I Pullback & orthogonal projector

ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,097
Reaction score
1,384
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...?
 
Last edited:
Physics news on Phys.org
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...!
 
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...
Back
Top