How do you show that this wedge derivative is equal to a certain derivative, $\dot\partial \wedge P(\dot{a}(x)) = P(\partial \wedge a)$?

Question

I would like to show that $\dot\partial \wedge P(\dot{a}) = P(\partial \wedge a)$, where $P$ is a linear function $P(a(x)) = a(x)\cdot I(x)I^{-1}(x)$. The dot is used to indicate that the derivative is acting only on $a$, not on the $I$'s.

Note that $I$ is a pseudoscalar of a lower dimensional manifold that lives in a larger dimensional ambient space. The vector $a$ is also in the lower dim manifold. In general, $\partial \wedge a(x)$ has a component that is outside the manifold, and a component that is inside the manifold. This fact might be coming into play. For more context, the claim in Doran and Lasenby (pg 210) is that

$$ \partial \wedge a = \partial \wedge (P(a)) = \dot \partial \wedge \dot{P}(a) + P(\partial \wedge a) = \dot{\partial} \wedge \dot{P}(a) + D\wedge a $$

What I would like to prove is the unspoken part where $\dot \partial \wedge P(\dot{a})=P(\partial \wedge a)$.

Attempt: \begin{align*} \dot\partial \wedge P(\dot{a}) &= \langle \dot\partial \wedge (\dot a\cdot II^{-1}) \rangle_2 \\ &= \langle \dot\partial (\dot{a}I - \dot{a}\wedge I) I^{-1} \rangle_2 \\ &= \langle \dot\partial \dot{a}I) I^{-1} \rangle_2 \\ &= \langle \dot \partial \dot{a}\rangle_2 \\ &= \partial \wedge a \quad \textrm{(wrong)} \end{align*}

Answer 1

Equation (6.199), page 204 of Doran and Lasenby: $$ P(\partial) = \partial. $$ This is because $\partial$ is the tangential derivative that only differentiates along directions tangential to the manifold. It then follows simply that $$ \dot\partial\wedge P(\dot a) = P(\dot\partial)\wedge P(\dot a) = P(\dot\partial\wedge\dot a). $$

Answer 2

First expand $\partial$ in a basis $\partial = e^i \partial_i$ and note that the scalar derivative operator is mobile. It can be moved to the vector that it is operating on.

\begin{equation*} \dot\partial \wedge P(\dot{a}) = \dot\partial \wedge (\dot{a}\cdot II^{-1}) = e^i \wedge ((\partial_i a)\cdot II^{-1}) \end{equation*}

$\partial_i a$ contains in-manifold and out-of-manifold terms. The out-of-manifold term is removed by the projection. So, we are free to add another projection operator over the overall expression. This in turn frees us to remove the internal projection that removes the out-of-manifold term from $\partial_i a$: For any in-manifold r-blade $A_r$ and arbitary-manifold s-blade $B_s$

\begin{align*} A_r \wedge P(B_s) &= P(A_r \wedge P(B_s)) \\ &=P(A_r) \wedge P(P(B_s)) \\ &= P(A_r) \wedge P(B_s) \\ &= P(A_r \wedge B_s) \end{align*} which extends to multivectors by linearity.

\begin{equation*} P(e^i \wedge (\partial_i a)) = P(\partial \wedge a) \end{equation*}