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 \(\{\mathbf{\hat e}^i\}_{i=1}^n\), a field \(F\) 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 \(\lambda \in F\).

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

Proof:

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.

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.

Proof:

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.

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 \(\mathbf{v}_{k+1}, \dots, \mathbf{v}_n \in V\).

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

\[ \begin{align*} \bm{\mu} \lrcorner \mathbf{u}_1 \lrcorner \dots \lrcorner \mathbf{u}_k &= u_1^{i_1} \cdots u_k^{i_k} (\bm{\mu} \lrcorner \mathbf{e}_{i_1} \lrcorner \dots \lrcorner \mathbf{e}_{i_k}), \\ &= u_1^{i_1} \cdots u_k^{i_k} \mu_{i_1 \dots i_n} (\mathbf{\hat e}^{i_{k+1}} \wedge \cdots \wedge \mathbf{\hat e}^{i_{n}}), \end{align*} \]

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

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

Proof:

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

\[ \mathbf{g}(\bm{\epsilon}) = \epsilon^{i_1 \dots i_n} \mathbf{e}_{i_1} \otimes \cdots \otimes \mathbf{e}_{i_n} = \epsilon^{|i_1 \dots i_n|} \mathbf{e}_{i_1} \wedge \cdots \wedge \mathbf{e}_{i_n}, \]

then we have \(\epsilon^{i_1 \dots i_n} = \frac{1}{\sqrt{g}} \mu^{i_1 \dots i_n}\) and \(\epsilon^{|i_1 \dots i_n|} = \frac{1}{\sqrt{g}}\).