We know every tangent bundles is orientable and have even dimension. I think these properties are not sufficient to a vector bunble be (isomorphic) to a tangent one. I want to find some geometric/topological property, which is only satisfied by tangent bundle to formulate something like:

Let $E$ a $2n$-dimensional orientable manifold that can be realised as the total space of some vector bundle over a manifold $M$. If (some condition), then $E$ can also be realised as a tangent bundle.

I'm trying to get some feeling about vector bundle based on the tangent one.

Edit: the comments show me it's not a precise question. Ok, let's change a little bit: Fixed a manifold $M$, I want to characterize $TM$ in the follow sense: in what conditions a VB $E$ over $M$ is (isomorphic to) the tangent one and I can formulate:

Let $M$ be a manifold and $E$ be orientable $2n-$dimensional VB of $M$. If (some conditions over $E$), then $E$ is VB-isomorphic to $TM$, $E \simeq TM$.

I want, at least, a thinner necessary condition to $E \simeq TM$.

  • 1
    $\begingroup$ I think your initial question has a certain flaw. Vector bundle is a manifold (or scheme) along with a specific map, satisfying a bunch of good local properties. Of course all this comes along with an underlying base space over which all the above can happen. Your question makes no sense to me. You can probably come up with something more specific, because the obvious answer is "no" since tangent bundles is a very specific class of vector bundles. $\endgroup$
    – user321268
    Sep 29, 2017 at 19:45
  • 1
    $\begingroup$ I doubt this question makes sense; it seems you want to consider vector bundles over a fixed (necessarily even dimensional) manifold which is isomorphic to the tangent bundle of the base manifold, and as it is stated there is only one such vector bundle. If you want to see if there exists only one vector bundle over any even dimenisonal manifold, then it is obviously false. $\endgroup$
    – cjackal
    Sep 29, 2017 at 19:45
  • 3
    $\begingroup$ What do you mean exactly with your statement? In order to have a vector bundle, you need a manifold as base space, and it is not necessarily the case that the vector bundle in question is the tangent bundle. $\endgroup$ Sep 29, 2017 at 19:45
  • 2
    $\begingroup$ @ThadeuHenriqueCosta Obviously not every bundle is tangent. It seems to me that you are asking about some "characteristic properties" of tangent bundles, how they are different from other bundles. Then it would be a very broad and soft question and I don't think there is any simple answer. Look at the spheres, for instance -- tangent bundles of $S^1$, $S^3$ and $S^7$ are trivial, but the others are increasingly complicated. Let's see if someone else offers some insight... $\endgroup$ Sep 29, 2017 at 20:01
  • 4
    $\begingroup$ Here is another possible interpretation of this question, an edited version of the current 2nd paragraph which differs somewhat from the interpretation of @IvoTerek. "Let $K$ be a $2n$-dimensional orientable manifold such that there exists a vector bundle $E \mapsto M$ and a diffeomorphism $K \mapsto E$. What condition on $K$ is sufficient to ensure the existence of a manifold $N$ and a diffeomorphism $K \mapsto TN$?" $\endgroup$
    – Lee Mosher
    Sep 30, 2017 at 13:29