Derivatives of differential form functions ed

Will be needed in field theory with differential forms

Table of Contents

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

a) \( f(\omega) = \omega \)
\[ 0 = \lim_{\epsilon \rightarrow 0} \frac{ \| \omega + \epsilon - \omega - f' \wedge \epsilon \| }{ \| \epsilon \| } \quad \Rightarrow \quad f' = 1 \]

b) \( f(\omega) = \star \omega \)
\[ 0 = \lim_{\epsilon \rightarrow 0} \frac{ \| \star \omega + \star \epsilon - \star \omega - f' \wedge \epsilon \| }{ \| \epsilon \| } \quad \Rightarrow \quad \text{ :-(} \]
(not differentiable, only in the weaker sense of a linear operator \(f' \circ = \star \circ \))

c) \( f(\omega) = \mu \wedge \omega \)
\[ 0 = \lim_{\epsilon \rightarrow 0} \frac{ \| \mu \wedge \omega + \mu \wedge \epsilon - \mu \wedge \omega - f' \wedge \epsilon \| }{ \| \epsilon \| } \quad \Rightarrow \quad f' = \mu \]

d) \( f(\omega) = \omega \wedge \star \omega = \langle \omega, \omega \rangle dV \)
\[ 0 = \lim_{\epsilon \rightarrow 0} \frac{ \| \dots + \omega \wedge \star \epsilon + \epsilon \wedge \star \omega - f' \wedge \epsilon \| }{ \| \epsilon \| } = \lim_{\epsilon \rightarrow 0} \frac{ \| 2 \langle \omega, \epsilon \rangle dV - f' \wedge \epsilon \| }{ \| \epsilon \| } \quad \Rightarrow \quad f' = 2 \star \omega \]

e) \( f(\omega) = \omega \wedge \omega \)
\[ 0 = \lim_{\epsilon \rightarrow 0} \frac{ \| \dots \omega \wedge \epsilon + \epsilon \wedge \omega - f' \wedge \epsilon \| }{ \| \epsilon \| } = \lim_{\epsilon \rightarrow 0} \frac{ \| \omega \wedge \epsilon + (-1)^k \omega \wedge \epsilon - f' \wedge \epsilon \| }{ \| \epsilon \| } \]
\( \Rightarrow \quad f' = 0\) 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 ===

Linearizing \(f(g(\omega))\):\[ \approx f \left( g(\omega_0) + \frac{\partial g}{\partial \omega}(\omega_0) \wedge \Delta \omega \right) \]\[ \approx f(g(\omega_0)) + \frac{\partial f}{\partial g}(g(\omega_0)) \wedge \frac{\partial g}{\partial \omega}(\omega_0) \wedge \Delta \omega \]So:\[ \begin{array}{|c|} \hline \\ \frac{\partial f(g(\omega))}{\partial \omega} = f' \wedge g' \\ \\ \hline \end{array} \]

Outer derivative ed

Since \(f(\omega)\) is a l-form, we can ask, what is \( d(f(\omega)) \)?

To be more precise, we need to write \(f(\omega(x))\) and linearize at the point \(x_0\). The outer differential at the point \(x_0\) is then\[ d_{|x=x_0}(f(\omega(x_0 + \Delta x))) \]\[ = d_{|x=x_0} \left( \underbrace{f(\omega(x_0))}_{const} + \underbrace{\frac{\partial f}{\partial \omega}(\omega(x_0))}_{const} \wedge \left( \omega(x_0 + \Delta x) - \underbrace{\omega(x_0)}_{const} \right) \right) \]\[ = d_{|x=x_0} \left(\frac{\partial f}{\partial \omega}(\omega(x_0)) \wedge \left( \omega(x_0 + \Delta x) \right) \right) \]Using the rule \(d(\alpha\wedge\beta) = d\alpha \wedge \beta + (-1)^{\operatorname{deg} \alpha} \alpha \wedge d\beta\)\[ = (-1)^{l-k} \frac{\partial f}{\partial \omega}(\omega(x_0)) \wedge d_{|x=x_0} \omega(x_0 + \Delta x) \]So we have\[ \begin{array}{|c|} \hline \\ d(f(\omega)) = (-1)^{l-k} \frac{\partial f}{\partial \omega} \wedge d\omega \\ \\ \hline \end{array} \]

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 \(\omega_1,\dots,\omega_s\).

Categories: Mathematik, Gedanken zur Mathematik