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I am reading the book "An Introduction to Contact Topology" by Geiges.

In the proof of Lemma 1.1 where he proves that any co-dimension 1 hyperplane field distribution is locally kernel of a 1-form, he obtains the line bundle $TM/\xi=\xi^\perp$ and states that $\xi^\perp$ is locally trivial. I can not see why this is true, why is $\xi^\perp$ locally trivial? Thank you for any help.

Here $M$ is the manifold, $TM$ its tangent bundle, $\xi$ is the hyperplane field decomposition and $\xi^\perp$ is the orthogonal complement of $\xi$ with respect to a Riemannian metric on M.

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Let $x\in M$ you can construct a vector field $X$ on a neighborhood $U$ of $x$ such that $X(x)$ is not $\xi_x$ (suppose that $U$ is contractible, $TM_{\mid U}$ is trivial, take any $u\in T_xM$ which is not in $\xi_x$, $y\rightarrow(y,v)$ is $X$). This implies that there exists a neighborhood $V\subset U$ such that $X(y)$ is not an element of $\xi_y$. There exists an isomorphism $V\times\mathbb{R}\rightarrow (TM/\xi)_V$ defined by $(y,u)\rightarrow p(uX(y))$ where $p:TM\rightarrow TM/\xi$ is the quotient map.

The results follows from the fact that $TM/\xi$ is isomorphic to $\xi^{\perp}$.

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  • $\begingroup$ Thanks @Tsemo Aristide . $\endgroup$ – selfmathish Jan 24 at 15:56

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