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Given a smooth manifold $M$, we can naturally associate to it a vector bundle ${\rm T}M\rightarrow M$ called the tangent bundle of $M$. This operation induces a functor $\rm T:\rm Diff\rightarrow\rm Vect(Diff)$ from the category of smooth manifolds to the category of real vector bundles over smooth manifolds.

Is it possible to give a precise meaning for the question 'what vector bundles are tangent bundles' by asking: what is the image of the functor $\rm T$ in the category $\rm Vect(Diff)$ ? Put this way, is there a standard answer for that question?

On the other hand, let ${\rm Gr}(k,V)$ denote the Grassmannian of $k$-dimensional subspaces of a $n$-dimensional vector space $V$. Then it is known (is it, indeed? I am still in the process of absorbing these matters, so any corrections shall be really appreciated, terminology included!) that for any real vector bundle $\pi:E\rightarrow M$ over a (compact?) smooth manifold $M$, there are $k,n\in\Bbb N$ and a smooth map $\mu:M\rightarrow{\rm Gr}(k,V)$, such that the bundle $\pi$ is isomorphic to the pull-back by $\mu$ of the tautological bundle of ${\rm Gr}(k,V)$.

Is it possible to describe 'what vector bundles are tangent bundles' in terms of Grassmannians? And finally: is there a standard translation between these two perspectives on the same subject?

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  • $\begingroup$ Basically you want to classify all vector bundles on $M$, then decide whether your bundle is (isomorphic to) the tangent bundle. $\endgroup$
    – user10354138
    Jul 28, 2020 at 5:07
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    $\begingroup$ Well, I tried to write the question in a 'fancier' way in order to open more possibilities for the answers. For example, what special properties do tangent bundles have (not in a direct reference to the extra structures they carry, but possibly in relation with them...) among vector bundles in general? In other words, I'm more interested in a 'characterization' than in a 'criterion'. Anyway, what you said would probably be part, even if implicitly, of such an answer. $\endgroup$
    – Dry Bones
    Jul 28, 2020 at 5:27
  • $\begingroup$ Maybe there is a better answer using the synthetic version? I don't know enough about synthetic differential geometry to answer that question though. $\endgroup$
    – user10354138
    Jul 28, 2020 at 7:34
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    $\begingroup$ I think you're mistaken about the Grassmanian : the universal bundle over $Gr(m,V)$ has rank $m$, so the rank of $E$ would be $m$, which is the dimension of $M$, not the rank of $E$. For a compact manifold $M$ of dimension $m$ and a real vector bundle of rank $n$, there is some $k$ and a vector space of dimension $k,V$ with a map $M\to Gr(n,V)$ classifying $E$. Not sure this is true if $M$ is not compact, though $\endgroup$
    – Maxime Ramzi
    Jul 28, 2020 at 10:25
  • $\begingroup$ @MaximeRamzi, you are right. I'll edit the question. $\endgroup$
    – Dry Bones
    Jul 28, 2020 at 12:07

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