17
$\begingroup$

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?

  • 6
    Looks like it: http://events.berkeley.edu/?event_ID=47133&date=2011-10-04&tab=all_events2011-11-20
  • 0
    @QiaochuYuan, is there any way to find out the content of the talk?2011-11-21
  • 0
    What's a natural transformation $I\to T$? (A not too serious computation tells me only the zero section works)2011-11-21
  • 0
    @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.2011-11-21
  • 3
    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.2011-11-21
  • 0
    @Mariano, you're right, the zero section is the unique natural transformation from $1_{\mathbf{Diff}}$ to $T$.2011-11-21

1 Answers 1

18

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.

  • 3
    This is awesome, I can't wait to see the paper!2011-11-21
  • 1
    Is your article available somewhere?2013-01-16
  • 0
    Apparently Benoit Jubin has completed his thesis about this monad, the title is "The Tangent Functor Monad and Foliations". Unfortunately, I cannot find it online or published.2013-01-27
  • 1
    Just spotted this: http://arxiv.org/abs/1401.09402014-01-07