# What vector bundles are tangent bundles of smooth manifolds?

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?

• Basically you want to classify all vector bundles on $M$, then decide whether your bundle is (isomorphic to) the tangent bundle. Jul 28, 2020 at 5:07
• 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. Jul 28, 2020 at 5:27
• Maybe there is a better answer using the synthetic version? I don't know enough about synthetic differential geometry to answer that question though. Jul 28, 2020 at 7:34
• 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 Jul 28, 2020 at 10:25
• @MaximeRamzi, you are right. I'll edit the question. Jul 28, 2020 at 12:07