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

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.

n-forms

Definition 1: let \bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\}, if

\bm{\mu}(\mathbf{e}_1, \dots, \mathbf{e}_n) = 1,

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}.

Proposition 1: the unit volume form \bm{\mu} \in \bigwedge_n(V) \backslash \{\mathbf{0}\} may be given by

\begin{align*} \bm{\mu} &= \mathbf{\hat e}^1 \wedge \dots \wedge \mathbf{\hat e}^n, \ &= \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \dots \otimes \mathbf{\hat e}^{i_n}, \end{align*}

with \mu_{i_1 \dots i_n} = [i_1, \dots, i_n].

??? note "Proof:"

Will be added later.

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.

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}).

For a vector space with an oriented volume (V, \bm{\mu}) we may write

\bm{\mu} = \mu_{i_1 \dots i_n} \mathbf{\hat e}^{i_1} \otimes \cdots \otimes \mathbf{\hat e}^{i_n},

or, equivalently

\bm{\mu} = \mu_{|i_1 \dots i_n|} \mathbf{\hat e}^{i_1} \wedge \cdots \wedge \mathbf{\hat e}^{i_n},

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.

Proposition 2: let (V, \bm{\mu}) be a vector space with an oriented volume form, then we have

\bm{\mu}(\mathbf{v}_1, \dots, \mathbf{v}_n) = \det \big(\mathbf{k}(\mathbf{\hat e}^i, \mathbf{v}_j) \big),

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.

??? note "Proof:"

Will be added later.

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.

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.

(n - k)-forms

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

\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.

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

\begin{align*} \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}), \ &= 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}}), \end{align*}

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].

If we have a unit volume form \bm{\mu} with respect to \{\mathbf{e}_i\} then

\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},

for k \in \mathbb{N}[k < n].

Levi-Civita form

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

\bm{\epsilon} = \sqrt{g} \bm{\mu},

with g \overset{\text{def}}{=} \det (G), the determinant of the Gram matrix.

Therefore, if we decompose the Levi-Civita tensor by

\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},

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}.

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

\mathbf{g}(\bm{\epsilon}) = \frac{1}{\sqrt{g}} \bm{\mu}.

??? note "Proof:"

Will be added later.

We may decompose the reciprocal Levi-Civita tensor by

\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}}.