I Pullback & orthogonal projector

ergospherical
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We have a map ##\phi : M \rightarrow N##, where ##N## has dimension ##n## and ##M## has dimension ##m=n-1##. So we consider the hypersurface ##\Sigma \equiv \phi(M)## picked out by the map. We also have an orthogonal projector, ##{\bot^a}_b \equiv \delta^a_b + n^a n_b##, where ##n## is the unit normal to ##\Sigma##.

The exercise is to just verify that the expected properties hold, i.e. that

(i) for a vector ##V##, that ##\phi_{\star} V = \bot (\phi_{\star} V)##
(ii) for a ##(0,s)## tensor ##\omega##, that ##(\phi^{\star} \omega) = \phi^{\star}(\bot \omega)##
(iii) for a ##(r,0)## tensor ##T##, that ##\phi_{\star} T = \bot(\phi_{\star} T)##

I can reason for (i) - i.e. that since ##V## is the tangent vector to a curve ##\lambda : I \rightarrow M##, and the push-forward curve ##\phi \circ \lambda : I \rightarrow N## must lie completely in ##\Sigma##, so ##\phi_{\star} V## must be tangent to ##\Sigma##. So ##n_b (\phi_{\star} V)^b = 0## and:$${\bot^a}_b (\phi_{\star} V)^b = (\phi_{\star} V)^a + n^a n_b (\phi_{\star} V)^b = (\phi_{\star} V)^a$$For (ii), I'm missing a step or two. I would start just from the basics,\begin{align*}
(\phi^{\star} \omega)(X_1, \dots, X_s) &= \omega(\phi_{\star} X_1, \dots, \phi_{\star} X_s) \\
&= \omega( \bot (\phi_{\star} X_1), \dots, \bot (\phi_{\star} X_s) )
\end{align*}What next? I can't use the same reasoning as before to argue that ##\phi^{\star} \omega## is tangent to ##\Sigma##, because ##\omega## isn't obtained from a push-forward of a curve (it's instead defined by its action on vectors). Any ideas...?
 
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ergospherical said:
We have a map ##\phi : M \rightarrow N##, where ##N## has dimension ##n## and ##M## has dimension ##m=n-1##. So we consider the hypersurface ##\Sigma \equiv \phi(M)## picked out by the map. We also have an orthogonal projector, ##{\bot^a}_b \equiv \delta^a_b + n^a n_b##, where ##n## is the unit normal to ##\Sigma##.
What kind of manifolds are M and N? In a Riemannian manifold I would expect a minus sign in the orthogonal projection. In a pseudo-Riemannian one I would expect it to depend on the nature of ##n##.

ergospherical said:
The exercise is to just verify that the expected properties hold, i.e. that

(i) for a vector ##V##, that ##\phi_{\star} V = \bot (\phi_{\star} V)##
(ii) for a ##(0,s)## tensor ##\omega##, that ##(\phi^{\star} \omega) = \phi^{\star}(\bot \omega)##
(iii) for a ##(r,0)## tensor ##T##, that ##\phi_{\star} T = \bot(\phi_{\star} T)##

I can reason for (i) - i.e. that since ##V## is the tangent vector to a curve ##\lambda : I \rightarrow M##, and the push-forward curve ##\phi \circ \lambda : I \rightarrow N## must lie completely in ##\Sigma##, so ##\phi_{\star} V## must be tangent to ##\Sigma##. So ##n_b (\phi_{\star} V)^b = 0## and:$${\bot^a}_b (\phi_{\star} V)^b = (\phi_{\star} V)^b + n^a n_b (\phi_{\star} V)^b = (\phi_{\star} V)^b$$
Yes.

ergospherical said:
For (ii), I'm missing a step or two. I would start just from the basics,\begin{align*}
(\phi^{\star} \omega)(X_1, \dots, X_s) &= \omega(\phi_{\star} X_1, \dots, \phi_{\star} X_s) \\
&= \omega( \bot (\phi_{\star} X_1), \dots, \bot (\phi_{\star} X_s) )
\end{align*}What next? I can't use the same reasoning as before to argue that ##\phi^{\star} \omega## is tangent to ##\Sigma##, because ##\omega## isn't obtained from a push-forward of a curve (it's instead defined by its action on vectors). Any ideas...?
For a one-form ##\omega## and a vector ##X## in ##N##, write out ##\omega(\perp X)## on component form. What do you notice? What happens if you do this with (0,s) tensor?
 
You're right that I didn't fully specify the problem - more generally ##{\bot^a}_b = \delta^a_b \pm n^a n_b## where +/- holds for ##\Sigma## space/time-like. In this case I have assumed a sub manifold which is everywhere space like.

r.e. your answer, just considering a one-form for brevity,\begin{align*}
\omega( \bot(\phi_{\star} X)) &= \omega({\bot^a}_b (\phi_{\star} X)^b e_a) \\
&= {\bot^a}_b \omega_a (\phi_{\star} X)^b \\
&= (\bot \omega)_b (\phi_{\star} X)^b \\
&= (\bot{\omega})(\phi_{\star} X) \\
&= \phi^{\star} (\bot \omega)(X)
\end{align*}

And the generalization to an (0,s) tensor is straightforward due to multi linearity. For some reason I was getting confused about setting ##(\bot \omega) \equiv \bot(\omega)##. Obviously that's how it's defined...!
 
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