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Assume we are given a smooth manifold $M$ and let $TM^{\otimes r}, r>1,$ be some tensor power of its tangent bundle.

Is there any general observation/result saying when there exists a manifold $N$ and a smooth map $f: M \rightarrow N$ such that the pull back bundle $f^*TN$ is isomorphic to $TM^{\otimes r}$?

This question is just a result of curiosity.

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the idea is the folk fact that the universal space and universal stable ** bundle can be taken to be a manifold and its tangent bundle [see proof below ]

take r=3 and let M have dim d then N should have dimension 3d

any [stable] bundle [like the tensor power] over a d dim space [like M] can be induced by a map into the universal space by a map which goes into the d skeleton of the universal space

the d skeleton of the universal space can be embedded in R^2d and thickened to a manifold nghd to obtain a parallizable manifold

the universal bundle over the d skeleton can be reduced to a d dim bundle using obstruction theory

the total space of this bundle pulled back to the thickened nghd will be the N of dim 3d

its tangent bundle is isomorphic to the universal bundle [because they are stably isomorphic]

the classifying map pulls the tangent bundle back to the required bundle

this proof works for r bigger but not immediately for r=2 [but mazur thickening theory should do this case]

dennis sullivan

**stable here means fibre dim bigger than dim base

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  • $\begingroup$ D.S., why $\mathbb R^{2\cdot d}\ $ (so good), rather than only $\ \mathbb R^{2\cdot d+1}\,$ ? $\endgroup$ Oct 5, 2014 at 5:25

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