Derivatives of differential form functions ed
Will be needed in field theory with differential forms
Starting point ed
\(\bigwedge V\) is the Graßmann-Algebra of a vector space \(V\) with \(\bigwedge^k V\) the subspace of k-(multi-)vectors. \(\Omega^k\) is the space of k-forms (Differentialformen). Our space dimension is n.
Let\[ f : \bigwedge^k V \rightarrow \bigwedge^l V\]be a function mapping k-vectors onto l-vectors. For a k-form \(\omega \in \Omega^k\), we have\[ f(\omega) \in \Omega^l \]so we also have a mapping \(f:\Omega^k \rightarrow \Omega^l\).
We could allow general functions \(f : \bigwedge V \rightarrow \bigwedge V\), but that will create problems later...
Definition ed
(Using a form of Fréchet derivative)
The function \(f\) is differentiable, if there exists a (l-k)-form \( A = \frac{\partial f}{\partial \omega} \), so that
\[ \lim_{\epsilon \rightarrow 0} \frac{ \| f(\omega + \epsilon) - f(\omega) - A \wedge \epsilon \| }{ \| \epsilon \| } = 0 \text{ with } \epsilon \in \bigwedge^k V \]or\[ \lim_{\epsilon \rightarrow 0} \frac{ \| f(\omega + \epsilon \phi) - f(\omega) - \epsilon A \wedge \phi \| }{ |\epsilon| \| \phi \| } = 0, \quad \forall \phi \in \bigwedge^k V \]
\(f\) can only be differentiable, if \(l \gt k\).
Not even all linear \(f\) will be differentiable (see examples). We could have used a weaker definition with \(A \epsilon\) instead of \(A \wedge \epsilon\), turning \( \frac{\partial f}{\partial \omega} \) into a linear operator \(\operatorname{Hom}(\bigwedge V, \bigwedge V)\). But again, that would lead to problems...
Examples ed
| \(f(\omega)\) | \(\omega\) | \(\mu \wedge \omega\) | \(\omega^p = \omega \wedge \dots \wedge \omega\) | \(\star \langle \omega, \omega \rangle\) | \(\star \omega\) |
|---|---|---|---|---|---|
| \(f'\) | \(1\) | \(\mu\) | \(\begin{cases}p \omega^{p-1} & k \le n/2 \text{ even} \dots \\ 0 & \dots\end{cases}\) | \(2 (-1)^{k(n-k)} \star \omega\) | \(\star\) |
- a) \( f(\omega) = \omega \)
- b) \( f(\omega) = \star \omega \)
- (not differentiable, only in the weaker sense of a linear operator \(f' \circ = \star \circ \))
- c) \( f(\omega) = \mu \wedge \omega \)
- d) \( f(\omega) = \omega \wedge \star \omega = \langle \omega, \omega \rangle dV \)
- e) \( f(\omega) = \omega \wedge \omega \)
- \( \Rightarrow \quad f' = 0\) for k odd and \(f' = 2 \omega\) for k even.
Rules ed
Linearization ed
By definition we have the linear approximation\[ f(w_0 + \epsilon) \approx f(\omega_0) + \frac{\partial f}{\partial \omega}(\omega_0) \wedge \epsilon \]
Product rule ed
Let \( f(\omega) = g(\omega) \wedge h(\omega) \) with \(g, h\) differentiable of degrees \(l, m\). Linearization yields\begin{align}f(\omega) & \approx (g(\omega_0) + g' \wedge \epsilon) \wedge (h(\omega_0) + h' \wedge \epsilon) \\& \approx g_0 \wedge h_0 + g_0 \wedge h' \wedge \epsilon + g' \wedge \epsilon \wedge h_0 \\& = g_0 \wedge h_0 + g_0 \wedge h' \wedge \epsilon + (-1)^{km} g' \wedge h_0 \wedge \epsilon\end{align}Resulting in\[ \begin{array}{|c|} \hline \\ f' = g \wedge h' + (-1)^{km} g' \wedge h \\ \\ \hline \end{array} \]
Chain rule ed
Linearizing \((f \circ g)(\omega) = f(g(\omega))\):\begin{align}&\approx f \left( g(\omega_0) + \frac{\partial g}{\partial \omega}(\omega_0) \wedge \epsilon \right) \\&\approx f(g(\omega_0)) + \frac{\partial f}{\partial g}(g(\omega_0)) \wedge \frac{\partial g}{\partial \omega}(\omega_0) \wedge \epsilon\end{align}leading to the simple formula\[ \begin{array}{|c|} \hline \\ (f \circ g)' = f' \wedge g' \\ \\ \hline \end{array} \]
Outer derivative ed
Since \(f(\omega(x))\) is an \(l\)-form on \(M\), we can compute \( d(f(\omega)) \). To find the value at a point \(x_0 \in M\) and define \(\bar \omega(x) := \omega(x) - \omega(x_0) = \omega - \omega_0\). This definition depends on an arbitrary choice of coordinate system but allows linearization around \(x_0\):\begin{align}f(\omega) &= f(\omega_0 + \bar \omega) \\&\approx f(\omega_0) + f'(\omega_0) \wedge \bar \omega \\\Rightarrow d_{|x_0}(f(\omega)) &= d_{|x_0}(f(\omega_0)) + d_{|x_0}\big(f'(\omega_0) \wedge \bar \omega\big) \\&= d_{|x_0}(f(\omega_0)) + d_{|x_0} f'(\omega_0) \wedge \bar \omega(x_0) + (-1)^{l-k} f'(\omega_0) \wedge d_{|x_0}\bar \omega\end{align}And since \(\bar \omega(x_0)=0\) and \(d\bar\omega = d\omega\):\[ \begin{array}{|c|} \hline \\ d(f(\omega)) = (df)(\omega) + (-1)^{l-k} \, f' \wedge d\omega \\ \\ \hline \end{array} \label{eq:form-outer-derivative} \]
Note: the notation \((df)(\omega)\) is a shorthand for \(d(f(\omega_0))\) with constant \(\omega_0\), which makes sense in any fixed coordinate system. Since both of the other terms in eq.~\eqref{eq:form-outer-derivative} are geometric objects, \((df)\) also has to be independent of the choice of coordinates.
More parameters ed
Let's generalize to functions \(f(\omega_1,\dots,\omega_s,x)\) with multiple parameters and point-dependence.
The only significant change is the outer derivative\[ d(f(\omega_1,\dots,\omega_s,x)) = \sum_i (-1)^{l-k_i} \frac{\partial f}{\partial \omega_i} \wedge d\omega_i + (df)(\omega_1,\dots,\omega_s,x) \]with \((df)(\dots)\) the outer derivative of \(f\) for constant \(\omega_1,\dots,\omega_s\).
Hodge Star Operator ed
Since the hodge star \(f(\omega) = \star \omega\) is only differentiable as an operator, but is showing up frequently, we have to make sure, it still fulfills most of the rules we just derived.
We will denote \(\star_k\) the restriction to \(k\)-forms. And call \((\star \omega)' = \star_k\) by abuse of notation.
Linearization ed
Since \(\star\) is linear, it can obviously be linearized:\begin{align}\star(\omega + \epsilon) &\approx \star \omega + \star \epsilon\end{align}
Chain rule ed
Having \(\star\) as the outer function is very straight forward:\begin{align}f(\star(\omega + \epsilon)) &= f(\star \omega + \star \epsilon ) \\&\approx f(\star \omega) + f'(\star \omega) \wedge \star \epsilon \\\Rightarrow (f \circ \star_k)' &= f' \wedge \star_k\end{align}
With an inner \(\star\), the rule still applies, but we have to clearify the notation:\begin{align}\star(f(\omega + \epsilon)) &\approx \star\left( f(\omega) + f' \wedge \epsilon \right) \\&= \star f(\omega) + \star \left(f' \wedge \epsilon \right) \\\Rightarrow (\star_l \circ f)' &= \star_l \wedge f'\end{align}Here we call \(\star \wedge\) the non-greedy star operator, i.e. \((\star \wedge \alpha) \wedge \beta = \star( \alpha \wedge \beta)\).
Product rule ed
%Having \(\star\) as the outer function is very straight forward:\((f\wedge \star \omega)'\) again follows the rules:\begin{align}f(\omega + \epsilon) \wedge \star (\omega + \epsilon) &\approx \big(f(\omega) + f' \wedge \epsilon \big) \wedge \big( \star \omega + \star \epsilon \big) \\&\approx f(\omega) \wedge \star \omega + f'\wedge \epsilon \wedge \star \omega + f \wedge \star \epsilon \\&= f(\omega) \wedge \star \omega + (-1)^{k(n-k)} f' \wedge \star \omega\wedge \epsilon + f \wedge \star \epsilon \\\Rightarrow (f \wedge \star_k)' &= f \wedge \star_k + (-1)^{k(n-k)} f \wedge \star_k\end{align}
\((\star \omega \wedge f)'\) needs clearifications...:\begin{align}\star (\omega + \epsilon) \wedge f(\omega + \epsilon) &\approx \big( \star \omega + \star \epsilon \big) \wedge \big(f(\omega) + f' \wedge \epsilon \big) \\&\approx \star \omega \wedge f(\omega) + \star \omega \wedge f'\wedge \epsilon + \star \epsilon \wedge f \\&\approx \star \omega \wedge f(\omega) + \star \omega \wedge f'\wedge \epsilon + (-1)^{(n-k)l} f \wedge \star \epsilon \\\end{align}% \Rightarrow (f \wedge \star_k)' &= f \wedge \star_k + (-1)^{k(n-k)} f \wedge \star_k