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Let $M$ be a manifold of dimension $n$ and $\mathcal D$ be a distribution of dimension $n-1$. We consider the quotient bundle $TM/\mathcal D = \bigsqcup_{p \in M} T_pM/\mathcal D_p$ with the surjective submersion $\pi : TM \rightarrow TM/\mathcal D$ and a global section $\sigma : M \rightarrow TM/D$. I am trying "to pullback" this global section $\sigma$ onto a (local?) section $X:M \rightarrow TM$ but I have really no idea how to do that.

I tried to use the fact that $\pi$ is a surjective submersion and find (locally) an $Y$ such that $\pi_*Y = \sigma$ but $\pi_*$ goes from $T(TM)$ to $T(TM/\mathcal D)$ so it does not make any sens...

Does anyone have an idea how to do that ?

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    $\begingroup$ In the $C^{\infty}$ category, every exact sequence of vector bundles split: there is a linear (non canonical) section $TM/\mathcal{D}\rightarrow TM$, which you can use to lift $\sigma$ to a global section of $TM$. $\endgroup$
    – abx
    Commented Nov 25, 2020 at 19:17
  • $\begingroup$ Do you have a reference where I can find more details about that? $\endgroup$
    – Falcon
    Commented Nov 25, 2020 at 20:39
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    $\begingroup$ Split exact sequence of vector bundles. $\endgroup$
    – abx
    Commented Nov 25, 2020 at 20:49

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abx is right. This kind of things is built through a partition of the unity. Or if you like better, put a Riemannian structure on $M$. The vector bundle $D^\bot$ orthogonal to $D$ in $TM$ is isomorphic to $TM/D$; precisely, $\pi$ admits a section $s:TM/D\to D^\bot$; hence $X=s\circ\sigma$ works...

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  • $\begingroup$ I don't really understand why there is an $X_i$ with $\pi(X_i) = \sigma|_{U_i}$ (in your last version of the answer) $\endgroup$
    – Falcon
    Commented Nov 30, 2020 at 8:33
  • $\begingroup$ I don't have any background in Riemannian structure.. $\endgroup$
    – Falcon
    Commented Nov 30, 2020 at 9:49

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