Hodge dual computations
by Stéphane Haussler
In this page, I present a straightforward algorithmic method to calculate the
inner product of k–vectors (respectively k–forms). While my article on
Hodge duality fundamentals provides geometric intuition, the discussion here provides
an alternative approach to compute both the inner product of k–forms and the
Hodge dual of k–forms, with a simple algorithmic method using the interior
product \(⌟\).
This method might be non-standard, and I have not investigated whether it is
established. However it certainly works as I intuitively expect it should and
provides fast and direct computations.
The following assumes a solid understanding of the exterior product and Élie
Cartan’s differential forms, as well as knowledge of how to apply the interior
product \(⌟\).
The top-dimensional differential form is called the pseudo-scalar.
Specifically, the pseudo-scalar is:
The term pseudo-scalar is chosen because the Hodge dual of these
top-dimensional differential forms are scalars. They are dual to scalar, but do
not behave like scalars.
Inner product on k–froms
Let us first 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 \(∂_ν\), dual covectors \(dx^μ\) are defined
as:
\[dx^μ \left( ∂_ν \right) = δ^μ_ν\]
Where \(δ^μ_ν\) is the kronecker delta.
The 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 then
apply 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
\[⋆ dt ∧ dx ∧ dy ∧ dz =\]