# 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?

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Looks like it: events.berkeley.edu/… – Qiaochu Yuan Nov 20 '11 at 19:31
@QiaochuYuan, is there any way to find out the content of the talk? – Alexei Averchenko Nov 21 '11 at 5:34
What's a natural transformation $I\to T$? (A not too serious computation tells me only the zero section works) – Mariano Suárez-Alvarez Nov 21 '11 at 5:51
@MarianoSuárez-Alvarez, I tried the zero section, but it didn't work wih either $\pi T$ or $T \pi$. However, if one introduces some sort of equivalence between the two, then I guess some weak form of monad will be recovered. I haven't worked out the details yet, however. – Alexei Averchenko Nov 21 '11 at 6:09
A natural transformation $\eta:I\to T$ is determined by its value at $\mathbb R$, which is a map $\eta_{\mathbb R}:\mathbb R\to T\mathbb R$. Identifying $T\mathbb R$ with $\mathbb R^2$,this is given by $\eta_{\mathbb R}(t)=(\alpha(t),\beta(t))$, and naturality with respect to all smooth maps $\mathbb R\to\mathbb R$ determines $\alpha$ and $\beta$ completely. – Mariano Suárez-Alvarez Nov 21 '11 at 6:23

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.