How do you show that the projection operator for a diffeomorphism is $\textsf P^\prime = \textsf {fP}\textsf{f}^{-1}?$

Question

The pseudoscalar $I(x)$ defines a projection operator that projects an arbitrary multivector onto the component that is intrinsic to the manifold,

\begin{equation} \textsf P(A(x)) = \begin{cases} A_r(x)\cdot I(x) I^{-1}(x) = A_r \cdot II^{-1} & r \leq n \\ 0 & r > n \end{cases} \end{equation}

The action of any linear function on a pseudoscalar satisfies

$$ \textsf f(I) = (\det \textsf f) I $$

If $\textsf f$ is a diffeomorphism mapping multivector fields to some primed manifold, we also have (equation 2.272, Doran and Lasenby)

$$ \textsf f(I) = (\det \textsf f)I^\prime $$

Thus $I=I^\prime$, isn't that a contradiction? Or are these identical pseudoscalars, merely represented in different coordinates?

Supposedly, we also have eq 6.287

$$ \textsf P^\prime = \textsf {fP} \textsf f^{-1} $$

which suggests that in fact $\textsf P^\prime \neq \textsf P$, meaning $A\cdot I^\prime {I^{\prime}}^{-1} \neq A\cdot I I^{-1}$. But if $I=I^\prime$, then wouldn't we get $\textsf P = \textsf P^\prime$?

Answer 1

I think what is happening is that indeed $I=I^\prime$, but $I^\prime$ may superficially "look" different because it is constructed from primed vectors.

As for $\textsf P^\prime$, I believe it is defined by $\textsf P^\prime (A^\prime) = A^\prime \cdot I^\prime {I^\prime}^{-1}$, thus we can write

\begin{align*} \textsf P^\prime(A^\prime) &= A^\prime\cdot I^\prime {I^\prime}^{-1} \\ &= A^\prime\cdot I {I}^{-1} \\ &= \textsf f \textsf f^{-1}(A^\prime\cdot I I^{-1}) \\ &= \textsf f \textsf f^{-1} \left( (A^\prime \rfloor I) \rfloor I^{-1} \right) \quad \textrm{(Dorst syntax)} \\ &= \textsf f \left( \overline {\textsf f^{-1}}^{-1} (A^\prime \rfloor I) \rfloor \textsf f^{-1}(I^{-1}) \right) \end{align*}

Now, using the general formula for an inverse $\textsf f^{-1}(B) = \frac{\overline{\textsf f} (B\rfloor I^{-1}) \rfloor I }{\det \textsf f}$, we can compute the factor

$$ \textsf f^{-1}(I^{-1}) = \frac{ \overline{\textsf{f}}(I^{-1}\rfloor I^{-1})\rfloor I }{\det \textsf f} = \frac{I }{\det \textsf f} $$

These functions have the property $\overline{\textsf g^{-1}}=\overline{\textsf g}^{-1}$, so $(\overline{\textsf{f}^{-1}})^{-1} = \overline{\textsf{f}}$.

Then

\begin{align*} \textsf P^\prime(A^\prime) &= \frac{1}{\det \textsf f} \textsf f \left( \overline{\textsf{f}} (A^\prime \rfloor I )\rfloor I^{-1} \right) \\ &= \frac{1}{\det \textsf f} \textsf f \left( \left(\overline{\overline{\textsf{f}}}^{-1} (A^\prime) \rfloor \overline{\textsf{f}}(I )\right) \rfloor I^{-1} \right) \\ &= \textsf f \left( \left(\textsf{f}^{-1} (A^\prime) \rfloor I \right) \rfloor I^{-1} \right) \\ &= \textsf{fP}\textsf{f}^{-1} (A^\prime) \end{align*}

so $\textsf P^\prime = \textsf {fPf}^{-1}$.

Answer 2

$ \newcommand\R{\mathbb R} \newcommand\diff\mathsf \newcommand\adj[1]{\overline{\diff{#1}}} $

Suppose the ambient space in which we define our $k$-dimensional vector manifold $M$ is $\R^n$. It is true that if $J$ is any $n$-vector, then for any linear $f : \R^n \to \R^n$ we have $f(J) = (\det f)J$.

But the pseudoscalar field $I(x)$ of $M$ is a $k$-vector field, so this does not apply unless $k = n$. Even then, we do not a priori have any reason to think of $M$ and $M' = f(M)$ as living in the same space, so comparing $I$ and $I'$ is meaningless. Think of the case that $k=2$; does it make sense to say that some tangent plane of $M$ and some other of $M'$ are parallel? It could given additional context, but it need not.

I think the way Doran and Lasenby introduce Eq. 6.272 is bad and extremely misleading. The equation $$ \diff f(I) = (\det\diff f)I' $$ is a definition of $\det(\diff f)$; $I$ and $I'$ have to be specified beforehand for this to make sense.

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As for $P$ and $P'$, the easiest way to get this is to use tangential derivatives $\partial$ and $\partial'$: $$ a\cdot\partial = P(a)\cdot\partial,\quad a\cdot\partial' = P'(a)\cdot\partial'. $$ This follows simply from the fact that $a\cdot\partial = 0$ when $a$ is orthogonal to $M$ (and similarly for $M'$). Using the fact that $\partial = \adj f(\partial')$ (which is just the chain rule), it follows that $$ P'(a) = a\cdot\partial'x' = a\cdot\adj f^{-1}(\partial) f(x) = P(\diff f^{-1}(a))\cdot\partial f(x) = \diff f(P(\diff f^{-1}(a))). $$