Undergrad Pullback & orthogonal projector

[ergospherical]

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...?

 

[Orodruin]

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 $n$.

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.

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?

 

[ergospherical]

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...!