Is every real vector bundle immersed in the tangent bundle of a Manifold?

294 Views Asked by At
  • For me "manifold" = "$C^\infty$-manifold" and also "vector bundle" = "smooth vector bundle"

There's not much context, im learning about Vector Bundles and I wonder if every vector bundle over a manifold can be inmerse in the tangent bundle of a Manifold. Even stronger, let $E$ be a vector bundle over a manifold $M$ with fibers of dimension equal to $dim (M)$, I would like to know if there is a diffeomorphism $f:M \rightarrow M$ whose push-out $f_* :E \rightarrow T(M)$ is a vector bundle isomorphism. If the answer to the last one is negative, are there some hypothesis over $M$ and/or $E$ to make it true? For example, I've just realised you can take many different vector bundles over $S^1$ not ishomorphic (because they make a different amount of Twists).

1

There are 1 best solutions below

1
On BEST ANSWER

Throughout, let $M$ be an $n$-dimensional manifold.

It will not generally be the case that every rank $n$ vector bundle over $M$ is isomorphic to $TM$. For instance, whenever there is more than one isomorphism class of rank $n$ vector bundle over $M$ there will be counterexamples, since at most one of them contains $TM$. Thus, for any nontrivial bundle $E\to M$, either $E$ or $M\times\mathbb{R}^n$ is not isomorphic to $TM$. For instance, we have the counterexamples

  • The Möbius strip $\Sigma\to S^1$
  • The trivial bundle over the sphere $S^2\times\mathbb{R}^2\to S^2$.
  • The "twisted" bundle over the torus $\Sigma\times S^1\times\mathbb{R}\to S^1\times S^1$ (where $\Sigma$ still denotes the Möbius strip)

and so on. It might be worth noting that whenever $M$ is contractible, all vector bundles over $M$ are isomorphic (and thus isomorphic to $TM$).

The question of what distinguishes $TM$ among rank $n$ vector bundles over $M$ is a bit more subtle. One can define a tangent structure to be a $(1,1)$-tensor field $J$ on the total space $E$ of a vector bundle $E\to M$ satisfying certain axioms. The tangent structure exists for all tangent bundles, and the existence of such a tensor field on an arbitrary vector bundle $E$ ensures that $E$ is isomorphic to $TM$. This wikipedia subsection gives a brief overview of the definition of this tensor field.