Is there a way to make tangent bundle a monad?

The tangent bundle functor $T: \mathbf{Diff} \to \mathbf{Diff}$ together with the bundle projection $\pi: T \Rightarrow 1_\mathbf{Diff}$ basically screams 'monad' at me, especially because both $\pi T$ and $T \pi$ satisfy the associativity axiom, but so far I couldn't find a proper unit for it (the zero section doesn't work out, although there is still a chance that it will up to a 3-equivalence thanks to the canonical involution between $\pi T$ and $T \pi$).

Is it possible to make $T$ a monad? Do $T$-algebras have a nice description then?

Solution 1:

Yes, there is a unique monad on the tangent endofunctor. Its unit is the zero section and its multiplication is $T \tau + \tau T$, the sum of the two projections from the second tangent bundle to the tangent bundle. It is straightforward to check that it is a monad and not too hard (using 'test functions' and naturality) to show that it is the only one. Also, there is no comonad on $T$ (roughly, because there is no natural connection on a manifold).

This is the starting point of the talk I gave that Qiaochu linked to. Most of the talk was about the study (in progress) of its $T$-algebras, and I'm currently writing a paper about this. I'll put a link here when it's available.