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Projecting push forward of a vector
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[QUOTE="Augbrah, post: 5470910, member: 593705"] [h2]Homework Statement [/h2] Say we have two manifolds N(dim d) and M(dim d-1). Let Φ: M →N be a diffeomorphism where Σ = Φ[M] is hypersurface in N. Let n be unit normal field (say timelike) on Σ and ⊥ projector (in N) defined by: [tex]⊥^a_b = \delta^a_b + n^a n_b[/tex] Where acting on (s, 0) tensor projection operator is: [itex]⊥T^{a_1 a_2 ... a_r}=⊥^{a_1} _{b_1}...⊥^{a_s} _{b_s}T^{b_1 ... b_s}[/itex] How to show that for a vector V at p on M, the pushforward of V satisfies: [tex]Φ_*(V)=⊥(Φ_*V)[/tex] (I can do that) And then generalize to (s, 0) tensor: [tex]Φ_*(T)=⊥(Φ_*T)[/tex] [h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2] [/B] In component form we would get: [itex] (⊥(Φ_*V))^a = ⊥^a_b (Φ_*V)^b=(\delta^a_b + n^a n_b)(Φ_*V)^b=(Φ_*V)^a + n^an_b(Φ_*V)^b[/itex]. So remains to show that the last bit vanishes. The last term vanishes since [itex]a_b[/itex] is normal to [itex](Φ_*V)^b[/itex], so we're done. Now to generalize for (s, 0) we have: [tex](⊥(Φ_*T))^{a_1 a_2 ... a_s} = ^{a_1} _{b_1}...⊥^{a_s} _{b_s}(Φ_*T)^{b_1 ... b_s} = (\delta^{a_1}_{b1} + n^{a_1} n_{b_1}) ... (\delta^{a_s}_{b_s} + n^{a_s} n_{b_s})(Φ_*T)^{b_1 ... b_s}[/tex] My problem is that I'm not quite sure what does it mean to act [itex]n_{b_s} (Φ_*T)^{b_1 ... b_s}[/itex], notion of orthogonality is easily understood for two vectors, but for vector and a tensor I'm not sure. Any ideas much appreciated! [/QUOTE]
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Projecting push forward of a vector
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