Projecting push forward of a vector

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SUMMARY

The discussion focuses on the mathematical concept of projecting the pushforward of a vector through a diffeomorphism Φ: M → N, where M and N are manifolds of dimensions d-1 and d, respectively. The user successfully demonstrates that the pushforward of a vector V at point p on M satisfies the equation Φ_*(V) = ⊥(Φ_*V) using the defined projector ⊥. The challenge lies in generalizing this result to (s, 0) tensors, specifically understanding the implications of orthogonality when applying the projector to tensors, which may involve interpreting tensors as sums of tensor products of basis vectors.

PREREQUISITES
  • Understanding of manifolds and diffeomorphisms
  • Familiarity with tensor notation and operations
  • Knowledge of orthogonal projections in differential geometry
  • Basic concepts of vector fields and normal fields
NEXT STEPS
  • Study the properties of diffeomorphisms in differential geometry
  • Learn about orthogonal projections and their applications in tensor analysis
  • Explore the concept of pushforward in the context of vector fields
  • Investigate the representation of tensors as sums of tensor products of basis vectors
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Mathematicians, physicists, and students studying differential geometry, particularly those interested in tensor analysis and the geometric interpretation of manifolds.

Augbrah
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Homework Statement


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:
⊥^a_b = \delta^a_b + n^a n_b

Where acting on (s, 0) tensor projection operator is: ⊥T^{a_1 a_2 ... a_r}=⊥^{a_1} _{b_1}...⊥^{a_s} _{b_s}T^{b_1 ... b_s}

How to show that for a vector V at p on M, the pushforward of V satisfies: Φ_*(V)=⊥(Φ_*V) (I can do that)

And then generalize to (s, 0) tensor: Φ_*(T)=⊥(Φ_*T)

Homework Equations

The Attempt at a Solution


[/B]
In component form we would get: (⊥(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. So remains to show that the last bit vanishes. The last term vanishes since a_b is normal to (Φ_*V)^b, so we're done.

Now to generalize for (s, 0) we have:

(⊥(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}

My problem is that I'm not quite sure what does it mean to act n_{b_s} (Φ_*T)^{b_1 ... b_s}, notion of orthogonality is easily understood for two vectors, but for vector and a tensor I'm not sure. Any ideas much appreciated!
 
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Would it help to think of the tensor in terms of (a sum of) tensor products of basis vectors?
 

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