$
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>
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> then $\epsilon_{i_1 \dots i_n} = \overline \epsilon_{i_1 \dots i_n}$ is an invariant holor.
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??? note "*Proof*:"
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Will be added later.
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# Volume forms
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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 $V.$
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## n forms
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> *Definition 1*: let $\bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$, if
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>
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> $$
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> \bm{\mu}(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1,
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> $$
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>
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> then $\bm{\mu}$ is the **unit volume form** with respect to the basis $\{\mathbf{e}_i\}$.
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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 $\lambda \in \mathbb{K}$.
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> *Proposition 1*: the unit volume form $\bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\}$ may be given by
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>
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> $$
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> \begin{align*}
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> \bm{\mu} &= \mathbf{\hat e}^1 \wedge \dots \wedge \mathbf{\hat e}^n, \\
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> &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \dots \otimes \mathbf{\hat e}^{i_n},
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> \end{align*}
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> $$
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>
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> with $\mu_{i_1 \dots i_n} = [i_1, \dots, i_n]$.
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??? note "*Proof*:"
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Will be added later.
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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.
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> *Definition 2*: let $(V, \bm{\mu})$ denote the vector space $V$ endowed with an **oriented volume form** $\bm{\mu}$. For $\bm{\mu} > 0$ we have a positive orientation of $(V, \bm{\mu})$ and for $\bm{\mu} < 0$ we have a negative orientation of $(V, \bm{\mu})$.
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For a vector space with an oriented volume $(V, \bm{\mu})$ we may write
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$$
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\bm{\mu} = \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n},
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$$
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or, equivalently
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$$
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\bm{\mu} = \mu_{|i_1 \dots i_n|} \mathbf{\hat e}^{i_1} \wedge \cdots \wedge \mathbf{\hat e}^{i_n},
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$$
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by convention, to resolve ambiguity with respect to the meaning of $\mu_{i_1 \dots i_n}$ without using another symbol or extra accents.
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Using theorem 2 in the section of [tensor symmetries]() we may state the following.
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> *Proposition 2*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form, then we have
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>
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> $$
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> \bm{\mu}(\mathbf{v}_1, \dots, \mathbf{v}_n) = \det \big(\mathbf{k}(\mathbf{\hat e}^i, \mathbf{v}_j) \big),
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> $$
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>
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> for all $\mathbf{v}_1, \dots, \mathbf{v}_n \in V$ with $(i,j)$ denoting the entry of the matrix over which the determinant is taken.
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??? note "*Proof*:"
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Will be added later.
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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.
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From proposition 2 it may also be observed that within a geometrical context the oriented volume form may represent the area of a parallelogram in $n=2$ or the volume of a parallelepiped in $n=3$, span by its basis.
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## n - k forms
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> *Definition 3*: let $(V, \bm{\mu})$ be a vector space with an oriented volume form and let $\mathbf{u}_1, \dots, \mathbf{u}_k \in V$ with $k \in \mathbb{N}[k < n]$. Let the $(n-k)$-form $\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k \in \bigwedge_{n-k}(V)$ be defined as
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>
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> $$
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> \bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k(\mathbf{v}_{k+1}, \dots, \mathbf{v}_n) = \bm{\mu}(\mathbf{u}_1, \dots, \mathbf{u}_k, \mathbf{v}_{k+1}, \dots, \mathbf{v}_n),
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> $$
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>
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> for all $\mathbf{v}_{k+1}, \dots, \mathbf{v}_n \in V$ with $\lrcorner$ the insert operator.
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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
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$$
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\begin{align*}
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\bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k &= u_1^{i_1} \cdots u_k^{i_k} (\bm{\mu} \lrcorner \mathbf{e}_{i_1} \lrcorner \dots \lrcorner \mathbf{e}_{i_k}), \\
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&= u_1^{i_1} \cdots u_k^{i_k} \mu_{i_1 \dots i_n} (\mathbf{\hat e}^{i_{k+1}} \wedge \cdots \wedge \mathbf{\hat e}^{i_{n}}),
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\end{align*}
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$$
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for $\mathbf{u}_1, \dots, \mathbf{u}_k \in V$ with $k \in \mathbb{N}[k < n]$ and decomposition by $\mathbf{u}_q = u_q^{i_q} \mathbf{e}_{i_q}$ for $q \in \mathbb{N}[q \leq k]$.
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If we have a unit volume form $\bm{\mu}$ with respect to $\{\mathbf{e}_i\}$ then
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$$
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\bm{\mu}\lrcorner\mathbf{e}_1 \lrcorner \dots \lrcorner \mathbf{e}_k = \mathbf{\hat e}^{i_{k+1}} \wedge \cdots \wedge \mathbf{e}^{i_n},
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$$
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for $k \in \mathbb{N}[k < n]$.
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## Levi-Civita form
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> *Definition 4*: let $(V, \bm{\mu})$ be a vector space with a unit volume form with invariant holor. Let $\bm{\epsilon} \in \bigwedge_n(V)$ be the **Levi-Civita tensor** which is the unique unit volume form of positive orientation defined as
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>
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> $$
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> \bm{\epsilon} = \sqrt{g} \bm{\mu},
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> $$
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>
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> with $g \overset{\text{def}}{=} |\det (G)|$, the absolute value of the determinant of the [Gram matrix]().
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Therefore, if we decompose the Levi-Civita tensor by
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$$
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\bm{\epsilon} = \epsilon_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \dots \otimes \mathbf{\hat e}^{i_n} = \epsilon_{|i_1 \dots i_n|} \mathbf{\hat e}^{i_1} \wedge \dots \wedge \mathbf{\hat e}^{i_n},
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$$
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then we have $\epsilon_{i_1 \dots i_n} = \sqrt{g} \mu_{i_1 \dots i_n}$ and $\epsilon_{|i_1 \dots i_n|} = \sqrt{g}$.
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> *Theorem 2*: let $(V, \bm{\mu})$ be a vector space with a unit volume form with invariant holor. Let $\mathbf{g}(\bm{\epsilon}) \in \bigwedge^n(V)$ be the **reciprocal Levi-Civita tensor** which is given by
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>
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> $$
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> \mathbf{g}(\bm{\epsilon}) = \frac{1}{\sqrt{g}} \bm{\mu}.
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> $$
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??? note "*Proof*:"
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Will be added later.
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We may decompose the reciprocal Levi-Civita tensor by
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$$
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\mathbf{g}(\bm{\epsilon}) = \epsilon^{i_1 \dots i_n} \mathbf{e}_{i_1} \otimes \cdots \otimes \mathbf{e}_{i_n} = \epsilon^{|i_1 \dots i_n|} \mathbf{e}_{i_1} \wedge \cdots \wedge \mathbf{e}_{i_n},
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$$
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then we have $\epsilon^{i_1 \dots i_n} = \frac{1}{\sqrt{g}} \mu^{i_1 \dots i_n}$ and $\epsilon^{|i_1 \dots i_n|} = \frac{1}{\sqrt{g}}$.
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