Derivatives of differential form functions ed

Will be needed in intuitive Hamiltonsche Feldmechanik

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

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

\[ f(w_0 + \Delta \omega) \approx f(\omega_0) + \frac{\partial f}{\partial \omega}(\omega_0) \wedge \Delta \omega \]

Product rule ed

Let \[ f(\omega) = g(\omega) \wedge h(\omega) \] with g,h differentiable of degrees l, m.

\[ 0 = \lim_{\epsilon \rightarrow 0} \frac{ \| g(\omega + \epsilon) \wedge h(\omega + \epsilon) - g(\omega) \wedge h(\omega) - f' \wedge \epsilon \| }{ \| \epsilon \| } \]
\[ = \lim_{\epsilon \rightarrow 0} \frac{ \| \left[g(\omega) + g'(\omega) \wedge \epsilon\right] \wedge \left[h(\omega) + h'(\omega) \wedge \epsilon\right] - g(\omega) \wedge h(\omega) - f' \wedge \epsilon \| }{ \| \epsilon \| } \]
\[ = \lim_{\epsilon \rightarrow 0} \frac{ \| g(\omega) \wedge h'(\omega) \wedge \epsilon + g'(\omega) \wedge \epsilon \wedge h(\omega) - f' \wedge \epsilon \| }{ \| \epsilon \| } \]
Since
\[ \epsilon \wedge h(\omega) = (-1)^{km} h(\omega) \wedge \epsilon \]
we have:
\[ \Rightarrow f' = g' \wedge h + (-1)^{km} g \wedge h' \]

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 \[x0\]. 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
\[ d(f(\omega)) = (-1)^{l-k} \frac{\partial f}{\partial \omega} \wedge d\omega \]

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