6 KiB
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 spaceV
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
withk \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}}
.