Hodge dual computations
by Stéphane Haussler
In this page, I present a straightforward algorithmic method to calculate the
inner product between k–vectors (or k–forms). The dicussion in the article on
Hodge duality fundamentals provides geometric intuition. Here, I offer an alternative
approach to compute the inner product of k–forms, as well as the Hodge dual of
k–forms, with a simple algorithmic method using the interior product
\(⌟\).
My use of the interior product might be non-standard, and I have not
investigated whether what follows is established. It certainly works as I
intuitively expect it should. You may also just find this method a usefull as
an efficient trick to determine the inner product between k–forms, or the
Hodge dual of k-forms.
In this page, I call the top-dimensional differential form the
pseudo-scalar. Specifically, the pseudo-scalar is:
The term pseudo-scalar is chosen because the Hodge dual of these
top-dimensional differential forms results in scalar quantities. This naming
convention highlights the relationship between these forms and scalar fields
through the Hodge duality.
Inner product on k–froms
To establish a common foundation, recall that the inner product of k–vectors
is equal to that of k–forms. For basis vectors and covectors, the inner
product is the metric. In flat spacetime, this yields the Minkowski
metric \(η\):
\[∂_μ · ∂_ν = dx^μ · dx^ν = η^{μν} = η_{μν}\]
Considering basis vectors, dual covectors are defined as:
\[dx^μ \left( ∂_ν \right) = δ^μ_ν\]
In this sense, 1–forms measure vectors, and the measure of a basis vector
\(∂_ν\) through it corresponding basis covector \(dx^μ\) is one. The
inner product, denoted with \(·\) or \(\braket{|}\), measures the shadow of one vector onto another.
The standard interior product is an operation between a vector and a form. It
consist of rearanging the form to bring the corresponding covector to the
front, and applying the vector to that front slot. This operation transforms a
k–form to a (k-1)–form. For example:
\[\newcommand{\⌟}{\:⌟\:}
∂_y \⌟ dx ∧ dy ∧ dz = ∂_y \⌟ (- dy ∧ dx ∧ dz) = - dy \left( ∂_y \right) ∧ dx ∧ dz = - dx ∧ dz\]
Here, we have applied the basis vector \(∂_ν\) to the basis form
\(dx^μ\).
We equate the interior product to the dot product for a vector pair to the
interior product \(⌟\):
\[\newcommand{\⌟}{\:⌟\:}
∂_ν \⌟ ∂_μ = ∂_μ · ∂_ν = η_{μν}\]
For covectors:
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
dx^ν \⌟ dx^μ = dx^μ · dx^ν = η^{μν} \\\end{split}\]
The interior product of a vector can be applied to a general k–vector. For a
basis vector acting on a basis 2–vector, we bring the corresponding basis
1–vector to the front and apply the vector to the first slot:
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
dt \⌟ dx ∧ dt &= dt \⌟ \left( - dt ∧ dx \right) \\
&= - ( dt · dt ) \; dx \\
&= - dx\end{split}\]
Simmilarly, the interior product of a basis vector can be taken with a
3–vector or a 4–vector. This provides a straightforward algorithmic approach to calculate the inner product with no thinking involved:
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
(dt ∧ dx) · (dt ∧ dx) &= dx \⌟ dt \⌟ dt ∧ dx \\
&= dx \⌟ (dt · dt) ∧ dx \\
&= dx \⌟ dx \\
&= dx · dx \\
&= -1\end{split}\]
Minkowski space
Inner product
We can systematicall apply the procedure to obtain the same result as above:
1–forms
\[\begin{split}\begin{array}{c|rrrr}
& dt & dx & dy & dz \\
\hline
dt & +1 & 0 & 0 & 0 \\
dx & 0 & -1 & 0 & 0 \\
dy & 0 & 0 & -1 & 0 \\
dz & 0 & 0 & 0 & -1 \\
\end{array}\end{split}\]
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
\begin{alignedat}{5}
dt · dt &=& dt &\⌟& dt & = +1 \\
dx · dt &=& dx &\⌟& dx & = -1 \\
dy · dt &=& dy &\⌟& dy & = -1 \\
dz · dt &=& dz &\⌟& dz & = -1 \\
\end{alignedat}\end{split}\]
2–forms
\[\begin{split}\begin{array}{c|rrrrrr}
& dt ∧ dx & dt ∧ dy & dt ∧ dz & dy ∧ dz & dz ∧ dx & dx ∧ dy \\
\hline
dt ∧ dx & -1 & 0 & 0 & 0 & 0 & 0 \\
dt ∧ dy & 0 & -1 & 0 & 0 & 0 & 0 \\ dt ∧ dz & 0 & 0 & -1 & 0 & 0 & 0 \\
dy ∧ dz & 0 & 0 & 0 & +1 & 0 & 0 \\
dz ∧ dx & 0 & 0 & 0 & 0 & +1 & 0 \\
dx ∧ dy & 0 & 0 & 0 & 0 & 0 & +1 \\
\end{array}\end{split}\]
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
\newcommand{\·}{\:·\:}
\begin{alignedat}{5}
(& dt ∧ dx &) \· (& dt ∧ dx &) =& dx &\⌟& dt &\⌟& dt ∧ dx &= + dx &\⌟& dx &= -1 \\
(& dt ∧ dy &) \· (& dt ∧ dy &) =& dy &\⌟& dt &\⌟& dt ∧ dy &= + dy &\⌟& dy &= -1 \\
(& dt ∧ dy &) \· (& dt ∧ dz &) =& dz &\⌟& dt &\⌟& dt ∧ dz &= + dz &\⌟& dz &= -1 \\
(& dy ∧ dz &) \· (& dy ∧ dz &) =& dz &\⌟& dy &\⌟& dy ∧ dz &= - dz &\⌟& dz &= +1 \\
(& dz ∧ dx &) \· (& dz ∧ dx &) =& dx &\⌟& dz &\⌟& dz ∧ dx &= - dx &\⌟& dx &= +1 \\ (& dx ∧ dy &) \· (& dx ∧ dy &) =& dy &\⌟& dx &\⌟& dx ∧ dy &= - dy &\⌟& dy &= +1 \\
\end{alignedat}\end{split}\]
3–forms
\[\begin{split}\begin{array}{c|rrrr}
& dx ∧ dy ∧ dz & dt ∧ dy ∧ dz & dt ∧ dz ∧ dx & dt ∧ dx ∧ dy \\
\hline
dx ∧ dy ∧ dz & -1 & 0 & 0 & 0 \\
dt ∧ dy ∧ dz & 0 & +1 & 0 & 0 \\
dt ∧ dz ∧ dx & 0 & 0 & +1 & 0 \\ dt ∧ dx ∧ dy & 0 & 0 & 0 & +1 \\
\end{array}\end{split}\]
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
\newcommand{\·}{\:·\:}
\small
\begin{alignedat}{5}
(& dx ∧ dy ∧ dz &) \· (& dx ∧ dy ∧ dz &) =& dz \⌟ dy \⌟ dx \⌟ dx ∧ dy ∧ dz &=& - dz \⌟ dy \⌟ dy ∧ dz &= + dz \⌟ d = -1 \\
(& dt ∧ dy ∧ dz &) \· (& dt ∧ dy ∧ dz &) =& dz \⌟ dy \⌟ dt \⌟ dt ∧ dy ∧ dz &=& + dz \⌟ dy \⌟ dy ∧ dz &= - dz \⌟ d = +1 \\
(& dt ∧ dz ∧ dx &) \· (& dt ∧ dz ∧ dx &) =& dx \⌟ dz \⌟ dt \⌟ dt ∧ dz ∧ dx &=& + dx \⌟ dz \⌟ dz ∧ dx &= - dx \⌟ d = +1 \\ (& dt ∧ dx ∧ dy &) \· (& dt ∧ dx ∧ dy &) =& dy \⌟ dx \⌟ dt \⌟ dt ∧ dx ∧ dy &=& + dy \⌟ dx \⌟ dx ∧ dy &= - dy \⌟ d = +1 \\
\end{alignedat}\end{split}\]
4–forms
\[\begin{split}\begin{array}{c|c}
& dt ∧ dx ∧ dy ∧ dz \\ \hline
dt ∧ dx ∧ dy ∧ dz & -1 \\
\end{array}\end{split}\]
\[\begin{split}\newcommand{\⌟}{\:⌟\:}
\newcommand{\·}{\:·\:}
(dt ∧ dx ∧ dy ∧ dz) \· (dt ∧ dx ∧ dy ∧ dz) &= dz \⌟ dy \⌟ dx \⌟ dt \⌟ dt ∧ dx ∧ dy ∧ dz \\
&= dz \⌟ dy \⌟ dx \⌟ dx ∧ dy ∧ dz \\
&= -1 dz \⌟ dy \⌟ ∧ dy ∧ dz \\
&= +1 dz \⌟ ∧ dz \\
&= -1\end{split}\]
Hodge duals
1-forms
\[\begin{split}⋆ dt &=& dx ∧ dy ∧ dz \\
⋆ dx &=& dt ∧ dy ∧ dz \\
⋆ dy &=& dt ∧ dz ∧ dx \\
⋆ dz &=& dt ∧ dx ∧ dy \\\end{split}\]
Calculations
Take the interior product with the pseudoscalar
\[\begin{split}⋆ dt &=& dt &\⌟ dt ∧ dx ∧ dy ∧ dz \\
⋆ dx &=& dx &\⌟ dt ∧ dx ∧ dy ∧ dz \\
⋆ dy &=& dy &\⌟ dt ∧ dx ∧ dy ∧ dz \\
⋆ dz &=& dz &\⌟ dt ∧ dx ∧ dy ∧ dz \\\end{split}\]
Reorder
\[\begin{split}⋆ dt &=& + dt &\⌟ dt ∧ dx ∧ dy ∧ dz \\
⋆ dx &=& - dx &\⌟ dx ∧ dt ∧ dy ∧ dz \\
⋆ dy &=& - dy &\⌟ dy ∧ dt ∧ dz ∧ dx \\
⋆ dz &=& - dz &\⌟ dz ∧ dt ∧ dx ∧ dy \\\end{split}\]
Apply the interior product
\[\begin{split}⋆& dt &=& + (& dt &\·& dt &) \: & dx ∧ dy ∧ dz \\
⋆& dx &=& - (& dx &\·& dx &) \: & dt ∧ dy ∧ dz \\
⋆& dy &=& - (& dy &\·& dy &) \: & dt ∧ dz ∧ dx \\
⋆& dz &=& - (& dz &\·& dz &) \: & dt ∧ dx ∧ dy \\\end{split}\]
Apply numerical values and conclude
\[\begin{split}⋆ dt &=& dx ∧ dy ∧ dz \\
⋆ dx &=& dt ∧ dy ∧ dz \\
⋆ dy &=& dt ∧ dz ∧ dx \\
⋆ dz &=& dt ∧ dx ∧ dy \\\end{split}\]
2-forms
\[\begin{split}⋆ dt ∧ dx &= \\
⋆ dt ∧ dy &= \\
⋆ dt ∧ dz &= \\
⋆ dy ∧ dz &= \\
⋆ dz ∧ dx &= \\
⋆ dx ∧ dy &= \\\end{split}\]
3-forms
\[\begin{split}⋆ dx ∧ dy ∧ dz &= \\
⋆ dt ∧ dy ∧ dz &= \\
⋆ dt ∧ dz ∧ dx &= \\
⋆ dt ∧ dx ∧ dy &= \\\end{split}\]
4-forms
\[⋆ dt ∧ dx ∧ dy ∧ dz =\]